- Identify the slope in the form \(mx+b\). - Decide whether the slope of a linear function changes from point to point. - A tangent line to a straight line is the line itself.

1. The function \(f(x) = 0.8x-5\) is linear. 2. A linear function of the form \(f(x) = mx+b\) has constant slope \(m\). 3. Here, \(m = 0.8\), so the tangent-line slope is \(0.8\) at every x-value.

a) \(0.8\) b) \(0.8\)

- Decide whether the information refers to one moment or to a span of time. - An average rate requires an interval with two different times. - Think about what a speedometer-like display reports right now.

1. The display refers to one exact moment, \(2{:}15\) p.m., rather than to an interval of time. 2. Therefore, \(18\,\text{mph}\) represents an instantaneous rate of change.

It is an instantaneous rate because it describes the bicycle's speed at one moment.

- Compare the change in the function values with the change in the inputs. - Keep the endpoint order the same in the numerator and denominator.

1. The change in output is \(10-4=6\). 2. The change in input is \(3-1=2\). 3. The average rate of change is \(\frac{6}{2}=3\).

\(3\)

- Check what each axis measures before using a slope. - Distance divided by elapsed time gives a speed unit.

1. The graph shows \(0\) miles at the start and \(9\) miles after \(3\) hours. 2. The average speed is \(\frac{9-0}{3-0}=3\,\frac{\text{mi}}{\text{h}}\).

\(3\,\text{mph}\)

- A curved graph can still have a straight secant through two points. - Compare the height at the right endpoint with the height at the left endpoint.

1. The curve passes through \(P=(0,4)\) and \(Q=(2,0)\). 2. The average rate is \(\frac{0-4}{2-0}=-2\), which is negative.

\(-2\); negative.

- Follow the curve from left to right just before and just after the marked point. - Imagine zooming in very closely at the point. Which direction would a tangent line follow?

1. Immediately before the highest point the curve rises; immediately after it the curve falls. 2. At the smooth highest point, the tangent is horizontal and its slope is zero.

Zero. The curve rises before the smooth peak and falls after it, making the tangent there horizontal.

- Which kind of rate could be calculated from an entire journey's distance and elapsed time? - Which reading could change from one second to the next without affecting a past trip's average?

1. Reading A covers a whole hour, so it describes average speed. 2. Reading B describes one moment, so it describes instantaneous speed.

A: Average rate. B: Instantaneous rate.

- Does this notation describe a distance, or how that distance is changing? - Think about what the input \(6\) and the two measurement units tell you.

1. The prime in \(s'(6)\) means the instantaneous rate at the input \(t=6\). 2. It describes how quickly the runner's distance from the start is changing at exactly \(6\,\text{s}\), in meters per second.

The runner's instantaneous rate of change of distance from the start at \(6\,\text{s}\), in \(\text{m/s}\).

- Think about what the sign of a vertical rate says about direction. - Interpret the magnitude separately from the negative sign.

1. A negative rate means the height is falling. 2. The size of the rate is \(8\,\text{ft/s}\), so its height is decreasing at \(8\) feet per second at that instant.

Falling at \(8\,\text{ft/s}\).

- What formula gives average speed from a change in distance and a change in time? - Find the rocket’s distance at the beginning and end of the time interval. - Think about the difference quotient.

1. Find the distance at \(t = 4\): \(s(4) = 3 \cdot 4^2 = 48\). 2. Find the distance at \(t = 6\): \(s(6) = 3 \cdot 6^2 = 108\). 3. Compute the change in distance: \(\Delta s = 108 - 48 = 60\). 4. Compute the change in time: \(\Delta t = 6 - 4 = 2\). 5. The average speed is \(\frac{\Delta s}{\Delta t} = \frac{60}{2} = 30\,\frac{\text{km}}{\text{min}}\).

\(30\,\frac{\text{km}}{\text{min}}\)

- Recall the difference-quotient formula for average rate of change. - How do you evaluate a function at a given input? - Be especially careful with signs when substituting negative values. - Geometrically, what does the result tell you about the secant line through the endpoint points?

1. Evaluate the endpoints: \(f(-2) = \frac{1}{2} \cdot (-2)^3 - 2 \cdot (-2) = -4 + 4 = 0\) and \(f(0) = 0\). Then \(\frac{f(0) - f(-2)}{0 - (-2)} = \frac{0 - 0}{2} = 0\). 2. Evaluate the endpoints: \(f(1) = \frac{1}{2} \cdot 1^3 - 2 \cdot 1 = -1.5\) and \(f(4) = \frac{1}{2} \cdot 4^3 - 2 \cdot 4 = 32 - 8 = 24\). Then \(\frac{f(4) - f(1)}{4 - 1} = \frac{24 - (-1.5)}{3} = \frac{25.5}{3} = 8.5\).

1. \(0\) 2. \(8.5\)

- Decide which endpoint values belong in the numerator and denominator of the difference quotient. - First calculate all needed values of \(h(x)\). - Work step by step and be careful when subtracting negative values.

1. For \([0, 2]\), \(h(0) = -5\) and \(h(2) = -2^2 + 6 \cdot 2 - 5 = 3\). Thus, \(\frac{h(2) - h(0)}{2 - 0} = \frac{3 - (-5)}{2} = 4\). 2. For \([3, 5]\), \(h(3) = -3^2 + 6 \cdot 3 - 5 = 4\) and \(h(5) = -5^2 + 6 \cdot 5 - 5 = 0\). Thus, \(\frac{h(5) - h(3)}{5 - 3} = \frac{0 - 4}{2} = -2\).

For \([0, 2]\), the average rate of change is \(4\). For \([3, 5]\), the average rate of change is \(-2\).

- Recall the connection between average rate of change and the slope of a secant line. - Which values belong in the difference quotient? - Check your function values at \(x = 2\), \(x = 4\), and \(x = 2.5\).

1. Evaluate the needed function values: \(f(2) = 2 \cdot 2^2 - 4 \cdot 2 = 0\), \(f(4) = 2 \cdot 4^2 - 4 \cdot 4 = 16\), and \(f(2.5) = 2 \cdot 2.5^2 - 4 \cdot 2.5 = 12.5 - 10 = 2.5\). 2. Over \([2, 4]\), \(\frac{f(4) - f(2)}{4 - 2} = \frac{16 - 0}{2} = 8\). 3. Over \([2, 2.5]\), \(\frac{f(2.5) - f(2)}{2.5 - 2} = \frac{2.5 - 0}{0.5} = 5\).

a) \(8\) b) \(5\)

- What is the difference-quotient formula over an interval \([a, b]\)? - Be careful with signs when substituting the function values. - A quick sketch of the reciprocal function can help you check whether the rates should be positive or negative.

1. Evaluate the needed function values: \(g(1) = 4\), \(g(4) = 1\), and \(g(1.6) = 2.5\). 2. Over \([1, 4]\), \(\frac{g(4) - g(1)}{4 - 1} = \frac{1 - 4}{3} = -1\). 3. Over \([1, 1.6]\), \(\frac{g(1.6) - g(1)}{1.6 - 1} = \frac{2.5 - 4}{0.6} = -2.5\).

Over \([1, 4]\), the average rate of change is \(-1\). Over \([1, 1.6]\), the average rate of change is \(-2.5\).

- What calculation gives the average change over an interval? - Choose the table entries at the two endpoints of each interval. - What does a negative result mean in the context of temperature? - Divide the change in temperature by the change in time.

1. Over \([0, 24]\), \(\frac{T(24) - T(0)}{24 - 0} = \frac{7 - 5}{24} = \frac{1}{12} \approx 0.083\,\frac{^\circ\text{C}}{\text{h}}\). 2. Over \([4, 8]\), \(\frac{T(8) - T(4)}{8 - 4} = \frac{6 - 2}{4} = 1\,\frac{^\circ\text{C}}{\text{h}}\). 3. Over \([8, 16]\), \(\frac{T(16) - T(8)}{16 - 8} = \frac{19 - 6}{8} = \frac{13}{8} = 1.625\,\frac{^\circ\text{C}}{\text{h}}\). 4. Over \([16, 24]\), \(\frac{T(24) - T(16)}{24 - 16} = \frac{7 - 19}{8} = -1.5\,\frac{^\circ\text{C}}{\text{h}}\).

a) \(\frac{1}{12}\,\frac{^\circ\text{C}}{\text{h}} \approx 0.083\,\frac{^\circ\text{C}}{\text{h}}\) b) \(1\,\frac{^\circ\text{C}}{\text{h}}\) c) \(1.625\,\frac{^\circ\text{C}}{\text{h}}\) d) \(-1.5\,\frac{^\circ\text{C}}{\text{h}}\)

- How are distance, time, and average speed related? - Recall the slope formula for a secant line through two points on a graph. - What units should the speed have here? - Compare your answers. What do they suggest about how the car’s speed changes?

1. Over \([0, 4]\), \(\frac{s(4) - s(0)}{4 - 0} = \frac{12.8 - 0}{4} = 3.2\,\frac{\text{m}}{\text{s}}\). 2. Over \([4, 8]\), \(\frac{s(8) - s(4)}{8 - 4} = \frac{51.2 - 12.8}{4} = 9.6\,\frac{\text{m}}{\text{s}}\). 3. Over \([8, 10]\), \(\frac{s(10) - s(8)}{10 - 8} = \frac{80.0 - 51.2}{2} = 14.4\,\frac{\text{m}}{\text{s}}\). 4. Over \([0, 10]\), \(\frac{s(10) - s(0)}{10 - 0} = \frac{80.0 - 0}{10} = 8.0\,\frac{\text{m}}{\text{s}}\).

a) \(3.2\,\frac{\text{m}}{\text{s}}\) b) \(9.6\,\frac{\text{m}}{\text{s}}\) c) \(14.4\,\frac{\text{m}}{\text{s}}\) d) \(8.0\,\frac{\text{m}}{\text{s}}\)

- How do you find the slope through two data points? - What does the result represent per unit of time? - Does the temperature rise or fall as time passes? - What phrase describes how quickly an object becomes cooler?

1. Over \([0, 4]\), \(\frac{T(4) - T(0)}{4 - 0} = \frac{76 - 92}{4} = -4\,\frac{^\circ\text{C}}{\text{min}}\). Over \([4, 20]\), \(\frac{T(20) - T(4)}{20 - 4} = \frac{38 - 76}{16} = -2.375\,\frac{^\circ\text{C}}{\text{min}}\). 2. The negative signs indicate that the drink’s temperature decreases over time. 3. In this context, the quantity is the average cooling rate, or average rate of temperature change.

1. Over \([0, 4]\): \(-4\,\frac{^\circ\text{C}}{\text{min}}\) Over \([4, 20]\): \(-2.375\,\frac{^\circ\text{C}}{\text{min}}\) 2. The negative signs mean that the drink is cooling. 3. The average cooling rate.

- The unit of a rate is the output unit divided by the input unit. - Read which variable each quantity depends on before writing the denominator unit. - Focus on units rather than specialized names for the rates.

1. Energy is measured in joules, and its rate with respect to seconds is measured in joules per second. 2. Water height is measured in inches, and its rate is inches per minute. 3. Rope mass is measured in pounds, and its rate with respect to length is pounds per foot. 4. Profit is measured in dollars, and its rate with respect to pie count is dollars per pie.

A: \(\text{J}\); \(\text{J/s}\). B: \(\text{in.}\); \(\text{in./min}\). C: \(\text{lb}\); \(\text{lb/ft}\). D: \(\$\); \(\$/\text{pie}\).

- Determine the units of each expression. - Distinguish between a time interval and one instant. - A function value gives an amount, an average rate describes an interval, and a derivative describes an instant.

1. The statement \(A(10) = 50\) means that \(10\) days after the first measurement, algae cover \(50\,\text{m}^2\) of the lake. 2. The quotient \(\frac{A(10)-A(0)}{10} = 4\) is the average rate of change over the first \(10\) days. The covered area grew by an average of \(4\,\text{m}^2\) per day. 3. The derivative value \(A'(10) = 6\) is the instantaneous growth rate on day \(10\). At that time, the covered area is growing at \(6\,\text{m}^2\) per day.

a) After \(10\) days, algae cover \(50\,\text{m}^2\). b) During the first \(10\) days, the covered area grew by an average of \(4\,\frac{\text{m}^2}{\text{day}}\). c) On day \(10\), the covered area is growing at \(6\,\frac{\text{m}^2}{\text{day}}\).

- How are distance traveled and speed related? - Distinguish between an average over an interval and a value at one instant. - Distance divided by time has speed units.

1. The statement \(s(4) = 40\) means that after \(4\) seconds, the vehicle has traveled \(40\) meters. 2. The quotient \(\frac{s(10)-s(0)}{10} = 25\) is the average speed over \([0, 10]\), so the vehicle averaged \(25\,\frac{\text{m}}{\text{s}}\) during the first \(10\) seconds. 3. The derivative value \(s'(10) = 45\) is the instantaneous speed at \(t = 10\), so the vehicle is traveling at \(45\,\frac{\text{m}}{\text{s}}\) at that instant.

a) After \(4\) seconds, the vehicle has traveled \(40\,\text{m}\). b) Its average speed during the first \(10\) seconds is \(25\,\frac{\text{m}}{\text{s}}\). c) Its instantaneous speed at \(t = 10\) is \(45\,\frac{\text{m}}{\text{s}}\).

- For part a), focus on the signs of the numerator and denominator of a secant slope rather than its exact value. - Compare the two endpoint function values. - For part b), use the average-rate-of-change formula after you have justified the sign.

1. Evaluate the endpoint values: \(f(1)=-1.5\) and \(f(3)=7.5\). 2. Since \(f(3)-f(1)>0\) and \(3-1>0\), the average rate of change is positive. 3. Compute the difference quotient: \(\frac{7.5-(-1.5)}{3-1}=\frac{9}{2}=4.5\).

a) Positive, because \(f(3)>f(1)\) while \(3-1>0\). b) \(4.5\)

- For positive inputs, compare \(a^2\) and \(b^2\) before taking reciprocals. - Determine the signs of the numerator and denominator of a general secant slope separately. - For the numerical interval, use the average-rate-of-change formula only after completing the structural sign argument.

1. Since \(0<a<b\), we have \(a^2<b^2\), so \(\frac{1}{a^2}>\frac{1}{b^2}\). Therefore, \(g(a)>g(b)\). In the secant slope \(\frac{g(b)-g(a)}{b-a}\), the numerator is negative and the denominator is positive, so every such secant slope is negative. 2. Evaluate \(g(1)=8\) and \(g(4)=\frac12\). The average rate of change is \(\frac{\frac12-8}{4-1}=\frac{-\frac{15}{2}}{3}=-\frac52\).

a) The average rate is negative because for \(0<a<b\), \(g(a)>g(b)\), so \(g(b)-g(a)<0\) while \(b-a>0\). b) \(-\frac52\)

- Identify what the input and output of the function represent. - A difference of output values represents a change in height. - Dividing a height change by a time change gives a growth rate. - The derivative represents an instantaneous rate of change.

1. \(h(t+7)-h(t)\) is the sunflower's change in height during the seven days after day \(t\). Its units are inches. 2. \(\frac{h(t+\Delta t)-h(t)}{\Delta t}\) is the sunflower's average growth rate from day \(t\) to day \(t+\Delta t\). Its units are inches per day. 3. \(h'(t)\) is the sunflower's instantaneous growth rate on day \(t\). Its units are inches per day.

a) The change in height over seven days, measured in inches b) The average growth rate from \(t\) to \(t+\Delta t\), measured in inches per day c) The instantaneous growth rate at time \(t\), measured in inches per day

- Compare displacement over an interval with motion at one instant. - A difference quotient gives an average rate over an interval. - A derivative gives an instantaneous rate at one time.

1. \(\frac{s(t_2)-s(t_1)}{t_2-t_1}\) is the vehicle's average velocity from time \(t_1\) to time \(t_2\), measured in meters per second. 2. \(s'(t)\) is the vehicle's instantaneous velocity at time \(t\), measured in meters per second.

a) The average velocity from \(t_1\) to \(t_2\), measured in meters per second b) The instantaneous velocity at time \(t\), measured in meters per second

- Determine the units of the numerator and denominator in each expression. - Distinguish between a change over a time interval and a rate at one instant. - Relate the derivative of height to vertical velocity.

1. \(h(t)-h(t_0)\) is the balloon's change in height from time \(t_0\) to time \(t\). Its units are feet. 2. \(\frac{h(t)-h(t_0)}{t-t_0}\) is the balloon's average vertical velocity from time \(t_0\) to time \(t\). Its units are feet per minute. 3. \(h'(t_0)\) is the balloon's instantaneous vertical velocity at time \(t_0\). Its units are feet per minute.

a) The change in height from \(t_0\) to \(t\), measured in feet b) The average vertical velocity from \(t_0\) to \(t\), measured in feet per minute c) The instantaneous vertical velocity at \(t_0\), measured in feet per minute

- What type of function has the same slope at every input? - Use the two given points to calculate the slope of the line. - Interpret the derivative as the change in the function value per unit change in \(x\). - One given point lies on the \(y\)-axis.

1. A function with a constant derivative is linear, and the derivative equals its slope. Using \(P\) and \(Q\), \(c=\frac{9-1.5}{5-0}=\frac{7.5}{5}=1.5\). 2. Write \(f(x)=mx+b\). Since \(m=1.5\) and \(P(0, 1.5)\) gives \(b=1.5\), the function is \(f(x)=1.5x+1.5\). 3. The slope gives the change in the function value for an increase of \(1\) in \(x\). Therefore, the value of \(f\) increases by \(1.5\) each time \(x\) increases by \(1\).

1. \(c=1.5\) 2. \(f(x)=1.5x+1.5\) 3. The function value increases by \(1.5\).

- Write the difference-quotient formula over an interval \([a, b]\). - Read the function values at the endpoints from the graph. - Find the change in the function values and divide by the change in \(x\). - Substitute the values carefully.

1. Read the endpoint values from the graph. For part a, \(f(-1) = -1\) and \(f(1) = 3\). For part b, \(f(0) = 1\) and \(f(2) = -1\). 2. Over \([-1, 1]\), \(\frac{f(1) - f(-1)}{1 - (-1)} = \frac{3 - (-1)}{2} = 2\). 3. Over \([0, 2]\), \(\frac{f(2) - f(0)}{2 - 0} = \frac{-1 - 1}{2} = -1\).

a) \(2\) b) \(-1\)

- Read the coordinates of the relevant points from the graph. - Divide the change in height by the elapsed time. - Recall the secant-slope formula. - What does a positive or negative rate mean for the drone’s motion?

1. From the graph, \(h(1) = 3\,\text{m}\) and \(h(5) = 17\,\text{m}\). 2. Over \([1, 5]\), \(\frac{h(5) - h(1)}{5 - 1} = \frac{17 - 3}{4} = 3.5\,\frac{\text{m}}{\text{min}}\). 3. From the graph, \(h(6) = 18\,\text{m}\) and \(h(8) = 10\,\text{m}\). 4. Over \([6, 8]\), \(\frac{h(8) - h(6)}{8 - 6} = \frac{10 - 18}{2} = -4\,\frac{\text{m}}{\text{min}}\). 5. The negative sign means that the drone’s altitude decreases during this interval; it descends by an average of \(4\) meters per minute.

a) \(3.5\,\frac{\text{m}}{\text{min}}\) b) \(-4\,\frac{\text{m}}{\text{min}}\). The negative sign means that the drone descends on average during this interval.

- The derivative at a point is the slope of the tangent line there. - Choose two grid points on each tangent line. - Use \(\frac{\Delta y}{\Delta x}\) to calculate slope. - A horizontal line has slope \(0\).

1. A derivative value equals the slope of the tangent line at the point of tangency. 2. In panel a), the tangent line passes through \((0,0)\) and \((2,2)\), so \(f'(2)=\frac{2-0}{2-0}=1\). 3. In panel b), the tangent line is horizontal, so \(g'(2)=0\). 4. In panel c), the tangent line passes through \((1,3)\) and \((2,1)\), so \(h'(2)=\frac{1-3}{2-1}=-2\).

a) \(f'(2)=1\) b) \(g'(2)=0\) c) \(h'(2)=-2\)

- Use the slope formula for the secant line through the endpoint graph points. - Read the required function values directly from the graph. - An instantaneous rate of change is the slope of a tangent line. - Choose two easy-to-read points on each tangent line.

1. From the graph, \(f(-1)=0\) and \(f(1)=3\). Therefore, the average rate over \([-1,1]\) is \(\frac{3-0}{1-(-1)}=\frac{3}{2}=1.5\). 2. From the graph, \(f(2)=6\) and \(f(4)=0\). Therefore, the average rate over \([2,4]\) is \(\frac{0-6}{4-2}=-3\). 3. The instantaneous rate is the slope of the corresponding tangent line. The line \(t_0\) passes through \((0,0)\) and \((1,2)\), so \(f'(0)=\frac{2-0}{1-0}=2\). The line \(t_2\) passes through \((1,4)\) and \((2,6)\), so \(f'(2)=\frac{6-4}{2-1}=2\).

a) \(1.5\) b) \(-3\) c) \(f'(0)=2\) and \(f'(2)=2\)

- The derivative at a point equals the slope of the tangent line there. - Choose two marked points on each tangent line. - Calculate each slope as rise divided by run. - Pay close attention to negative coordinates when subtracting.

1. On \(t_{-2}\), use \((-3,1)\) and \((-1,3)\): \(f'(-2)=\frac{3-1}{-1-(-3)}=1\). 2. On \(t_0\), use \((-1,2)\) and \((1,-2)\): \(f'(0)=\frac{-2-2}{1-(-1)}=-2\). 3. On \(t_2\), use \((1,-3)\) and \((3,-1)\): \(f'(2)=\frac{-1-(-3)}{3-1}=1\).

a) \(f'(-2)=1\) b) \(f'(0)=-2\) c) \(f'(2)=1\)

- The instantaneous rate of change equals the slope of the tangent line. - Choose two grid points on each tangent line. - Use slope \(=\frac{\text{rise}}{\text{run}}\). - Check the vertical scale in panel c).

1. In panel a), the tangent line passes through \((2, 0)\) and \((3, 2)\). Its slope is \(\frac{2-0}{3-2} = 2\). 2. In panel b), the tangent line passes through \((2, 1)\) and \((3, 0)\). Its slope is \(\frac{0-1}{3-2} = -1\). 3. In panel c), the tangent line passes through \((1, 1)\) and \((3, 2)\). Its slope is \(\frac{2-1}{3-1} = 0.5\).

a) \(2\) b) \(-1\) c) \(0.5\)

- First read the water levels at the interval endpoints. - Use \(\frac{h(b) - h(a)}{b - a}\). - Divide the water-level unit by the time unit. - What does a positive rate mean for the water level?

1. From the graph, \(h(0) = 1\,\text{m}\), \(h(6) = 7\,\text{m}\), \(h(2) = 2.5\,\text{m}\), and \(h(8) = 5.5\,\text{m}\). 2. Over \([0, 6]\), \(\frac{h(6) - h(0)}{6 - 0} = \frac{7 - 1}{6} = 1\,\frac{\text{m}}{\text{h}}\). 3. Over \([2, 8]\), \(\frac{h(8) - h(2)}{8 - 2} = \frac{5.5 - 2.5}{6} = 0.5\,\frac{\text{m}}{\text{h}}\). 4. During the first six hours, the water level rises by an average of \(1\) meter per hour.

a) \(h(0) = 1\,\text{m}\), \(h(6) = 7\,\text{m}\), and the average rate is \(1\,\frac{\text{m}}{\text{h}}\). b) \(h(2) = 2.5\,\text{m}\), \(h(8) = 5.5\,\text{m}\), and the average rate is \(0.5\,\frac{\text{m}}{\text{h}}\). c) The water level rises by an average of \(1\) meter per hour from \(t = 0\) to \(t = 6\).

- Read the function values at the interval endpoints. - Use the average-rate-of-change formula. - When is a fraction equal to zero? - Look for two graph points with the same y-coordinate.

1. Read the values from the graph: \(f(-2) = 1\), \(f(-1) = 3\), \(f(0) = 1\), \(f(1) = -1\), and \(f(2) = 1\). 2. Over \([-2, -1]\), \(\frac{f(-1) - f(-2)}{-1 - (-2)} = \frac{3 - 1}{1} = 2\). 3. Over \([-1, 1]\), \(\frac{f(1) - f(-1)}{1 - (-1)} = \frac{-1 - 3}{2} = -2\). 4. An average rate of \(0\) occurs when the endpoint function values are equal. Since \(f(-2) = f(0) = f(2) = 1\), possible intervals include \([-2, 0]\), \([0, 2]\), and \([-2, 2]\).

a) \([-2,-1]\): \(2\); \([-1,1]\): \(-2\) b) One possible interval is \([-2,0]\).

- Recall the difference-quotient formula. - Read the volume values at the interval endpoints. - Divide the change in volume by the length of the time interval. - The rate unit is volume divided by time.

1. From the graph, \(V(0) = 0\,\text{m}^3\), \(V(4) = 12\,\text{m}^3\), and \(V(8) = 40\,\text{m}^3\). 2. Over \([0, 4]\), \(\frac{V(4) - V(0)}{4 - 0} = \frac{12 - 0}{4} = 3\,\frac{\text{m}^3}{\text{h}}\). 3. Over \([4, 8]\), \(\frac{V(8) - V(4)}{8 - 4} = \frac{40 - 12}{4} = 7\,\frac{\text{m}^3}{\text{h}}\).

a) \(3\,\frac{\text{m}^3}{\text{h}}\) b) \(7\,\frac{\text{m}^3}{\text{h}}\)

- Use the difference quotient over each interval. - Read each function value at the interval endpoints. - The marked points help you read exact coordinates. - Calculate each function separately.

1. From the graphs, \(f(2) = 3\), \(f(3) = 6\), \(f(4) = 3\), \(f(5) = 0\), and \(f(6) = 3\). Also, \(g(2) = 0\), \(g(3) = 3\), \(g(4) = 4\), \(g(5) = 3\), and \(g(6) = 0\). 2. For \(f\): over \([2, 4]\), the rate is \(\frac{3 - 3}{2} = 0\); over \([3, 5]\), \(\frac{0 - 6}{2} = -3\); over \([4, 6]\), \(\frac{3 - 3}{2} = 0\); and over \([2, 6]\), \(\frac{3 - 3}{4} = 0\). 3. For \(g\): over \([2, 4]\), the rate is \(\frac{4 - 0}{2} = 2\); over \([3, 5]\), \(\frac{3 - 3}{2} = 0\); over \([4, 6]\), \(\frac{0 - 4}{2} = -2\); and over \([2, 6]\), \(\frac{0 - 0}{4} = 0\).

a) \(f\): \(0\); \(g\): \(2\) b) \(f\): \(-3\); \(g\): \(0\) c) \(f\): \(0\); \(g\): \(-2\) d) \(f\): \(0\); \(g\): \(0\)

- A derivative value equals the slope of the tangent line. - Choose two points on the tangent line whose coordinates can be read exactly. - Use rise divided by run.

1. The derivative \(f'(1)\) equals the slope of tangent line \(t\). 2. Two readable points on the tangent line are \((1,-1)\) and \((2,0)\). 3. Calculate the slope: \(\frac{0-(-1)}{2-1}=1\). 4. Therefore, \(f'(1)=1\).

\(f'(1) = 1\)

- A decreasing graph has negative slope; an increasing graph has positive slope. - Imagine a tangent line at each marked point. - A steep downward tangent has a very negative slope. - Compare the steepness at the two points where the graph is increasing.

1. A graph has negative slope where it decreases from left to right. The graph is decreasing at \(A\) and \(D\), so those points have negative slopes. 2. At \(D\), the graph decreases most steeply, so its slope is the smallest. 3. At \(A\), the graph decreases less steeply, so its slope is greater than the slope at \(D\) but still negative. 4. The graph increases at \(B\) and \(C\). It is steeper at \(C\), so the slope at \(C\) is greater than the slope at \(B\). 5. Therefore, the order is \(D, A, B, C\).

a) \(A\) and \(D\) b) \(D, A, B, C\)

- A derivative value equals the slope of the tangent line. - Use two readable grid points on each drawn tangent line. - A horizontal tangent has slope \(0\). - The derivative is positive where the graph rises and negative where it falls.

1. At \(x=0\), the tangent line passes through \((0,-1)\) and \((1,-3)\). Its slope is \(\frac{-3-(-1)}{1-0}=-2\), so \(f'(0)=-2\). 2. At \(x=2\), the tangent line is horizontal, so \(f'(2)=0\). 3. At \(x=4\), the tangent line passes through \((4,-1)\) and \((5,1)\). Its slope is \(\frac{1-(-1)}{5-4}=2\), so \(f'(4)=2\). 4. The graph is decreasing at \(x=-1\) and \(x=1\), so \(f'(-1)<0\) and \(f'(1)<0\). 5. The graph is increasing at \(x=5\), so \(f'(5)>0\).

a) \(f'(0)=-2\), \(f'(2)=0\), and \(f'(4)=2\) b) \(f'(-1)<0\), \(f'(1)<0\), and \(f'(5)>0\)

- Read the start and end depth for each requested interval. - Average rate is the depth change divided by elapsed hours. - For the maximum, read the highest point of the curve and check the height scale.

1. The graph gives \(h(0)=0\), \(h(2)=0.5\), \(h(6)=2.7\), \(h(10)=2.5\), and \(h(12)=0\), all in meters. 2. The rates are \(\frac{0.5-0}{2}=0.25\,\text{m/h}\), \(\frac{2.7-0.5}{4}=0.55\,\text{m/h}\), and \(\frac{0-2.5}{2}=-1.25\,\text{m/h}\). 3. The full-interval rate is \(\frac{h(12)-h(0)}{12}=0\,\text{m/h}\), because the starting and ending depths match. 4. The highest point is at \(t=8\,\text{h}\), with depth \(h(8)=3.2\,\text{m}\), read to the requested precision.

a) 1. \(0.25\,\text{m/h}\) 2. \(0.55\,\text{m/h}\) 3. \(-1.25\,\text{m/h}\) b) \(0\,\text{m/h}\); the net depth change is zero. c) At \(t=8\,\text{h}\); maximum depth about \(3.2\,\text{m}\).

- The derivative at a point is the slope of the graph there. - Imagine a tangent line at \(x = 1\) in each panel. - A local maximum has derivative \(0\). - Compare the direction and steepness of the other three graphs.

1. Graph \(A\) is increasing steeply at \(x = 1\), with tangent slope \(2\). Thus, \(A\) matches (2). 2. Graph \(B\) has a local maximum at \(x = 1\), so its tangent is horizontal and the derivative is \(0\). Thus, \(B\) matches (1). 3. Graph \(C\) is increasing moderately at \(x = 1\), with tangent slope \(1\). Thus, \(C\) matches (4). 4. Graph \(D\) is decreasing steeply at \(x = 1\), with tangent slope \(-2\). Thus, \(D\) matches (3).

\(A\rightarrow (2)\) \(B\rightarrow (1)\) \(C\rightarrow (4)\) \(D\rightarrow (3)\)

- The derivative is the slope of the tangent line. - A horizontal tangent has slope \(0\). - Choose readable grid points on each shown tangent line. - Check the sign of each slope.

1. At \(x_1=-2\), the tangent line is horizontal. Therefore, \(f'(-2)=0\). 2. At \(x_2=0\), the tangent line passes through \((0,2)\) and \((1,0)\). Its slope is \(\frac{0-2}{1-0}=-2\), so \(f'(0)=-2\). 3. At \(x_3=2\), the tangent line passes through \((1,0)\) and \((2,-4)\). Its slope is \(\frac{-4-0}{2-1}=-4\), so \(f'(2)=-4\).

\(f'(-2) = 0\) \(f'(0) = -2\) \(f'(2) = -4\)

- Average rate of change is the slope of the secant line through the two endpoint points. - Use \(\frac{f(b)-f(a)}{b-a}\). - Check whether the graph rises or falls on each interval.

1. Read the endpoint values from the graph: \(f(-5) = 3\), \(f(0) = -2\), and \(f(5) = 3\). 2. On \([-5, 0]\), the average rate of change is \(\frac{f(0)-f(-5)}{0-(-5)} = \frac{-2-3}{5} = -1\). 3. On \([0, 5]\), the average rate of change is \(\frac{f(5)-f(0)}{5-0} = \frac{3-(-2)}{5} = 1\). 4. The rates have the same absolute value but opposite signs. This is consistent with the graph being symmetric about the y-axis: it decreases on the first interval and increases on the second.

On \([-5, 0]\), the average rate of change is \(-1\). On \([0, 5]\), the average rate of change is \(1\). The rates have equal absolute values and opposite signs.

- When do two endpoints make a horizontal secant? - Look for pairs of points at equal heights. - Which line divides this graph into mirror-image halves?

1. The graph is symmetric about the y-axis, so \(f(-4)=f(4)\). The average rate of change on \([-4,4]\) is \(\frac{f(4)-f(-4)}{8}=0\). 2. Any interval \([-c,c]\) inside the shown domain has equal endpoint values. For example, \([-2,2]\) also has average rate \(0\).

a) \(0\) b) One example is \([-2,2]\); the two endpoint heights are equal.

- Identify what each axis represents. - Average rate of change is change in water level divided by change in time. - In this context, what does a negative rate indicate?

1. Read the values from the graph: \(h(2) = 4.8\,\text{m}\) and \(h(6) = 3.2\,\text{m}\). 2. The average rate of change is \(\frac{h(6)-h(2)}{6-2} = \frac{3.2-4.8}{4} = \frac{-1.6}{4} = -0.4\,\frac{\text{m}}{\text{h}}\). 3. The negative sign means that the water level decreased over the interval. On average, it fell by \(0.4\,\text{m}\) each hour.

\(-0.4\,\frac{\text{m}}{\text{h}}\). The negative sign means the water level was decreasing during the interval.

- Identify the two graph points at the endpoints of each interval. - Use the slope formula for the secant line. - Pay attention to the scale on each axis.

1. For \([0, 2]\), the graph gives \(f(0) = 4\) and \(f(2) = 2\). 2. The average rate of change is \(\frac{f(2)-f(0)}{2-0} = \frac{2-4}{2} = -1\). 3. For \([-2, 0]\), the graph gives \(f(-2) = 2\) and \(f(0) = 4\). 4. The average rate of change is \(\frac{f(0)-f(-2)}{0-(-2)} = \frac{4-2}{2} = 1\).

a) \(-1\) b) \(1\)

- Only the two endpoint heights determine the average rate, even if the curve turns between them. - For the other interval, locate both marked heights relative to the labeled horizontal grid lines. - Divide the change in height by the interval length.

1. On \([0,2]\), \(g(0)=g(2)=0\), so the average rate is \(0\). 2. Read \(g(1)=-0.75\) and \(g(3)=3.75\). The average rate on \([1,3]\) is \(\frac{3.75-(-0.75)}{3-1}=2.25\).

a) \(0\) b) \(2.25\)

- Substitute each interval endpoint into the function to find the corresponding y-values. - Be careful with signs when substituting or subtracting negative numbers. - Use \(\frac{h(b)-h(a)}{b-a}\).

1. For \([1, 3]\), evaluate the endpoints: \(h(1) = 3(1)^2 - 6(1) + 2 = -1\) and \(h(3) = 3(3)^2 - 6(3) + 2 = 11\). 2. The average rate of change is \(\frac{h(3)-h(1)}{3-1} = \frac{11-(-1)}{2} = 6\). 3. For \([-1, 1]\), evaluate the endpoints: \(h(-1) = 3(-1)^2 - 6(-1) + 2 = 11\) and \(h(1) = -1\). 4. The average rate of change is \(\frac{h(1)-h(-1)}{1-(-1)} = \frac{-1-11}{2} = -6\).

a) \(6\) b) \(-6\)

- First calculate \(k(0)\), \(k(2)\), and \(k(4)\). - Substitute the values into the average-rate-of-change formula. - Use the graph’s symmetry to explain the relationship between the answers.

1. Evaluate the needed values: \(k(0) = 0\), \(k(2) = 4\), and \(k(4) = 0\). 2. On \([0, 2]\), the average rate of change is \(\frac{k(2)-k(0)}{2-0} = \frac{4-0}{2} = 2\). 3. On \([2, 4]\), the average rate of change is \(\frac{k(4)-k(2)}{4-2} = \frac{0-4}{2} = -2\). 4. The rates have the same absolute value but opposite signs. The parabola is symmetric about \(x = 2\), increasing on the first interval and decreasing on the second.

a) \(2\) b) \(-2\) The rates have equal absolute values and opposite signs.

- Use rise divided by run on each drawn tangent line. - The derivative is positive where the graph rises and negative where it falls. - At a local maximum or minimum, the tangent is horizontal. - Estimate an undrawn tangent from the local direction of the curve.

1. At \(x = -1\), the tangent rises \(1\) unit for each \(1\) unit of run, so \(f'(-1) = 1\). 2. At \(x = 2\), the tangent falls \(2\) units for each \(1\) unit of run, so \(f'(2) = -2\). 3. At \(x = 0\), the graph has a maximum, so \(f'(0) = 0\). 4. The graph is decreasing at \(x = 3\), so \(f'(3)<0\). It is increasing at \(x = -2\), so \(f'(-2)>0\). 5. Near \(x = 1\), the graph falls about \(1\) unit for each \(1\) unit of run, so \(f'(1)\approx -1\).

a) \(f'(-1) = 1\) and \(f'(2) = -2\) b) \(f'(0) = 0\), \(f'(3)<0\), and \(f'(-2)>0\) c) \(f'(1)\approx -1\)

- A derivative value equals the slope of the tangent line. - Use two readable points on each drawn tangent. - Estimate the direction and steepness of the graph at \(A\). - A tangent at a minimum is horizontal.

1. The tangent at \(x = 1\) passes through \((1, -1.5)\) and \((2.5, 0)\). Its slope is \(\frac{0-(-1.5)}{2.5-1} = 1\), so \(f'(1) = 1\). 2. The tangent at \(x = -2\) passes through \((-2, 0)\) and \((-1, -2)\). Its slope is \(\frac{-2-0}{-1-(-2)} = -2\), so \(f'(-2) = -2\). 3. At \(A(2, 0)\), the graph rises about \(2\) units for each unit of run, so the slope is \(\approx 2\). 4. Point \(B(0, -2)\) is a minimum, so its tangent is horizontal and the slope is \(0\).

a) \(f'(1) = 1\) and \(f'(-2) = -2\) b) At \(A\): \(\approx 2\); at \(B\): \(0\)

- The average rate tells how many inches the water height rises per minute, on average. - Identify the graph points at the endpoints of each interval. - Include the correct units in each answer.

1. Read the graph values: \(h(0) = 0\), \(h(5) = 5\), and \(h(15) = 10\). 2. On \([0, 5]\), the average rate is \(\frac{5-0}{5-0} = 1\,\frac{\text{in.}}{\text{min}}\). 3. On \([5, 15]\), the average rate is \(\frac{10-5}{15-5} = \frac{5}{10} = 0.5\,\frac{\text{in.}}{\text{min}}\). 4. On \([0, 15]\), the average rate is \(\frac{10-0}{15-0} = \frac{2}{3}\,\frac{\text{in.}}{\text{min}} \approx 0.67\,\frac{\text{in.}}{\text{min}}\).

a) \(1\,\frac{\text{in.}}{\text{min}}\) b) \(0.5\,\frac{\text{in.}}{\text{min}}\) c) \(\frac{2}{3}\,\frac{\text{in.}}{\text{min}} \approx 0.67\,\frac{\text{in.}}{\text{min}}\)

- The derivative at a point is the slope of the tangent line. - Use two grid points to calculate each slope. - A horizontal line has slope \(0\).

1. At \(x = 0\), the tangent passes through \((0, 1)\) and \((1, 3)\). Its slope is \(\frac{3-1}{1-0} = 2\), so \(f'(0) = 2\). 2. At \(x = 2\), the tangent is horizontal, so \(f'(2) = 0\). 3. At \(x = 4\), the tangent passes through \((4, 1)\) and \((5, -1)\). Its slope is \(\frac{-1-1}{5-4} = -2\), so \(f'(4) = -2\).

\(f'(0) = 2\), \(f'(2) = 0\), and \(f'(4) = -2\)

- The derivative equals the tangent-line slope. - Horizontal tangent lines have slope \(0\). - Use two readable grid points on the tangent at \(B\).

1. At \(A(-2, 4)\), the tangent line is horizontal, so \(g'(-2) = 0\). 2. At \(B(0, 0)\), the tangent passes through \((0, 0)\) and \((1, -3)\). Its slope is \(\frac{-3-0}{1-0} = -3\), so \(g'(0) = -3\). 3. At \(C(2, -4)\), the tangent line is horizontal, so \(g'(2) = 0\).

\(g'(-2) = 0\), \(g'(0) = -3\), and \(g'(2) = 0\)

- Average rate of change is the slope of the secant line through the endpoint points. - Read the y-values at the given x-values first. - Use \(\frac{y_2-y_1}{x_2-x_1}\).

1. Read the graph values: \(h(0) = 0\), \(h(2) = -2\), and \(h(4) = 0\). 2. On \([0, 2]\), the average rate of change is \(\frac{h(2)-h(0)}{2-0} = \frac{-2-0}{2} = -1\). 3. On \([2, 4]\), the average rate of change is \(\frac{h(4)-h(2)}{4-2} = \frac{0-(-2)}{2} = 1\). 4. On \([0, 4]\), the average rate of change is \(\frac{h(4)-h(0)}{4-0} = \frac{0-0}{4} = 0\).

a) \(-1\) b) \(1\) c) \(0\)

- Average rate of change is the slope of the line through the two endpoint points. - Pay attention to the scale on each axis. - A rate of \(0\) means the endpoint function values are equal, not necessarily that the graph is constant between them.

1. Read the graph values: \(k(0) = 0\), \(k(2) = 2\), \(k(4) = 0\), and \(k(6) = -2\). 2. On \([0, 2]\), the average rate of change is \(\frac{k(2)-k(0)}{2-0} = \frac{2-0}{2} = 1\). 3. On \([0, 4]\), the average rate of change is \(\frac{k(4)-k(0)}{4-0} = \frac{0-0}{4} = 0\). 4. On \([2, 6]\), the average rate of change is \(\frac{k(6)-k(2)}{6-2} = \frac{-2-2}{4} = -1\).

a) \(1\) b) \(0\) c) \(-1\)

- Instantaneous rate of change is the tangent-line slope. - Choose two easy-to-read points on the line labeled \(t\).

1. Choose two points on the tangent line, such as \((1, 3)\) and \((3, 1)\). 2. The change in \(y\) is \(1-3 = -2\), and the change in \(x\) is \(3-1 = 2\). 3. The tangent slope is \(\frac{-2}{2} = -1\). Therefore, the instantaneous rate of change at \(P\) is \(-1\).

The instantaneous rate of change is \(-1\).

- Use the tangent line's y-intercept and point \(P\) as two easy-to-read points. - Compare the vertical change with the horizontal change.

1. Two points on the tangent line are \((0, 1)\) and \(P(4, 2)\). 2. The rise is \(2-1 = 1\), and the run is \(4-0 = 4\). 3. The slope is \(\frac{1}{4} = 0.25\).

The slope is \(0.25\).

- The large axis values do not change the slope formula. - Check how much each grid interval represents on both axes. - Use rise divided by run.

1. Two readable points on the tangent line are \((25, 0)\) and \((75, 50)\). 2. The rise is \(50-0 = 50\), and the run is \(75-25 = 50\). 3. The slope is \(\frac{50}{50} = 1\).

The slope is \(1\).

- Imagine the tangent line at each marked point. - Relate increasing, decreasing, and horizontal behavior to the sign of slope. - Compare a negative value, zero, and a positive value.

1. At \(A\), the graph is decreasing, so the tangent slope and derivative are negative. 2. At \(B\), the graph has a minimum, so the tangent is horizontal and the derivative is \(0\). 3. At \(C\), the graph is increasing, so the tangent slope and derivative are positive. 4. A negative number is less than \(0\) and less than a positive number, so the derivative is smallest at \(A\).

The derivative is smallest at \(A\), where the graph has a negative tangent slope.

- Average rate of change is the slope of a secant line. - Read the height at each interval endpoint. - What does a negative rate mean for the balloon’s motion? - Use \(\frac{h(b)-h(a)}{b-a}\).

1. Read the values from the graph: \(h(0) = 100\,\text{m}\), \(h(1) = 160\,\text{m}\), and \(h(4) = 100\,\text{m}\). 2. On \([0, 1]\), the average rate of change is \(\frac{h(1)-h(0)}{1-0} = \frac{160-100}{1} = 60\,\frac{\text{m}}{\text{min}}\). 3. On \([1, 4]\), the average rate of change is \(\frac{h(4)-h(1)}{4-1} = \frac{100-160}{3} = -20\,\frac{\text{m}}{\text{min}}\). The negative sign indicates descent.

a) \(60\,\frac{\text{m}}{\text{min}}\) b) \(-20\,\frac{\text{m}}{\text{min}}\); the balloon descends at an average rate of \(20\,\frac{\text{m}}{\text{min}}\).

- Read \(V(2)\) and \(V(8)\) from the graph. - Average rate of change is the slope of the secant line. - Divide a change in volume by a change in time. - Interpret the sign and units in context.

1. Read the endpoint values from the graph: \(V(2) = 10\,\text{L}\) and \(V(8) = 16\,\text{L}\). 2. The average rate of change is \(\frac{V(8)-V(2)}{8-2} = \frac{16-10}{6} = 1\,\frac{\text{L}}{\text{min}}\). 3. Over this interval, the volume increased by an average of \(1\) liter per minute.

\(1\,\frac{\text{L}}{\text{min}}\). The tank’s water volume increased by an average of \(1\) liter per minute.

- Interpret average rate of change as the slope of a secant line. - Read the function values at the interval endpoints. - When is a secant line horizontal? - Look for two graph points with the same y-coordinate.

1. Read the graph values: \(f(0) = 4\), \(f(1) = 3\), and \(f(-2) = 0\). 2. On \([0, 1]\), the average rate of change is \(\frac{f(1)-f(0)}{1-0} = \frac{3-4}{1} = -1\). 3. On \([-2, 0]\), the average rate of change is \(\frac{f(0)-f(-2)}{0-(-2)} = \frac{4-0}{2} = 2\). 4. An average rate of change of \(0\) occurs when the endpoint function values are equal. For example, \(f(-1) = f(1) = 3\), so \([-1, 1]\) works.

a) On \([0, 1]\): \(-1\); on \([-2, 0]\): \(2\) b) One possible interval is \([-1, 1]\).

- Average rate of change is the slope of the secant line through two graph points. - Read the y-values at the given x-values carefully. - Use \(\frac{y_2-y_1}{x_2-x_1}\). - For a rate of \(0\), find two points with the same y-coordinate.

1. Read the graph values: \(g(-1) = 2\), \(g(1) = -2\), and \(g(2) = 2\). 2. On \([-1, 1]\), the average rate of change is \(\frac{g(1)-g(-1)}{1-(-1)} = \frac{-2-2}{2} = -2\). 3. On \([1, 2]\), the average rate of change is \(\frac{g(2)-g(1)}{2-1} = \frac{2-(-2)}{1} = 4\). 4. For an average rate of change of \(0\), the endpoint function values must be equal. Since \(g(-2) = g(1) = -2\), the interval \([-2, 1]\) works.

a) On \([-1, 1]\): \(-2\); on \([1, 2]\): \(4\) b) One possible interval is \([-2, 1]\).

- As \(x\) increases, what happens to the values of this function? What does that suggest about the sign of the rate? - Evaluate the function at each interval endpoint. - Divide the change in the function value by the change in \(x\).

1. Evaluate the needed values: \(g(1) = 6\), \(g(3) = 2\), \(g(2) = 3\), and \(g(5) = \frac{6}{5} = 1.2\). 2. On \([1, 3]\), the average rate of change is \(\frac{g(3)-g(1)}{3-1} = \frac{2-6}{2} = -2\). 3. On \([2, 5]\), the average rate of change is \(\frac{g(5)-g(2)}{5-2} = \frac{1.2-3}{3} = -0.6\).

On \([1, 3]\): \(-2\) On \([2, 5]\): \(-0.6\)

- Read the temperature at the beginning and end of each interval. - Calculate the slope of the secant line through those two points. - The units are degrees Fahrenheit per minute. - What does the negative sign mean for the tea’s temperature?

1. On \([0, 20]\), read \(T(0) = 140\,{}^\circ\text{F}\) and \(T(20) = 100\,{}^\circ\text{F}\). The average rate of change is \(\frac{100-140}{20-0} = -2\,\frac{{}^\circ\text{F}}{\text{min}}\). 2. On \([20, 40]\), read \(T(20) = 100\,{}^\circ\text{F}\) and \(T(40) = 80\,{}^\circ\text{F}\). The average rate of change is \(\frac{80-100}{40-20} = -1\,\frac{{}^\circ\text{F}}{\text{min}}\). 3. The tea cools faster during the first interval because the first rate has the greater absolute value.

a) \(-2\,\frac{{}^\circ\text{F}}{\text{min}}\) b) \(-1\,\frac{{}^\circ\text{F}}{\text{min}}\)

- Read the plant’s height at \(t = 0\), \(t = 10\), and \(t = 20\). - Find the change in height over each period. - Divide the height change by the number of weeks. - Include the units inches per week.

1. On \([0, 10]\), read \(h(0) = 20\,\text{in.}\) and \(h(10) = 80\,\text{in.}\). The average growth rate is \(\frac{80-20}{10-0} = 6\,\frac{\text{in.}}{\text{week}}\). 2. On \([10, 20]\), read \(h(10) = 80\,\text{in.}\) and \(h(20) = 100\,\text{in.}\). The average growth rate is \(\frac{100-80}{20-10} = 2\,\frac{\text{in.}}{\text{week}}\). 3. The sunflower grew faster, on average, during the first \(10\) weeks.

a) \(6\,\frac{\text{in.}}{\text{week}}\) b) \(2\,\frac{\text{in.}}{\text{week}}\)

- A line through the origin makes the slope calculation direct. - Check whether the tangent rises or falls from left to right. - Use rise divided by run.

1. The derivative \(f'(0)\) equals the slope of the tangent line through the origin. 2. The tangent line passes through \((0, 0)\) and \((1, -1)\). 3. Its slope is \(\frac{-1-0}{1-0} = -1\). 4. Therefore, \(f'(0) = -1\).

\(f'(0) = -1\)

- The derivative at a point equals the slope of the tangent line there. - Read two clear points on the tangent, paying attention to the axis scales.

1. The instantaneous rate of change is the slope of the tangent line at \(x=2\). 2. Two points on the tangent are \((2, 1)\) and \((4, 0)\). 3. The slope is \(\frac{0-1}{4-2}=-\frac{1}{2}=-0.5\).

\(f^{\prime}(2)=-0.5\)

- Use the tangent line to determine the slope at the marked point. - Choose two clear grid points on the tangent that are reasonably far apart.

1. The derivative is the slope of the tangent line. 2. Two points on the tangent are \((1, 2)\) and \((2, 3)\). 3. Therefore, the slope is \(\frac{3-2}{2-1}=1\).

\(f^{\prime}(x_0)=1\)

- Pay close attention to signs when using negative x-coordinates. - A line that falls from left to right has negative slope. - Use rise divided by run.

1. The derivative \(f'(-2)\) equals the slope of the tangent line at \((-2, 0)\). 2. Two points on the tangent line are \((-2, 0)\) and \((-3, 2)\). 3. The slope is \(\frac{2-0}{-3-(-2)} = -2\). 4. Therefore, \(f'(-2) = -2\).

\(f'(-2) = -2\)

- Find two integer-coordinate points on the tangent line. - Since the tangent decreases from left to right, its slope must be negative.

1. The derivative at \(x=1\) is the slope of the tangent at \((1,2)\). 2. The tangent also passes through \((2,0)\). 3. Its slope is \(\frac{0-2}{2-1}=-2\).

\(f'(1)=-2\)

- How do you calculate the average slope between two points? - Find the height at each given x-value first. - Use the difference quotient for the two endpoint points. - What does a negative slope mean for the direction of the track?

1. Evaluate the endpoint heights: \(h(1) = -0.2(1)^2+5 = 4.8\,\text{ft}\) and \(h(4) = -0.2(4)^2+5 = 1.8\,\text{ft}\). 2. The average slope is \(\frac{h(4)-h(1)}{4-1} = \frac{1.8-4.8}{3} = -1\). 3. The negative sign means the track descends. On average, the height drops \(1\) foot for every \(1\) foot of horizontal distance over this section.

The average slope is \(-1\). The track drops an average of \(1\) foot vertically for each \(1\) foot of horizontal distance.

- Find slopes on the two drawn tangent lines. - A horizontal tangent means zero instantaneous change. - Which physical quantity describes a change in velocity over time?

1. The tangent from the start has slope \(\frac{4-0}{4-0}=1\,\frac{\text{m}}{\text{s}^2}\); this is the right-hand rate at \(t=0\). 2. The tangent approaching the stop has slope \(\frac{0-4}{20-16}=-1\,\frac{\text{m}}{\text{s}^2}\); this is the left-hand rate at \(t=20\). 3. The rate is zero at the top, \(t=10\), when the speed is \(5\,\frac{\text{m}}{\text{s}}\). 4. The rate of change of velocity is acceleration; here it is negative as the car slows down.

a) Just after \(t=0\): \(1\,\frac{\text{m}}{\text{s}^2}\); just before \(t=20\): \(-1\,\frac{\text{m}}{\text{s}^2}\) b) At \(t=10\,\text{s}\); speed \(5\,\frac{\text{m}}{\text{s}}\). c) Acceleration

- Locate the graph points at the endpoints of each interval. - Average rate of change is the slope of the secant line through those points. - Use \(\frac{y_2-y_1}{x_2-x_1}\). - Be careful when subtracting negative function values.

1. On \([1, 3]\), read \(g(1) = -2\) and \(g(3) = 4\). The average rate of change is \(\frac{g(3)-g(1)}{3-1} = \frac{4-(-2)}{2} = 3\). 2. On \([-2, 0]\), read \(g(-2) = 4\) and \(g(0) = 1\). The average rate of change is \(\frac{g(0)-g(-2)}{0-(-2)} = \frac{1-4}{2} = -1.5\).

a) \(3\) b) \(-1.5\)

- Speed is distance divided by time. - Compare the distance gained during each four-hour interval. - A steeper distance-time graph indicates a greater average speed. - A secant line through the first and last points represents an average over the full interval.

1. For the entire hike, the average speed is \(\frac{14\,\text{mi}}{8\,\text{h}} = 1.75\,\text{mph}\). 2. On \([0, 4]\), the hiker travels \(8\) miles, so the average speed is \(\frac{8}{4} = 2\,\text{mph}\). On \([4, 8]\), the hiker travels \(14-8 = 6\) miles, so the average speed is \(\frac{6}{4} = 1.5\,\text{mph}\). The hiker was faster on average during \([0, 4]\), which also appears as the steeper portion of the distance-time graph. 3. The secant slope through the start and end points is the average speed for the entire hike, \(1.75\,\text{mph}\).

1. \(1.75\,\text{mph}\) 2. \([0, 4]\); the average speeds are \(2\,\text{mph}\) and \(1.5\,\text{mph}\), respectively. 3. The secant slope represents the hiker’s average speed over all eight hours.

- Recall how the slope of a tangent line is related to the derivative at the point of tangency. - Compare the rise and run between two grid points on the tangent line. - Choose two points on the line that lie exactly on grid intersections.

1. Point \(P(2, 4)\) is the point of tangency. 2. Two convenient points on tangent line \(t\) are \((1, 3)\) and \((2, 4)\). 3. The tangent slope is \(m = \frac{4-3}{2-1} = 1\). 4. The derivative equals the tangent slope at the point, so \(f'(2) = 1\).

\(f'(2) = 1\)

- Which two outputs mark the ends of the interval? - How do you compare those outputs to the change in input?

1. The endpoints are \((1,3)\) and \((3,7)\). 2. The secant slope is \(\frac{7-3}{3-1}=2\). 3. Average rate depends on the endpoints, so it uses the secant line rather than the changing slopes along the curve.

\(2\); the secant uses the changes between the two endpoints, whereas a tangent gives the slope at just one point of the curve.

- The slope of a straight line does not depend on which two points you choose on it. - Compare the rise along equal horizontal distances in the two panels.

1. In a), the tangent rises by \(3\) units for a run of \(1\), so its slope is \(3\). 2. In b), the tangent rises by \(1\) unit for the same run, so its slope is \(1\). 3. The local slope is greater at the marked point in a).

a); tangent slopes \(3\) in a) and \(1\) in b).

- Use the output change divided by the input change for each interval. - Compare the rates as the right endpoint moves closer to the starting input.

1. From \(1\) to \(1.5\), the rate is \(\frac{2.25-1}{1.5-1}=2.5\). 2. From \(1\) to \(1.1\), the rate is \(\frac{1.21-1}{1.1-1}=2.1\). 3. From \(1\) to \(1.01\), the rate is \(\frac{1.0201-1}{0.01}=2.01\). 4. The shorter intervals give rates approaching \(2\), so the estimated instantaneous slope is \(2\).

Rates: \(2.5\), \(2.1\), \(2.01\), in order of decreasing interval length. Estimated tangent slope: \(2\).

- Follow the graph from left to right on each side of the corner. - Ask whether the line direction stays the same as you pass through the corner.

1. Just left of zero, the graph falls with slope \(-1\) as x increases. 2. Just right of zero, it rises with slope \(1\). 3. Because the two sides have different slopes, one straight tangent cannot match both directions.

No. The slope is \(-1\) from the left and \(1\) from the right.

- What change in money does the average rate describe over all four hours? - Keep the amount already in the jar separate from the new tips.

1. The change is average rate times elapsed time: \(3\cdot4=12\) dollars. 2. Add the original \(\$9\): \(9+12=21\) dollars.

\(\$21\)

- Compare the starting temperature with the target temperature. - Use the meaning of the hourly rate to connect that change with an elapsed time.

1. The temperature change is \(80-65=15\,\text{°F}\). 2. At \(5\,\text{°F/h}\), the time is \(15\div5=3\) hours. 3. Three hours after 2 p.m. is 5 p.m.

5 p.m.

- Read the curve's heights at the start, middle, and end. - Find each interval's change in output divided by its change in input. - Compare the signs of the short-interval rates with the shape of the curve.

1. Read \(f(0)=1\), \(f(1)=0\), and \(f(2)=1\). 2. The average rates are \(\frac{0-1}{1-0}=-1\) on \([0,1]\), \(\frac{1-0}{2-1}=1\) on \([1,2]\), and \(\frac{1-1}{2-0}=0\) on \([0,2]\). 3. The negative first rate matches the decreasing curve and the positive second rate matches the increasing curve. The zero rate over the whole interval does not mean the curve is flat.

\([0,1]\): \(-1\); \([1,2]\): \(1\); \([0,2]\): \(0\). The curve decreases first and increases afterward, even though its full-interval average rate is zero.

- How do you find the slope between two points on a curve? - Track the units: volume changes over a time interval. - What does comparing the magnitudes of the two rates tell you about how quickly the process occurs? - As the amount of water in the tank decreases, what might happen to the outflow rate?

1. Evaluate the volume at the interval endpoints: \(V(0) = 200\), \(V(10) = 0.5 \cdot 10^2 - 20 \cdot 10 + 200 = 50\), and \(V(20) = 0.5 \cdot 20^2 - 20 \cdot 20 + 200 = 0\). 2. Over \([0, 10]\), the average rate of change is \(\frac{V(10) - V(0)}{10 - 0} = \frac{50 - 200}{10} = -15\,\frac{\text{L}}{\text{min}}\). 3. Over \([10, 20]\), the average rate of change is \(\frac{V(20) - V(10)}{20 - 10} = \frac{0 - 50}{10} = -5\,\frac{\text{L}}{\text{min}}\). 4. The volume decreases by an average of \(15\,\frac{\text{L}}{\text{min}}\) in the first interval but only \(5\,\frac{\text{L}}{\text{min}}\) in the second. The smaller magnitude means the tank drains more slowly later in the process.

a) Over \([0, 10]\), \(-15\,\frac{\text{L}}{\text{min}}\); over \([10, 20]\), \(-5\,\frac{\text{L}}{\text{min}}\). b) The outflow rate decreases in magnitude. On average, \(15\) liters per minute leave the tank during the first half, compared with \(5\) liters per minute during the second half.

- Which input would make the denominator equal to zero? - How do you find the slope of a secant line through two points on a graph? - Which function values must you calculate before applying the average-rate-of-change formula?

1. The denominator cannot equal zero: \(x - 2 \ne 0\), so \(x \ne 2\). Therefore, the domain is \(\mathbb{R} \setminus \{2\}\). 2. Evaluate the endpoint values: \(f(-4) = \frac{6}{-4 - 2} = -1\), \(f(-1) = \frac{6}{-1 - 2} = -2\), and \(f(1) = \frac{6}{1 - 2} = -6\). 3. Over \(I_1\), \(\frac{f(-1) - f(-4)}{-1 - (-4)} = \frac{-2 - (-1)}{3} = -\frac{1}{3}\). 4. Over \(I_2\), \(\frac{f(1) - f(-1)}{1 - (-1)} = \frac{-6 - (-2)}{2} = -2\). 5. Over \(I_3\), \(\frac{f(1) - f(-4)}{1 - (-4)} = \frac{-6 - (-1)}{5} = -1\).

\(D_f\): \(\mathbb{R} \setminus \{2\}\) \(I_1\): \(-\frac{1}{3}\) \(I_2\): \(-2\) \(I_3\): \(-1\)

- What condition must the expression under a square root satisfy? - Recall the difference quotient \(\frac{f(b) - f(a)}{b - a}\). - First write the function values at \(x = 0\), \(x = 3\), and \(x = 8\).

1. The radicand must be nonnegative: \(x + 1 \ge 0\), so \(x \ge -1\). Therefore, the domain is \([-1, \infty)\). 2. Evaluate the endpoint values: \(f(0) = 1\), \(f(3) = 2\), and \(f(8) = 3\). 3. Over \(I_1\), \(\frac{f(3) - f(0)}{3 - 0} = \frac{2 - 1}{3} = \frac{1}{3}\). 4. Over \(I_2\), \(\frac{f(8) - f(3)}{8 - 3} = \frac{3 - 2}{5} = \frac{1}{5} = 0.2\). 5. Over \(I_3\), \(\frac{f(8) - f(0)}{8 - 0} = \frac{3 - 1}{8} = \frac{1}{4} = 0.25\).

\(D_f\): \([-1, \infty)\) \(I_1\): \(\frac{1}{3}\) \(I_2\): \(\frac{1}{5} = 0.2\) \(I_3\): \(\frac{1}{4} = 0.25\)

- How does a quantity change over time under constant percent growth? - What calculation represents the average change over an interval? - What does the slope of a secant line through two points represent? - What units should the average rate have?

1. The annual growth factor is \(1.04\), so \(K(t) = 1000 \cdot 1.04^t\). 2. Over \(I_1\), \(K(0) = 1000\) and \(K(10) = 1000 \cdot 1.04^{10} \approx 1480.24\). Thus, \(\frac{K(10) - K(0)}{10 - 0} \approx \frac{1480.24 - 1000}{10} \approx 48.02\,\frac{\text{dollars}}{\text{year}}\). 3. Over \(I_2\), \(K(20) = 1000 \cdot 1.04^{20} \approx 2191.12\) and \(K(30) = 1000 \cdot 1.04^{30} \approx 3243.40\). Thus, \(\frac{K(30) - K(20)}{30 - 20} \approx \frac{3243.40 - 2191.12}{10} \approx 105.23\,\frac{\text{dollars}}{\text{year}}\). 4. Each value is the average annual increase in dollars during that interval. Compound interest applies the same percentage to a larger balance in later years, so the later absolute increase is greater.

a) \(K(t) = 1000 \cdot 1.04^t\) b) (1) Approximately \(\$48.02\) per year (2) Approximately \(\$105.23\) per year c) The values are average annual increases in dollars. Because the account balance grows over time, the same \(4\%\) rate produces larger dollar increases in later years.

- Recall the difference-quotient formula. - What does a negative rate of change say about the graph? - Compare the absolute amounts lost during the two intervals. - Does the amount lost depend on how much is still present?

1. Over \(I_1\), \(C(0) = 200\) and \(C(2) = 200 \cdot 0.8^2 = 128\). Therefore, \(\frac{C(2) - C(0)}{2 - 0} = \frac{128 - 200}{2} = -36\,\frac{\text{mg}}{\text{L} \cdot \text{h}}\). 2. Over \(I_2\), \(C(4) = 200 \cdot 0.8^4 = 81.92\) and \(C(6) = 200 \cdot 0.8^6 \approx 52.43\). Therefore, \(\frac{C(6) - C(4)}{6 - 4} \approx \frac{52.43 - 81.92}{2} \approx -14.75\,\frac{\text{mg}}{\text{L} \cdot \text{h}}\). 3. The negative signs indicate that the concentration decreases over time. 4. The first rate has greater magnitude because exponential decay removes a fixed percentage of the amount present. More of the active ingredient is present at the beginning, so the absolute decrease per hour is larger then.

a) \(I_1\): \(-36\,\frac{\text{mg}}{\text{L} \cdot \text{h}}\) \(I_2\): Approximately \(-14.75\,\frac{\text{mg}}{\text{L} \cdot \text{h}}\) b) Each negative sign indicates that the concentration is decreasing. c) The concentration decreases faster in absolute terms during \(I_1\). Because \(20\%\) of the remaining concentration is lost each hour, the amount lost per hour becomes smaller as the concentration decreases.

- Convert all time measurements to the same unit before using the velocity. - What does a difference quotient represent graphically when two points on a graph are connected? - Must an object move at its average velocity at every instant in the interval?

1. Convert the elapsed time to seconds: \(\Delta t = (28 - 12) \cdot 60 = 960\,\text{s}\). 2. Find the change in height: \(\Delta h = 3.5\,\frac{\text{m}}{\text{s}} \cdot 960\,\text{s} = 3360\,\text{m}\). 3. Subtract this change from the final height: \(h(t_1) = 7450 - 3360 = 4090\,\text{m}\). 4. Geometrically, the average upward velocity is the slope of the secant line through the two measured points. If time is measured in minutes on the graph, the secant slope is \(210\,\frac{\text{m}}{\text{min}}\), equivalent to \(3.5\,\frac{\text{m}}{\text{s}}\). 5. An average velocity does not require the velocity to be constant. Without the actual height function, the path of the graph between the two endpoints is unknown, so the height at \(20\) minutes cannot be determined exactly.

a) \(4090\,\text{m}\) b) The average upward velocity is the slope of the secant line through the two endpoint measurements. With time measured in minutes, that slope is \(210\,\frac{\text{m}}{\text{min}}\). The balloon’s velocity may vary between the endpoints, so its height at \(20\) minutes cannot be found from the average velocity alone.

- How are average rate of change, elapsed time, and total change related? - Recall the definition of a difference quotient. - What is the graphical difference between a secant line and a tangent line?

1. The time interval has length \(15 - 5 = 10\,\text{min}\). 2. The total temperature change is \(\Delta T = -1.8\,\frac{\text{K}}{\text{min}} \cdot 10\,\text{min} = -18\,\text{K}\). 3. Therefore, \(T(5) = T(15) - \Delta T = 42.5 - (-18) = 60.5\,^\circ\text{C}\). A temperature difference has the same numerical value in kelvins and degrees Celsius. 4. The difference quotient is \(\frac{T(15) - T(5)}{15 - 5} = \frac{42.5 - 60.5}{10} = -1.8\,\frac{\text{K}}{\text{min}}\). 5. This quotient gives the average cooling rate over the full interval, or the slope of a secant line. The value \(T'(10)\) gives the instantaneous cooling rate at \(t = 10\), or the slope of the tangent line there. Without the temperature function, the instantaneous rate cannot be computed and may or may not equal the average rate.

a) \(60.5\,^\circ\text{C}\) b) \(\frac{T(15) - T(5)}{15 - 5} = -1.8\,\frac{\text{K}}{\text{min}}\). This is the secant slope over \([5, 15]\), while \(T'(10)\) is the tangent slope at \(t = 10\). The instantaneous rate cannot be determined without knowing \(T\).

- Recall the difference-quotient formula. - Substitute both interval endpoints into the function. - Expand \((2 + h)^2\) carefully. - Factor the numerator so that you can simplify the quotient.

1. Evaluate the function at \(x = 2\): \(f(2) = 2 \cdot 2^2 - 2 = 6\). 2. Evaluate the function at \(x = 2 + h\): \(f(2 + h) = 2(2 + h)^2 - (2 + h) = 2(4 + 4h + h^2) - 2 - h = 2h^2 + 7h + 6\). 3. Form the difference quotient: \(\frac{f(2 + h) - f(2)}{(2 + h) - 2} = \frac{2h^2 + 7h}{h}\). 4. Since \(h > 0\), divide by \(h\): \(\frac{h(2h + 7)}{h} = 2h + 7\).

\(2h + 7\)

- A secant slope is the average rate of change over the corresponding interval. - How do you combine the fractions in the numerator? - Can you factor the numerator so that a common factor cancels? - Be careful with signs when distributing.

1. Evaluate the function at the two inputs: \(g(1) = 3\) and \(g(1 + a) = \frac{3}{1 + a}\). 2. Form the secant slope: \(m_{\text{sec}} = \frac{g(1 + a) - g(1)}{(1 + a) - 1} = \frac{\frac{3}{1 + a} - 3}{a}\). 3. Combine the terms in the numerator: \(\frac{3}{1 + a} - 3 = \frac{3 - 3(1 + a)}{1 + a} = \frac{-3a}{1 + a}\). 4. Divide by \(a\): \(m_{\text{sec}} = \frac{-3a}{1 + a} \cdot \frac{1}{a} = -\frac{3}{1 + a}\).

\(-\frac{3}{1 + a}\)

- Use the difference quotient \(\frac{h(t_2) - h(t_1)}{t_2 - t_1}\). - What quantity results when a change in height is divided by elapsed time? - For part 3, set the average-rate-of-change expression equal to \(10\). - Can you divide the numerator by \(b\) after simplifying?

1. Evaluate the endpoints: \(h(0.5) = -5 \cdot 0.5^2 + 20 \cdot 0.5 + 2 = 10.75\) and \(h(1.5) = -5 \cdot 1.5^2 + 20 \cdot 1.5 + 2 = 20.75\). Thus, \(\frac{h(1.5) - h(0.5)}{1.5 - 0.5} = \frac{20.75 - 10.75}{1} = 10\,\frac{\text{m}}{\text{s}}\). 2. This value is the ball’s average vertical velocity over the interval, measured in meters per second. 3. Set \(\frac{h(b) - h(0)}{b - 0} = 10\). Since \(h(0) = 2\), \(\frac{-5b^2 + 20b + 2 - 2}{b} = 10\). Because \(b > 0\), divide by \(b\): \(-5b + 20 = 10\). Solving gives \(b = 2\,\text{s}\).

1. \(10\,\frac{\text{m}}{\text{s}}\) 2. The ball’s average vertical velocity, in meters per second. 3. \(b = 2\,\text{s}\)

- What does a difference quotient represent in a real-world context? - What is the name of the line through two points on a function’s graph? - Compare the rates for the two time intervals.

1. Evaluate the needed values: \(h(0) = 5\), \(h(4) = -0.1 \cdot 4^2 + 2 \cdot 4 + 5 = 11.4\), and \(h(8) = -0.1 \cdot 8^2 + 2 \cdot 8 + 5 = 14.6\). Over \([0, 4]\), the average growth rate is \(\frac{11.4 - 5}{4} = 1.6\,\frac{\text{cm}}{\text{week}}\). Over \([4, 8]\), it is \(\frac{14.6 - 11.4}{4} = 0.8\,\frac{\text{cm}}{\text{week}}\). 2. The first average growth rate is the slope of the secant line through \((0, 5)\) and \((4, 11.4)\) on the graph. 3. The average growth rate decreases from \(1.6\) to \(0.8\) centimeters per week, so the plant’s growth slows over time in this model.

1. First 4 weeks: \(1.6\,\frac{\text{cm}}{\text{week}}\) Weeks 4 through 8: \(0.8\,\frac{\text{cm}}{\text{week}}\) 2. It is the slope of the secant line through the graph points at \(t = 0\) and \(t = 4\). 3. The growth is slowing because the average growth rate decreases.

- Since the table tracks total route distance, the change over any interval is the distance driven during that interval. - For each average speed, divide the change in miles by the number of elapsed minutes. - Compare the first and last requested intervals rather than using the full trip totals.

1. Over \([0,80]\), the average speed is \(\frac{48-0}{80-0}=0.6\,\frac{\text{mi}}{\text{min}}\). 2. From 0 to 30 minutes, it is \(\frac{15-0}{30}=0.5\,\frac{\text{mi}}{\text{min}}\). 3. From 30 to 80 minutes, it is \(\frac{48-15}{50}=0.66\,\frac{\text{mi}}{\text{min}}\). 4. During the first 10 minutes it is \(\frac{9}{10}=0.9\,\frac{\text{mi}}{\text{min}}\). During the final 15 minutes, it is \(\frac{48-33}{15}=1\,\frac{\text{mi}}{\text{min}}\), which is greater.

a) \(0.6\,\frac{\text{mi}}{\text{min}}\) b) First 30 minutes: \(0.5\,\frac{\text{mi}}{\text{min}}\); remaining 50 minutes: \(0.66\,\frac{\text{mi}}{\text{min}}\). c) First 10 minutes: \(0.9\,\frac{\text{mi}}{\text{min}}\); final 15 minutes: \(1\,\frac{\text{mi}}{\text{min}}\). The final interval is faster.

- Average acceleration is the average rate of change of velocity. - Use \(\frac{\Delta v}{\Delta t}\). - Identify the correct endpoint times and speeds for each interval. - What average acceleration results when velocity does not change over an interval?

1. Over \([0, 20]\), the average acceleration is \(\frac{v(20) - v(0)}{20 - 0} = \frac{30 - 0}{20} = 1.5\,\frac{\text{m}}{\text{s}^2}\). 2. Over \([0, 10]\), it is \(\frac{12 - 0}{10} = 1.2\,\frac{\text{m}}{\text{s}^2}\). 3. Over \([10, 20]\), it is \(\frac{30 - 12}{10} = 1.8\,\frac{\text{m}}{\text{s}^2}\). 4. Over \([0, 4]\), it is \(\frac{12 - 0}{4} = 3\,\frac{\text{m}}{\text{s}^2}\). Over \([18, 20]\), it is \(\frac{30 - 25}{2} = 2.5\,\frac{\text{m}}{\text{s}^2}\). Therefore, the average acceleration is greater over \([0, 4]\).

a) \(1.5\,\frac{\text{m}}{\text{s}^2}\) b) First half: \(1.2\,\frac{\text{m}}{\text{s}^2}\) Second half: \(1.8\,\frac{\text{m}}{\text{s}^2}\) c) \([0, 4]\), because \(3\,\frac{\text{m}}{\text{s}^2} > 2.5\,\frac{\text{m}}{\text{s}^2}\).

- Recall the slope formula for a line through two points. - Can you factor a common coefficient from the numerator? - After factoring, do you recognize a difference of squares? - What factor cancels from the numerator and denominator?

1. The secant slope is \(m_{\text{sec}} = \frac{f(x) - f(2)}{x - 2} = \frac{0.5x^2 + 2 - 4}{x - 2} = \frac{0.5x^2 - 2}{x - 2}\), where \(x \ne 2\). 2. Factor the numerator: \(\frac{0.5(x^2 - 4)}{x - 2} = \frac{0.5(x - 2)(x + 2)}{x - 2} = 0.5(x + 2) = 0.5x + 1\), where \(x \ne 2\). 3. For \(x = 2.2\), \(m_{\text{sec}} = 0.5(2.2 + 2) = 0.5 \cdot 4.2 = 2.1\).

a) \(m_{\text{sec}} = \frac{0.5x^2 - 2}{x - 2}\), where \(x \ne 2\) b) \(m_{\text{sec}} = 0.5x + 1\), where \(x \ne 2\) c) \(2.1\)

- Substitute \(x_0 + h\) and \(x_0\) into the function. - Be careful when subtracting the second function value. - Expand \((x_0 + h)^2\). - Factor \(h\) from the numerator before simplifying.

1. Form the difference quotient: \(\frac{g(x_0 + h) - g(x_0)}{h} = \frac{(x_0 + h)^2 - 5(x_0 + h) - (x_0^2 - 5x_0)}{h}\). 2. Expand and combine like terms: \(\frac{x_0^2 + 2x_0h + h^2 - 5x_0 - 5h - x_0^2 + 5x_0}{h} = \frac{2x_0h + h^2 - 5h}{h}\). 3. Factor and divide by \(h\): \(\frac{h(2x_0 + h - 5)}{h} = 2x_0 + h - 5\). 4. For \(x_0 = 3\) and \(h = 0.1\), the rate is \(2 \cdot 3 + 0.1 - 5 = 1.1\).

a) \(\frac{(x_0 + h)^2 - 5(x_0 + h) - (x_0^2 - 5x_0)}{h}\) b) \(2x_0 + h - 5\) c) \(1.1\)

- Use the slope formula for a line through two points. - Substitute the expression for the point \(Q\) into the formula. - Be careful when subtracting a negative function value. - Factor the numerator to simplify the quotient.

1. Use the slope formula: \(m_{\text{sec}} = \frac{f(x) - f(-1)}{x - (-1)}\). 2. Substitute the function values: \(m_{\text{sec}} = \frac{(2x^2 - 3) - (-1)}{x + 1} = \frac{2x^2 - 2}{x + 1}\). 3. Factor the numerator: \(\frac{2(x^2 - 1)}{x + 1} = \frac{2(x - 1)(x + 1)}{x + 1}\). 4. Since \(x \ne -1\), simplify to \(m_{\text{sec}} = 2(x - 1) = 2x - 2\).

\(m_{\text{sec}} = 2x - 2\), where \(x \ne -1\)

- Use the difference quotient for the slope between two graph points. - Recall how to factor a difference of cubes such as \(x^3 - a^3\). - Polynomial division is another option when the denominator corresponds to a zero of the numerator.

1. Form the difference quotient: \(m_{\text{sec}} = \frac{g(x) - g(2)}{x - 2} = \frac{(x^3 + 2) - 10}{x - 2} = \frac{x^3 - 8}{x - 2}\). 2. Factor the difference of cubes: \(x^3 - 8 = (x - 2)(x^2 + 2x + 4)\). 3. Since \(x \ne 2\), cancel \(x - 2\) to obtain \(m_{\text{sec}} = x^2 + 2x + 4\).

\(m_{\text{sec}} = x^2 + 2x + 4\), where \(x \ne 2\)

- What does a secant slope represent in this context? - Track the units of each rate. - Divide the change in temperature by the change in time. - A negative rate means that the temperature is decreasing.

1. Over \([0, 10]\), \(\frac{T(10) - T(0)}{10 - 0} = \frac{62.0 - 85.0}{10} = -2.3\,\frac{^\circ\text{C}}{\text{min}}\). 2. Over \([20, 50]\), \(\frac{T(50) - T(20)}{50 - 20} = \frac{23.0 - 46.0}{30} = -\frac{23}{30} \approx -0.767\,\frac{^\circ\text{C}}{\text{min}}\). 3. The rates for the consecutive intervals are: \([0, 5]\): \(-2.5\,\frac{^\circ\text{C}}{\text{min}}\); \([5, 10]\): \(-2.1\,\frac{^\circ\text{C}}{\text{min}}\); \([10, 20]\): \(-1.6\,\frac{^\circ\text{C}}{\text{min}}\); \([20, 30]\): \(-1.1\,\frac{^\circ\text{C}}{\text{min}}\); \([30, 50]\): \(-0.6\,\frac{^\circ\text{C}}{\text{min}}\). The greatest cooling rate in magnitude occurs over \([0, 5]\).

a) \(-2.3\,\frac{^\circ\text{C}}{\text{min}}\) b) \(-\frac{23}{30}\,\frac{^\circ\text{C}}{\text{min}} \approx -0.767\,\frac{^\circ\text{C}}{\text{min}}\) c) \([0, 5]\), with an average rate of \(-2.5\,\frac{^\circ\text{C}}{\text{min}}\)

- Divide the change in the population by the number of years in the interval. - What does a negative result mean for the population? - Compare the rates for all consecutive measurement intervals. - The time gaps between measurements are not all equal.

1. From 2015 to 2023, \(\frac{2100 - 1200}{2023 - 2015} = \frac{900}{8} = 112.5\,\frac{\text{animals}}{\text{year}}\). 2. From 2017 to 2018, \(\frac{1820 - 1550}{1} = 270\,\frac{\text{animals}}{\text{year}}\). From 2018 to 2020, \(\frac{1740 - 1820}{2} = -40\,\frac{\text{animals}}{\text{year}}\). 3. The negative rate means that the population decreased on average during 2018–2020. 4. The consecutive-interval rates are: 2015–2017, \(175\,\frac{\text{animals}}{\text{year}}\); 2017–2018, \(270\,\frac{\text{animals}}{\text{year}}\); 2018–2020, \(-40\,\frac{\text{animals}}{\text{year}}\); and 2020–2023, \(120\,\frac{\text{animals}}{\text{year}}\). The greatest average growth occurred from 2017 to 2018.

a) \(112.5\,\frac{\text{animals}}{\text{year}}\) b) 2017–2018: \(270\,\frac{\text{animals}}{\text{year}}\) 2018–2020: \(-40\,\frac{\text{animals}}{\text{year}}\) c) The population decreased during that interval. d) 2017–2018, with \(270\,\frac{\text{animals}}{\text{year}}\)

- Rewrite \(|x|\) separately for positive and negative values of \(x\). - Simplify the resulting linear expression in each case. - Read the slope from the coefficient of \(x\).

1. When \(x > 0\), \(|x| = x\). Therefore, \(g(x) = x+2x = 3x\), which has slope \(3\). 2. When \(x < 0\), \(|x| = -x\). Therefore, \(g(x) = -x+2x = x\), which has slope \(1\). 3. Thus, the tangent-line slope is \(3\) when \(a > 0\) and \(1\) when \(a < 0\).

a) \(3\) b) \(1\)

- A forward difference measures average motion just after the chosen instant. - To find the peak without differentiation rules, rewrite the quadratic as a constant minus a squared term. - At the peak, try the difference quotient with a short time change h in either direction.

1. For \(\Delta t=0.001\), the forward difference quotient simplifies to \(-32t+64-0.016\). 2. At \(t=1,2,3\), the rates are \(31.984\), \(-0.016\), and \(-32.016\) feet per second, respectively. 3. The positive rate at \(t=1\) means rising; the negative rate at \(t=3\) means falling. The tiny negative rate at \(t=2\) comes from averaging over a short interval just after the peak. 4. Complete the square: \(H(t)=-16(t-2)^2+67\). Thus the height is greatest at \(t=2\), when \(H(2)=67\,\text{ft}\). 5. At the vertex, \(\frac{H(2+h)-H(2)}{h}=\frac{-16h^2}{h}=-16h\) for \(h\ne0\), and this approaches \(0\) as \(h\to0\). The exact instantaneous velocity there is zero.

a) At \(t=1\): \(31.984\,\text{ft/s}\); at \(t=2\): \(-0.016\,\text{ft/s}\); at \(t=3\): \(-32.016\,\text{ft/s}\). b) Rising at \(t=1\), falling at \(t=3\); the small negative forward rate at \(t=2\) spans the interval after the peak. c) \(H(t)=-16(t-2)^2+67\). At \(t=2\,\text{s}\), the exact rate is \(0\,\text{ft/s}\) and the maximum height is \(67\,\text{ft}\).

- A difference of cost values represents a change in total cost. - Dividing a cost change by an amount change gives cost per unit of amount. - In economics, the derivative of a cost function is called marginal cost.

1. \(C(x)-C(x_0)\) is the change in total cost when the amount of pollutants removed changes from \(x_0\) kilograms to \(x\) kilograms. Its units are dollars. 2. \(\frac{C(x)-C(x_0)}{x-x_0}\) is the average change in cost per kilogram between \(x_0\) and \(x\). Its units are dollars per kilogram. 3. \(C'(x_0)\) is the marginal cost at \(x_0\). It approximates the additional cost of removing one more kilogram when \(x_0\) kilograms have already been removed. Its units are dollars per kilogram.

a) The change in total cost, measured in dollars b) The average change in cost per kilogram from \(x_0\) to \(x\), measured in dollars per kilogram c) The marginal cost at \(x_0\), measured in dollars per kilogram

- A constant rate of change means the same rise over run everywhere. - Use the given point to find the missing intercept in a line equation. - Compare the vertical differences between points before and after an upward shift.

1. Constant rate means constant slope, so the graph is a decreasing straight line. 2. Write \(h(x)=-0.5x+b\). Using \(h(4)=2\) gives \(2=-2+b\), so \(b=4\) and \(h(x)=-0.5x+4\). 3. Moving every point up by 10 leaves all rises between pairs of points unchanged. The shifted graph also has slope \(-0.5\).

1. A decreasing straight line. 2. \(h(x)=-0.5x+4\) 3. No. Its rate remains \(-0.5\) because a vertical shift does not change slope.

- How is average velocity defined using distance traveled and elapsed time? - Write the difference quotient for \(s(t)\). - Use the difference-of-squares factorization to simplify. - Substitute the values only after simplifying the general expression.

1. Form the difference quotient: \(\frac{s(t_2) - s(t_1)}{t_2 - t_1} = \frac{\frac{1}{2}at_2^2 - \frac{1}{2}at_1^2}{t_2 - t_1}\). 2. Factor the numerator: \(\frac{\frac{1}{2}a(t_2^2 - t_1^2)}{t_2 - t_1} = \frac{\frac{1}{2}a(t_2 - t_1)(t_2 + t_1)}{t_2 - t_1}\). 3. Since \(t_1 < t_2\), cancel \(t_2 - t_1\): \(v_{\text{avg}} = \frac{1}{2}a(t_1 + t_2)\). 4. Substitute the given values: \(v_{\text{avg}} = \frac{1}{2} \cdot 2.4 \cdot (5 + 15) = 24\,\frac{\text{m}}{\text{s}}\).

a) \(v_{\text{avg}} = \frac{1}{2}a(t_1 + t_2)\) b) \(24\,\frac{\text{m}}{\text{s}}\)

- How are secant slope and average rate of change related? - Use the slope formula with the two given graph points. - For part b, write an equation with \(x_2\) as the unknown. - Simplify the quotient before solving the equation.

1. Evaluate the function values: \(f(2) = 0.2 \cdot 2^2 + 3 = 3.8\) and \(f(7) = 0.2 \cdot 7^2 + 3 = 12.8\). 2. The secant slope is \(\frac{f(7) - f(2)}{7 - 2} = \frac{12.8 - 3.8}{5} = 1.8\). 3. For part b, set \(\frac{f(x_2) - f(0)}{x_2 - 0} = 1.6\). Since \(f(0) = 3\), \(\frac{0.2x_2^2}{x_2} = 1.6\). 4. Because \(x_2 > 0\), divide by \(x_2\): \(0.2x_2 = 1.6\), so \(x_2 = 8\).

a) \(1.8\) b) \(x_2 = 8\)

- How is a percent increase converted to a growth factor? - What does \(a\) represent in an exponential model? - Use the secant-slope formula between the endpoint values. - What units should an area growth rate have?

1. The initial value is \(a = 2\), and the weekly growth factor is \(b = 1 + 0.25 = 1.25\). Therefore, \(A(t) = 2 \cdot 1.25^t\). 2. \(A(4) = 2 \cdot 1.25^4 \approx 4.8828\,\text{m}^2\), and \(A(8) = 2 \cdot 1.25^8 \approx 11.9209\,\text{m}^2\). 3. Over \([0, 4]\), the average rate is \(\frac{A(4) - A(0)}{4} \approx \frac{4.8828 - 2}{4} \approx 0.7207\,\frac{\text{m}^2}{\text{week}}\). 4. Over \([4, 8]\), the average rate is \(\frac{A(8) - A(4)}{4} \approx \frac{11.9209 - 4.8828}{4} \approx 1.7595\,\frac{\text{m}^2}{\text{week}}\).

a) \(A(t) = 2 \cdot 1.25^t\) b) After \(4\) weeks: approximately \(4.88\,\text{m}^2\) After \(8\) weeks: approximately \(11.92\,\text{m}^2\) c) First \(4\) weeks: approximately \(0.72\,\frac{\text{m}^2}{\text{week}}\) Weeks \(4\) through \(8\): approximately \(1.76\,\frac{\text{m}^2}{\text{week}}\)

- What factor represents a decrease by a fixed percent? - What does a negative rate of change mean in this context? - Use the average-rate-of-change formula over each time interval. - Why does the same percent loss produce a smaller dollar loss later?

1. The initial value is \(a = 5000\), and the annual decay factor is \(b = 1 - 0.15 = 0.85\). Thus, \(V(t) = 5000 \cdot 0.85^t\). 2. \(V(0) = 5000\) and \(V(2) = 5000 \cdot 0.85^2 = 3612.50\). Over \([0, 2]\), the average rate of change is \(\frac{3612.50 - 5000}{2} = -693.75\,\frac{\text{dollars}}{\text{year}}\). 3. \(V(5) = 5000 \cdot 0.85^5 \approx 2218.53\) and \(V(7) = 5000 \cdot 0.85^7 \approx 1602.89\). Over \([5, 7]\), the average rate of change is \(\frac{1602.89 - 2218.53}{2} \approx -307.82\,\frac{\text{dollars}}{\text{year}}\). 4. The later rate has smaller magnitude because the same \(15\%\) loss is applied to a lower market value.

a) \(V(t) = 5000 \cdot 0.85^t\) b) \(-\$693.75\) per year c) Approximately \(-\$307.82\) per year. The later annual dollar loss is smaller because the bicycle has a lower value by years \(5\) through \(7\).

- What factor represents an annual decrease of \(4\%\)? - Use the difference quotient over each requested interval. - How does the base amount to which the \(4\%\) applies change over time?

1. The initial value is \(W_0 = 15{,}000\), and the annual decay factor is \(a = 1 - 0.04 = 0.96\). Thus, \(W(t) = 15{,}000 \cdot 0.96^t\). 2. For \([0, 5]\), \(W(0) = 15{,}000\) and \(W(5) = 15{,}000 \cdot 0.96^5 \approx 12{,}230.59\). The average rate is \(\frac{12230.59 - 15000}{5} \approx -553.88\,\frac{\text{dollars}}{\text{year}}\). 3. For \([10, 15]\), \(W(10) = 15{,}000 \cdot 0.96^{10} \approx 9972.49\) and \(W(15) = 15{,}000 \cdot 0.96^{15} \approx 8131.30\). The average rate is \(\frac{8131.30 - 9972.49}{5} \approx -368.24\,\frac{\text{dollars}}{\text{year}}\). 4. The later rate has smaller magnitude because the constant \(4\%\) loss is applied to a smaller remaining value.

a) \(W(t) = 15{,}000 \cdot 0.96^t\) b) \([0, 5]\): approximately \(-\$553.88\) per year \([10, 15]\): approximately \(-\$368.24\) per year c) The absolute loss becomes smaller because the same \(4\%\) is applied to a decreasing value.

- In a difference quotient, what do the denominator values tell you about the interval? - Read the corresponding function values from each graph. - Check which graph contains the two points represented by each quotient. - After matching, simplify each quotient.

1. Difference quotient I uses the points \((1, -1.5)\) and \((3, 2.5)\), which lie on the graph of \(f\). It represents the average rate of change of \(f\) over \([1, 3]\). 2. Difference quotient II uses the points \((0, 1)\) and \((2, 0)\), which lie on the graph of \(g\). It represents the average rate of change of \(g\) over \([0, 2]\). 3. For \(f\), \(\frac{2.5 - (-1.5)}{3 - 1} = \frac{4}{2} = 2\). 4. For \(g\), \(\frac{0 - 1}{2 - 0} = -\frac{1}{2} = -0.5\).

1. Quotient I matches \(f\) over \([1,3]\), using \((1,-1.5)\) and \((3,2.5)\); Quotient II matches \(g\) over \([0,2]\), using \((0,1)\) and \((2,0)\). 2. \(f\): \(2\); \(g\): \(-0.5\)

- For the two drawn tangents, read rise over run from grid points. - Use symmetry to compare the tangent directions at opposite x-values. - For the undrawn tangent at x=1, focus on its direction rather than an unsupported exact number.

1. Tangent line \(t_1\) has slope \(\frac{-2-0}{-2-(-4)}=-1\), so \(f'(-2)=-1\). 2. Tangent line \(t_2\) has slope \(\frac{2-0}{1-0}=2\), so \(f'(0)=2\). 3. By the curve's symmetry, its tangent near \(x=2\) slopes downward about \(1\) per unit of run, so \(f'(2)\approx-1\). 4. The curve is rising at \(x=1\), so its tangent slope there is positive. An exact decimal slope is not readable from this graph.

a) \(f'(-2)=-1\) and \(f'(0)=2\) b) \(f'(2)\approx-1\); \(f'(1)\) is positive.

- Read the endpoints of each secant from the graph to compute its average slope. - Follow the curve from left to right to decide when it rises or falls. - A rate at one input is not the same as the average rate over a long interval.

1. Read the endpoint values \(f(-1)=-3\), \(f(0)=1\), \(f(3)=1\), \(f(4)=2\), and \(f(5)=6\). 2. The average slopes are a) \(\frac{1-(-3)}{3-(-1)}=1\); b) \(\frac{2-1}{4}=0.25\); c) \(\frac{6-1}{2}=2.5\). 3. Statement (1) is true: the graph falls at \(x=2\). 4. Statement (2) is true: the smooth curve has a peak at \(x=1\), so its tangent is horizontal. 5. Statement (3) is false: the graph falls from \(x=1\) to \(x=3\).

**Part 1:** a) \(1\) b) \(0.25\) c) \(2.5\) **Part 2:** (1) True; the curve slopes downward at \(x=2\). (2) True; the tangent is horizontal at the smooth peak. (3) False; the curve falls between \(x=1\) and \(x=3\).

- Read each quotient as \(\frac{y_2 - y_1}{x_2 - x_1}\). - Identify the two points represented by each card. - Find the graph that contains both points. - Pay close attention to negative coordinates.

1. Card A represents the points \((0, 1)\) and \((4, 1)\). Only graph 1, the graph of \(f\), contains both points. Therefore, graph 1 matches Card A. 2. Card B represents the points \((-1, -1)\) and \((3, 1)\). Only graph 2, the graph of \(g\), contains both points. Therefore, graph 2 matches Card B. 3. Card C represents the points \((0, 0)\) and \((2, -1)\). Only graph 3, the graph of \(h\), contains both points. Therefore, graph 3 matches Card C.

Graph 1: Card A, using \((0,1)\) and \((4,1)\); Graph 2: Card B, using \((-1,-1)\) and \((3,1)\); Graph 3: Card C, using \((0,0)\) and \((2,-1)\)

- Use elevation change divided by horizontal distance, then convert to a percent. - Read the halfway elevation from the graph. - Compare the first-half average with the grade over a shorter part of that half.

1. Average grade is the elevation change divided by the horizontal distance, as a percent. 2. Over the full trail, the grade is \(\frac{1200-1300}{1600}\cdot100\%=-6.25\%\). 3. Over the first \(800\) meters, it is \(\frac{1240-1300}{800}\cdot100\%=-7.5\%\). Over the second \(800\) meters, it is \(\frac{1200-1240}{800}\cdot100\%=-5\%\). The first half is steeper downhill on average. 4. Between \(400\) and \(800\) meters, the elevation drops from \(1280\) to \(1240\) meters. Its grade is \(\frac{1240-1280}{800-400}\cdot100\%=-10\%\), steeper than the first half's average.

a) \(-6.25\%\) b) First half \(-7.5\%\); second half \(-5\%\). The first half has the steeper average downhill grade. c) No. From \(400\) to \(800\) meters the grade is \(-10\%\), steeper than the first-half average.

- Instantaneous growth rate is represented by a tangent-line slope. - Use two readable grid points on the shown tangent line to estimate its slope. - Average growth rate is change in height divided by change in time. - Observe how the steepness of the curve changes from left to right.

1. The shown tangent line at day \(4\) passes through the grid points \((1, 1)\) and \((5, 2)\). 2. Its slope is \(\frac{2-1}{5-1} = \frac{1}{4} = 0.25\). The instantaneous growth rate on day \(4\) is \(0.25\,\frac{\text{in.}}{\text{day}}\). 3. Evaluate the endpoint heights: \(H(0) = 0.5\) and \(H(4) = 1.75\). 4. The average growth rate is \(\frac{H(4)-H(0)}{4-0} = \frac{1.75-0.5}{4} = 0.3125\,\frac{\text{in.}}{\text{day}}\). 5. The average rate over the first four days is greater than the instantaneous rate on day \(4\). 6. The graph becomes less steep from left to right, so the slopes of its tangent lines decrease as time increases.

a) \(0.25\,\frac{\text{in.}}{\text{day}}\) b) \(0.3125\,\frac{\text{in.}}{\text{day}}\); the average rate is greater than the instantaneous rate on day \(4\). c) The curve becomes progressively flatter, so its tangent slopes decrease over time.

- Identify what each axis and unit represent. - Use the slope between the endpoint points of each interval. - What does a negative vertical rate mean? - Does an average velocity have to equal the velocity at every moment? - Can you assume a curved graph is linear between its endpoints?

1. From \(0\) to \(20\) minutes, the balloon rises from \(100\,\text{m}\) to \(400\,\text{m}\), first slowly, then more quickly, and then more slowly. From \(20\) to \(40\) minutes, it remains at \(400\,\text{m}\). From \(40\) to \(60\) minutes, it descends to \(100\,\text{m}\), with a changing descent rate. 2. Over \([0, 20]\), the average rate is \(\frac{400 - 100}{20} = 15\,\frac{\text{m}}{\text{min}}\). Over \([20, 40]\), it is \(\frac{400 - 400}{20} = 0\,\frac{\text{m}}{\text{min}}\). Over \([40, 60]\), it is \(\frac{100 - 400}{20} = -15\,\frac{\text{m}}{\text{min}}\). 3. The graph does show \(h(10) = 250\,\text{m}\), so the numerical conclusion happens to be correct. However, using the average rate to assume constant linear motion is invalid because the graph is curved. The agreement occurs because of the shape and symmetry of this particular graph.

a) \([0,20]\): the balloon rises from \(100\,\text{m}\) to \(400\,\text{m}\) at a changing rate; \([20,40]\): it stays at \(400\,\text{m}\); \([40,60]\): it descends from \(400\,\text{m}\) to \(100\,\text{m}\) at a changing rate. b) \([0,20]\): \(15\,\frac{\text{m}}{\text{min}}\); \([20,40]\): \(0\,\frac{\text{m}}{\text{min}}\); \([40,60]\): \(-15\,\frac{\text{m}}{\text{min}}\). c) The graph gives \(h(10)=250\,\text{m}\), but the observer’s reasoning is invalid because an average velocity does not imply constant velocity.

- Read a function value by matching an x-coordinate to its y-coordinate on the graph. - Average rate of change is the slope of the secant line through the endpoint points. - Instantaneous rate of change is the slope of the tangent line. - Use two visible points on the tangent line to calculate its slope.

1. From the graph, \(f(0)=1\) and \(f(6)=4\). 2. Over \([0,6]\), the average rate is \(\frac{f(6)-f(0)}{6-0}=\frac{4-1}{6}=0.5\). Geometrically, this is the slope of the secant line through \((0,1)\) and \((6,4)\). 3. The tangent line at \(x=2\) passes through \((0,2)\) and \((2,4)\), so \(f'(2)=\frac{4-2}{2-0}=1\). Geometrically, this is the slope of the tangent line to the graph at \((2,4)\).

a) \(f(0)=1\) and \(f(6)=4\) b) \(0.5\), the slope of the secant line through \((0,1)\) and \((6,4)\) c) \(f'(2)=1\), the slope of the tangent line at \((2,4)\)

- Interpret slope as change in water level divided by change in time. - Use the drawn tangent line at \(A\) to estimate rise over run. - A local maximum has a horizontal tangent. - Use the drawn tangent at \(D\), and pay attention to both the sign and magnitude of its slope.

1. The instantaneous rate of change describes how quickly the water level is rising or falling at a specific time. Its units are feet per hour. 2. At \(A\), the tangent line has slope approximately \(3.14\,\frac{\text{ft}}{\text{h}}\), so the water level is rising at about \(3.1\,\frac{\text{ft}}{\text{h}}\). 3. Point \(B\) is a maximum, so its tangent is horizontal and the instantaneous rate is \(0\,\frac{\text{ft}}{\text{h}}\). 4. At \(D\), the tangent line has slope approximately \(-2.72\,\frac{\text{ft}}{\text{h}}\), so the water level is falling at about \(2.7\,\frac{\text{ft}}{\text{h}}\). 5. The negative sign confirms that the level is falling, but the magnitude is about \(2.7\), not \(4\). Noémie’s estimate is too high.

a) The rate tells how quickly the water level is rising or falling at one time, measured in feet per hour. b) At \(A\): approximately \(3.1\,\frac{\text{ft}}{\text{h}}\) At \(B\): \(0\,\frac{\text{ft}}{\text{h}}\) c) The claim is incorrect. At \(D\), the tangent-line estimate is approximately \(-2.7\,\frac{\text{ft}}{\text{h}}\), so the water level is falling at about \(2.7\) feet per hour.

- Average rate of change is a secant-line slope. - A horizontal tangent represents an instantaneous rate of \(0\). - Compare the steepness of the three dashed tangent lines before calculating a slope. - For the steepest tangent, use two clear grid points on that line to calculate rise over run.

1. From the graph, \(h(0)=2\) and \(h(4)=10\). The average rate on \([0,4]\) is \(\frac{10-2}{4-0}=2\,\frac{\text{m}}{\text{h}}\). 2. Also, \(h(6)=2\). The average rate on \([4,6]\) is \(\frac{2-10}{6-4}=-4\,\frac{\text{m}}{\text{h}}\). 3. At \(t=4\), the graph has a horizontal tangent, so the instantaneous rate is \(0\,\frac{\text{m}}{\text{h}}\). The water level is momentarily neither rising nor falling. 4. Comparing the three dashed tangent lines, the tangent at \(t=2\) is the steepest upward line. 5. That tangent passes through \((0,0)\) and \((2,6)\), so its slope is \(\frac{6-0}{2-0}=3\,\frac{\text{m}}{\text{h}}\).

a) On \([0,4]\): \(2\,\frac{\text{m}}{\text{h}}\); on \([4,6]\): \(-4\,\frac{\text{m}}{\text{h}}\) b) \(0\,\frac{\text{m}}{\text{h}}\); the water level is momentarily constant. c) \(t=2\) hours; \(3\,\frac{\text{m}}{\text{h}}\)

- Instantaneous rate of change is the slope of a tangent line. - Imagine a tangent line at each marked point. - Compare how steeply the graph rises or falls. - Use the sign of each listed value to eliminate mismatches.

1. At \(x_1 = -1\), the graph rises gently. The tangent slope is \(0.5\), which is Value B. 2. At \(x_2 = 0\), the graph rises steeply. The tangent slope is \(3\), which is Value C. 3. At \(x_3 = 2\), the graph falls. The tangent slope is \(-1\), which is Value A.

\(x_1\): Value B, \(x_2\): Value C, \(x_3\): Value A

- Convert kilometers to meters before calculating a percent grade. - Grade is change in elevation divided by horizontal distance. - What does a negative grade mean? - Can uphill and downhill changes cancel in a net average?

1. Convert each \(4\)-kilometer horizontal interval to \(4000\,\text{m}\), and convert the full \(12\)-kilometer course to \(12{,}000\,\text{m}\). 2. From the start to \(B\), the elevation change is \(260 - 100 = 160\,\text{m}\). The average grade is \(\frac{160}{4000} = 0.04 = 4\%\). 3. From \(B\) to \(C\), the elevation change is \(180 - 260 = -80\,\text{m}\). The average grade is \(\frac{-80}{4000} = -0.02 = -2\%\). 4. From \(C\) to the finish, the elevation change is \(420 - 180 = 240\,\text{m}\). The average grade is \(\frac{240}{4000} = 0.06 = 6\%\). 5. Over the full course, the elevation change is \(420 - 100 = 320\,\text{m}\). The average grade is \(\frac{320}{12000} = \frac{2}{75} \approx 0.0267 \approx 2.67\%\). 6. The overall average can understate the effort because uphill and downhill sections offset each other, while the runner still experiences the separate climbs and descents.

a) 1. \(4\%\) 2. \(-2\%\) 3. \(6\%\) b) Approximately \(2.67\%\). This single average does not represent the effort well because the climbs and descents partly cancel.

- The rate depends on changes in elevation, not on whether the elevation itself is positive or negative. - Compare endpoint heights for each average slope. - For a statement about an instant, check whether the curve rises or falls near that time.

1. Read \(h(0)=3\), \(h(1)=2\), \(h(5)=4\), \(h(6)=3\), and \(h(8)=-5\) meters from the graph, relative to the reference mark. 2. On \([1,5]\), the rate is \(\frac{4-2}{4}=0.5\,\text{m/h}\); on \([0,6]\), it is \(\frac{3-3}{6}=0\,\text{m/h}\); on \([5,8]\), it is \(\frac{-5-4}{3}=-3\,\text{m/h}\). 3. Statement (1) is true: the graph rises locally at \(t=3\). 4. Statement (2) is false: the curve stops falling and begins rising before \(t=3\). 5. Statement (3) is true: the plotted curve rises throughout \([2,4]\).

a) (1) \(0.5\,\text{m/h}\) (2) \(0\,\text{m/h}\) (3) \(-3\,\text{m/h}\) b) (1) True; the graph rises locally at \(t=3\). (2) False; the surface stops falling before \(t=3\). (3) True; the curve rises throughout the interval.

- Average speed is distance divided by time. - Convert minutes to hours before reporting miles per hour. - Instantaneous speed is the slope of the distance-time graph. - Use two convenient points on the tangent line at the steepest point.

1. Over the full ride, the cyclist travels \(2.5\,\text{mi}\) in \(10\) minutes, or \(\frac{1}{6}\) hour. The average speed is \(\frac{2.5}{1/6} = 15\,\text{mph}\). 2. From \(t = 2\) to \(t = 8\), the cyclist travels approximately \(2.12 - 0.38 \approx 1.74\,\text{mi}\) in \(6\) minutes, or \(0.1\) hour. The average speed is approximately \(\frac{1.74}{0.1} \approx 17.4\,\text{mph}\). 3. The tangent at \(t = 5\) passes through approximately \((1, 0)\) and \((9, 2.5)\). Its slope is \(\frac{2.5 - 0}{9 - 1} \approx 0.3125\,\frac{\text{mi}}{\text{min}}\). Converting gives \(0.3125 \cdot 60 \approx 18.75\,\text{mph}\). Since this is the maximum instantaneous speed, the cyclist never exceeds \(20\,\text{mph}\).

a) \(15\,\text{mph}\) b) Approximately \(17.4\,\text{mph}\) c) No. The maximum instantaneous speed is approximately \(18.75\,\text{mph}\).

- Start by finding the two input values for each requested interval. - Read the corresponding output heights from the graph and its scale. - Divide the output difference by the input difference before selecting a card.

1. Graph 1 has values \(f(-2)=-0.2\) and \(f(3)=0.8\), giving average rate \(\frac{0.8-(-0.2)}{3-(-2)}=0.2\). It matches B. 2. Graph 2 has \((1,2)\) and \((4,0.5)\), giving \(\frac{0.5-2}{4-1}=-0.5\). It matches C. 3. Graph 3 passes through \((0,0)\) and \((2,2)\), giving \(\frac{2-0}{2-0}=1\). It matches A.

Graph 1: Card B; \(\frac{0.8-(-0.2)}{3-(-2)}=0.2\). Graph 2: Card C; \(\frac{0.5-2}{4-1}=-0.5\). Graph 3: Card A; \(\frac{2-0}{2-0}=1\).

- Locate each specified pair of input values on the corresponding graph. - Read their heights, then calculate rise divided by run. - Use the completed calculations to check your card matches.

1. Graph 1 has endpoints \((0,-1)\) and \((2,2)\); its rate is \(\frac{2-(-1)}{2-0}=1.5\), matching Card C. 2. Graph 2 has endpoints \((0,3)\) and \((2,1)\); its rate is \(\frac{1-3}{2-0}=-1\), matching Card D. 3. Graph 3 has endpoints \((1,0)\) and \((3,0)\); its rate is \(\frac{0-0}{3-1}=0\), matching Card A. 4. Graph 4 has endpoints \((-1,0)\) and \((3,2)\); its rate is \(\frac{2-0}{3-(-1)}=0.5\), matching Card B.

Graph 1: Card C; \(\frac{2-(-1)}{2-0}=1.5\). Graph 2: Card D; \(\frac{1-3}{2-0}=-1\). Graph 3: Card A; \(\frac{0-0}{3-1}=0\). Graph 4: Card B; \(\frac{2-0}{3-(-1)}=0.5\).

- A line that falls from left to right has negative slope. - Use two readable points on each drawn tangent line. - Imagine tangent lines at \(C\) and \(D\) to compare their steepness. - Check that the estimated slope at \(C\) is more negative than at \(D\).

1. The tangent at \(P\) passes through \((0, 4)\) and \((4, 0)\). Its slope is \(\frac{0-4}{4-0} = -1\), so \(f'(2) = -1\). 2. The tangent at \(Q\) passes through \((-4, 0)\) and \((0, -4)\). Its slope is \(\frac{-4-0}{0-(-4)} = -1\), so \(f'(-2) = -1\). 3. Near \(C(1, 4)\), the graph falls about \(4\) units for each unit of run, so the slope is \(\approx -4\). 4. Near \(D(4, 1)\), the graph falls about \(1\) unit for every \(4\) units of run, so the slope is \(\approx -0.25\).

a) \(f'(2) = -1\) and \(f'(-2) = -1\) b) At \(C\): \(\approx -4\); at \(D\): \(\approx -0.25\)

- Use a slope triangle on the dashed tangent. - Think about whether a vertical translation changes a graph’s steepness. - A horizontal translation moves the entire pattern of slopes by the same amount.

1. Along the tangent, a run of \(1\) corresponds to a rise of \(1\), so the slope of \(f\) at \(x=3\) is \(1\). 2. A vertical translation changes every y-value by the same constant but does not change tangent slopes. Therefore, \(g\) also has slope \(1\) at \(x=3\). 3. Function \(h\) is \(f\) shifted \(2\) units to the right. Its entire slope pattern shifts right by \(2\), so the corresponding slope occurs at \(x=5\).

1. The slope is \(1\). 2. A vertical translation does not change slopes at corresponding x-values. 3. \(x=5\)

- The average rate of change is the slope of the line through the two endpoint points. - Make a table of the values of both functions at \(x = -2\), \(x = 0\), and \(x = 4\). - Be careful when subtracting negative values.

1. For \(f\), read \(f(-2) = 0\), \(f(0) = -1\), and \(f(4) = 3\). 2. On \([-2, 0]\), the average rate for \(f\) is \(\frac{-1-0}{0-(-2)} = -0.5\). On \([0, 4]\), it is \(\frac{3-(-1)}{4-0} = 1\). 3. For \(g\), read \(g(-2) = -2\), \(g(0) = 2\), and \(g(4) = -2\). 4. On \([-2, 0]\), the average rate for \(g\) is \(\frac{2-(-2)}{0-(-2)} = 2\). On \([0, 4]\), it is \(\frac{-2-2}{4-0} = -1\).

a) \(f: -0.5\); \(g: 2\) b) \(f: 1\); \(g: -1\)

- In a distance-time graph, what does slope represent? - Calculate the slope through the two endpoint points for each interval. - Compare the steepness of the graph over the two intervals. - Pay attention to the axis scales.

1. On \([0, 10]\), the graph gives \(s(0) = 0\,\text{m}\) and \(s(10) = 5\,\text{m}\). The average speed is \(\frac{5-0}{10-0} = 0.5\,\frac{\text{m}}{\text{s}}\). 2. On \([10, 20]\), the graph gives \(s(10) = 5\,\text{m}\) and \(s(20) = 20\,\text{m}\). The average speed is \(\frac{20-5}{20-10} = 1.5\,\frac{\text{m}}{\text{s}}\). 3. The second average speed is three times the first. The graph is steeper on \([10, 20]\), and its upward curvature indicates that the car is speeding up.

a) \(0.5\,\frac{\text{m}}{\text{s}}\) b) \(1.5\,\frac{\text{m}}{\text{s}}\) c) The second value is larger because the distance-time graph is steeper on \([10, 20]\). The upward curvature shows that the car is speeding up.

- Average rate of change is the slope of a secant line. - The units are volume per unit of time. - Compare the slopes of the graph over the four intervals. - What does decreasing water volume mean physically?

1. Over the full interval, the volume changes by \(400-50 = 350\,\text{L}\) in \(10\) minutes. The average inflow rate is \(\frac{350}{10} = 35\,\frac{\text{L}}{\text{min}}\). 2. On \([0, 2]\), the average rate is \(\frac{120-50}{2-0} = 35\,\frac{\text{L}}{\text{min}}\). 3. On \([2, 5]\), the average rate is \(\frac{200-120}{5-2} = \frac{80}{3}\,\frac{\text{L}}{\text{min}} \approx 26.67\,\frac{\text{L}}{\text{min}}\). 4. On \([5, 8]\), the average rate is \(\frac{350-200}{8-5} = 50\,\frac{\text{L}}{\text{min}}\). 5. On \([8, 10]\), the average rate is \(\frac{400-350}{10-8} = 25\,\frac{\text{L}}{\text{min}}\). Therefore, the greatest rate occurs on \([5, 8]\). 6. A negative slope would mean the water volume is decreasing, so the outflow and other losses would have to exceed the inflow.

a) \(35\,\frac{\text{L}}{\text{min}}\) b) \([5, 8]\), with an average rate of \(50\,\frac{\text{L}}{\text{min}}\) c) The volume would need to decrease; for example, outflow and other losses could exceed inflow.

- Compare endpoint heights, not the tallest point reached between them. - Calculate the three signed slopes before deciding their order. - Can a curve return to its starting height after rising or falling?

1. On Graph 1, the endpoints \((0,0)\) and \((5,5)\) give \(\frac{5-0}{5-0}=1\). 2. On Graph 2, the endpoints \((-2,2)\) and \((2,2)\) give \(\frac{2-2}{2-(-2)}=0\). 3. On Graph 3, the endpoints \((1,4)\) and \((4,1)\) give \(\frac{1-4}{4-1}=-1\). 4. The ranking is Graph 1, Graph 2, Graph 3. Graph 2's endpoints have the same height even though its curve rises and falls between them.

Graph 1: \(\frac{5-0}{5-0}=1\). Graph 2: \(\frac{2-2}{2-(-2)}=0\). Graph 3: \(\frac{1-4}{4-1}=-1\). Greatest to least: Graph 1, Graph 2, Graph 3. Graph 2's endpoints have equal heights although the curve rises and falls between them.

- A rate may be negative even when all graphed outputs are positive. - Work out each rate before comparing them. - Account for the minus sign when subtracting one rate from the other.

1. Graph 1 has endpoints \((2,1.5)\) and \((4,6)\). Its average rate is \(\frac{6-1.5}{4-2}=2.25\). 2. Graph 2 has endpoints \((0,2)\) and \((3,1)\). Its average rate is \(\frac{1-2}{3-0}=-\frac{1}{3}\). 3. Graph 1 has the greater rate by \(2.25-(-\frac{1}{3})=\frac{31}{12}\).

Graph 1: \(\frac{6-1.5}{4-2}=2.25\) from \((2,1.5)\) to \((4,6)\). Graph 2: \(\frac{1-2}{3-0}=-\frac{1}{3}\) from \((0,2)\) to \((3,1)\). Graph 1 has the greater rate by \(\frac{31}{12}\).

- Identify what the horizontal and vertical axes represent. - Average rate of change is the slope of a secant line through two graph points. - Subtract the endpoint water depths in the numerator and the endpoint times in the denominator. - What do positive, zero, and negative rates mean for the amount of water?

1. During the first \(4\) hours, the water depth rises from \(4\) inches to \(16\) inches. It remains constant from hour \(4\) through hour \(8\), then falls to \(8\) inches by hour \(12\). 2. On \([0, 4]\), the average rate is \(\frac{16-4}{4-0} = 3\,\frac{\text{in.}}{\text{h}}\). 3. On \([4, 8]\), the average rate is \(\frac{16-16}{8-4} = 0\,\frac{\text{in.}}{\text{h}}\). 4. On \([8, 12]\), the average rate is \(\frac{8-16}{12-8} = -2\,\frac{\text{in.}}{\text{h}}\). 5. The positive rate suggests net inflow, such as rain. The zero rate means no net change; inflow and outflow may have balanced. The negative rate indicates net outflow, possibly from use, drainage, or a leak.

a) The water depth rises, stays constant, and then falls. b) \([0, 4]\): \(3\,\frac{\text{in.}}{\text{h}}\); \([4, 8]\): \(0\,\frac{\text{in.}}{\text{h}}\); \([8, 12]\): \(-2\,\frac{\text{in.}}{\text{h}}\) c) The positive rate suggests rain or other net inflow; the zero rate suggests balanced inflow and outflow; the negative rate suggests use, drainage, or a leak.

- Average speed is the average rate of change of distance with respect to time. - Convert miles per minute to miles per hour by multiplying by \(60\). - A steeper distance-time graph indicates a greater speed. - For part c, decide whether the graph supports a constant pace.

1. On \([0, 20]\), the average speed is \(\frac{2.25-0}{20-0} = 0.1125\,\frac{\text{mi}}{\text{min}}\). Converting to miles per hour gives \(0.1125 \cdot 60 = 6.75\,\text{mph}\). 2. On \([20, 30]\), the average speed is \(\frac{3-2.25}{30-20} = 0.075\,\frac{\text{mi}}{\text{min}}\), or \(4.5\,\text{mph}\). 3. On \([30, 40]\), the average speed is \(\frac{4-3}{40-30} = 0.1\,\frac{\text{mi}}{\text{min}}\), or \(6\,\text{mph}\). 4. The runner was fastest on average during \([0, 20]\). The secant line through that interval’s endpoints is the steepest of the three. 5. The friend’s statement is only a projection based on the overall average speed. The graph shows that the runner’s speed varies, so maintaining the same average pace for another \(4\) miles is not guaranteed.

a) \([0, 20]\): \(6.75\,\text{mph}\); \([20, 30]\): \(4.5\,\text{mph}\); \([30, 40]\): \(6\,\text{mph}\) b) \([0, 20]\), because its endpoint secant line is the steepest. c) The claim is not guaranteed. It assumes the overall average pace will remain the same even though the graph shows changing speeds.

- At a local maximum or minimum, the tangent is horizontal. - Compare where the curve rises and falls most steeply. - A steep downward slope is a large negative number and is therefore small.

1. At \(Q\), the graph is increasing most steeply, so it has the greatest positive slope. 2. At \(R\), the graph has a local maximum, so its tangent is horizontal and its slope is \(0\). 3. At \(P\), the graph is decreasing only slightly, so its slope is a small negative value. 4. At \(S\), the graph is decreasing most steeply, so it has the most negative and therefore smallest slope. 5. From greatest to smallest slope, the order is \(Q, R, P, S\).

a) \(S\) b) \(Q, R, P, S\) c) The slope at \(R\) is \(0\).

- Instantaneous rate of change is estimated with a tangent-line slope. - Use two convenient grid points close to the drawn tangent line. - Interpret the negative sign in the temperature context. - Compare how steep the graph is early and late in the process.

1. The drawn tangent line represents the instantaneous rate of change at \(t = 10\). 2. It passes near the readable grid points \((0, 160)\) and \((20, 80)\). 3. Its slope is approximately \(\frac{80-160}{20-0} = -4.0\,\frac{{}^\circ\text{F}}{\text{min}}\). Thus, the instantaneous temperature rate is \(\approx -4.0\,\frac{{}^\circ\text{F}}{\text{min}}\). 4. The negative value means the coffee is cooling at about \(4.0\) degrees Fahrenheit per minute after \(10\) minutes. 5. The graph becomes flatter as time increases, so the magnitude of its negative tangent slopes decreases. Therefore, the cooling rate slows over time.

a) The instantaneous temperature rate is \(\approx -4.0\,\frac{{}^\circ\text{F}}{\text{min}}\), so the coffee is cooling at about \(4.0\) degrees Fahrenheit per minute. b) The graph becomes flatter over time, so the magnitude of the negative slope decreases.

- Use the drawn tangent line to estimate the rate at one time. - Calculate the tangent slope from two readable points. - Observe whether the curve becomes steeper or flatter from left to right.

1. The instantaneous inflow rate at \(t = 4\) is the slope of the drawn tangent line at \((4, 4)\). 2. The tangent line passes through \((0, 2)\) and \((12, 8)\). 3. Its slope is \(\frac{8-2}{12-0} = \frac{6}{12} = 0.5\,\frac{\text{gal}}{\text{min}}\). 4. The graph becomes flatter as time increases, so its tangent slopes decrease. Therefore, the inflow rate decreases over time.

a) \(\approx 0.5\,\frac{\text{gal}}{\text{min}}\) b) The inflow rate decreases because the graph becomes flatter over time.

- Locate the graph points at the requested times. - The difference quotient is the slope of a secant line. - When time is on the horizontal axis, slope describes change per unit of time. - For the instantaneous rate, use two readable points on the shown tangent line to estimate its slope.

1. From the graph, \(T(0) = 40\,{}^\circ\text{F}\) and \(T(6) = 64\,{}^\circ\text{F}\). 2. The difference quotient is \(\frac{64-40}{6-0} = \frac{24}{6} = 4\). 3. From 6:00 a.m. to noon, the temperature increased by an average of \(4\,{}^\circ\text{F}\) per hour. 4. The shown tangent line passes through readable points such as \((0, 42)\) and \((2, 52)\), so its slope is \(\frac{52-42}{2-0} = 5\,\frac{{}^\circ\text{F}}{\text{h}}\). Thus the instantaneous rate of change at \(t = 2\) is approximately \(5\,\frac{{}^\circ\text{F}}{\text{h}}\).

1. \(T(0) = 40\,{}^\circ\text{F}\); \(T(6) = 64\,{}^\circ\text{F}\) 2. \(4\) 3. The average warming rate was \(4\,\frac{{}^\circ\text{F}}{\text{h}}\) from 6:00 a.m. to noon. 4. Approximately \(5\,\frac{{}^\circ\text{F}}{\text{h}}\)

- Average rate of change uses only the endpoints of the interval. - Look for a special feature of the graph at \(x = 3\). - Recall the geometric meanings of secant and tangent slopes. - Find where the graph rises most steeply.

1. Read the endpoint values: \(g(0) = 3\) and \(g(5) = 5\). 2. The average rate of change is \(\frac{g(5)-g(0)}{5-0} = \frac{5-3}{5} = 0.4\). 3. At \(x = 3\), the graph has a local minimum, so the tangent is horizontal and \(g'(3) = 0\). 4. The value \(0.4\) is the slope of the secant line through \((0, 3)\) and \((5, 5)\). The value \(0\) is the slope of the tangent line at \(x = 3\). 5. The graph is increasing most steeply at the right endpoint, \(x = 5\), so the instantaneous rate of change is greatest there.

1. \(0.4\) 2. \(0\) 3. Part 1 is a secant-line slope; part 2 is a tangent-line slope. 4. \(x = 5\)

- A tangent-line slope gives the instantaneous rate of change. - First decide whether each slope is positive, negative, or nearly zero. - Then compare the steepness of the graph at the four marked points. - One listed value will not be used.

1. At \(t_1 = 0\), the graph is decreasing with a tangent slope of about \(-1.0\,\frac{\text{m}}{\text{min}}\). 2. At \(t_2 = 2\), the graph is decreasing more steeply, with a tangent slope of about \(-1.8\,\frac{\text{m}}{\text{min}}\). 3. At \(t_3 = 4\), the graph is nearly horizontal but still decreasing, with a tangent slope of about \(-0.2\,\frac{\text{m}}{\text{min}}\). 4. At \(t_4 = 5\), the graph is increasing, with a tangent slope of about \(1.5\,\frac{\text{m}}{\text{min}}\). 5. The value \(0.5\,\frac{\text{m}}{\text{min}}\) is not used.

\(t_1=0 \rightarrow -1.0\,\frac{\text{m}}{\text{min}}\); \(t_2=2 \rightarrow -1.8\,\frac{\text{m}}{\text{min}}\); \(t_3=4 \rightarrow -0.2\,\frac{\text{m}}{\text{min}}\); \(t_4=5 \rightarrow 1.5\,\frac{\text{m}}{\text{min}}\); \(0.5\,\frac{\text{m}}{\text{min}}\) is not used.

- Use \(P\) as one endpoint each time. - Divide the change in output by the change in input. - Look for a trend as the other endpoint moves closer to \(P\).

1. The secant slope is \(\frac{f(a)-1.1}{a-1}\). 2. At \(a=2\), it is \(\frac{1.8-1.1}{2-1}=0.7\). 3. At \(a=1.5\), it is \(\frac{1.3375-1.1}{0.5}=0.475\). 4. At \(a=1.1\), it is \(\frac{1.1331-1.1}{0.1}=0.331\). 5. The slopes move toward about \(0.3\), consistent with the limiting direction shown by the enlarged secant.

a) \(0.7\), \(0.475\), \(0.331\) b) Approximately \(0.3\). The secant slopes move toward this value as the second point approaches \(P\).

- A slope measured in feet per mile cannot be a percentage until both distances use the same unit. - Find the change in height and divide by the horizontal distance along the tangent. - A percent grade compares the vertical and horizontal changes in the same unit. - Think about the tangent direction at the start and the highest point.

1. The local rate describes how fast elevation changes per mile of horizontal travel. 2. The tangent passes through \((0,40)\) and \((2,520)\), so its slope is \(\frac{520-40}{2}=240\,\frac{\text{ft}}{\text{mi}}\). 3. The grade is \(\frac{240}{5280}\cdot100\%\approx4.5\%\). 4. The graph is horizontal at the start and at its highest point, \(s=4\). Both rates are \(0\,\frac{\text{ft}}{\text{mi}}\).

a) The elevation gain per mile at \(P\). b) \(240\,\frac{\text{ft}}{\text{mi}}\), or approximately \(4.5\%\). c) \(0\,\frac{\text{ft}}{\text{mi}}\) at \(s=0\) (from the right) and \(s=4\), where the curve is locally horizontal.

- Average rate compares the change in volume with the elapsed time. - Find two readable points on the tangent line. - Compare the directions and steepness of the curve at the two times.

1. The average rate on \([0,20]\) is \(\frac{40-0}{20}=2\,\frac{\text{gal}}{\text{min}}\). 2. The tangent at \(t=10\) passes through \((0,5)\) and \((20,45)\), giving \(\frac{45-5}{20}=2\,\frac{\text{gal}}{\text{min}}\). 3. The graph rises more steeply at \(t=5\) than at \(t=25\). Water enters faster at \(t=5\).

a) \(2\,\frac{\text{gal}}{\text{min}}\) b) Approximately \(2\,\frac{\text{gal}}{\text{min}}\) c) At \(t=5\), because the graph is steeper there.

- Use each drawn tangent-line slope to estimate instantaneous temperature change. - The maximum is the highest point on the graph. - The fastest cooling occurs where the graph falls most steeply. - At the top of the graph, the tangent is horizontal.

1. The graph reaches its highest point at \(t = 2\), where \(T = 45\,{}^\circ\text{C}\). 2. At \(t = 4\), the tangent line passes through \((4, 40)\) and \((9, 25)\), so its slope is \(\frac{25-40}{9-4} = -3\,\frac{{}^\circ\text{C}}{\text{min}}\). 3. At \(t = 6\), the tangent line passes through \((6, 35)\) and \((11, 25)\), so its slope is \(\frac{25-35}{11-6} = -2\,\frac{{}^\circ\text{C}}{\text{min}}\). 4. The substance cools most rapidly where the graph has its steepest downward tangent, at about \(t \approx 3.5\) minutes. 5. At \(t = 2\), the tangent is horizontal, so the instantaneous rate is \(0\). The temperature is at a local maximum and is momentarily neither increasing nor decreasing.

a) \(45\,{}^\circ\text{C}\) at \(t = 2\,\text{min}\) b) At \(t = 4\): \(\approx -3\,\frac{{}^\circ\text{C}}{\text{min}}\); at \(t = 6\): \(\approx -2\,\frac{{}^\circ\text{C}}{\text{min}}\) c) At \(t \approx 3.5\,\text{min}\) d) The rate is \(0\), marking the maximum temperature.

- Instantaneous rate is represented by the slope of the tangent line at that time. - For part a), choose two clear grid points on the \(t=5\) tangent. - For part b), compare the steepness of all three dashed tangents rather than focusing on only one line.

1. The tangent at \(t=5\) passes through \((3,2)\) and \((7,8)\). Its slope is \(\frac{8-2}{7-3}=1.5\,\frac{\text{m}}{\text{h}}\). 2. Comparing the three shown tangents, the tangent at \(t=5\) is steeper than the tangents at \(t=3\) and \(t=7\). Therefore, among the three shown times, the instantaneous rate is greatest at \(t=5\). 3. A rate of \(0\,\frac{\text{m}}{\text{h}}\) means the water depth is momentarily constant.

a) \(1.5\,\frac{\text{m}}{\text{h}}\) b) \(t=5\) hours; its tangent has the greatest slope of the three shown. c) The water depth is momentarily neither rising nor falling.

- Compare endpoint elevations for each interval. - Convert the horizontal distance to feet before calculating a percent. - When comparing a short section with a long one, give each its own horizontal distance in the rate calculation.

1. Average grade is \(100\%\) times elevation change divided by horizontal distance in feet. 2. From mile \(0\) to mile \(2\), it is \(\frac{900-500}{2\cdot5280}\cdot100\%\approx3.79\%\). 3. From mile \(2\) to mile \(3.5\), it is \(\frac{700-900}{1.5\cdot5280}\cdot100\%\approx-2.53\%\). 4. From mile \(3.5\) to mile \(5.5\), it is \(\frac{1550-700}{2\cdot5280}\cdot100\%\approx8.05\%\). 5. From mile \(4.5\) to mile \(5\), the elevation rises by \(300\,\text{ft}\) in \(0.5\,\text{mi}\), so the short stretch has average grade \(\frac{300}{0.5\cdot5280}\cdot100\%\approx11.36\%\), greater than the last section's overall rate.

a) Approximately \(3.79\%\), \(-2.53\%\), and \(8.05\%\), in the requested order. b) For example, mile \(4.5\) to mile \(5\) has grade about \(11.36\%\), steeper than the whole last section.

- Read the graph's temperature values at the interval endpoints. - Divide the change in temperature by the time elapsed. - For the fastest change, compare changes over intervals of the same length.

1. From 4:00 a.m. to noon, temperature rises from \(46\,{}^\circ\text{F}\) to \(74\,{}^\circ\text{F}\), giving \(\frac{74-46}{12-4}=3.5\,\frac{{}^\circ\text{F}}{\text{h}}\). 2. From noon to 8:00 p.m., it falls from \(74\,{}^\circ\text{F}\) to \(58\,{}^\circ\text{F}\), giving \(\frac{58-74}{20-12}=-2\,\frac{{}^\circ\text{F}}{\text{h}}\). 3. From 8:00 a.m. to 10:00 a.m., temperature rises by \(10\,{}^\circ\text{F}\) in two hours, or \(5\,\frac{{}^\circ\text{F}}{\text{h}}\). Every other marked two-hour interval changes by at most \(8\,{}^\circ\text{F}\) in magnitude.

a) \(3.5\,\frac{{}^\circ\text{F}}{\text{h}}\) and \(-2\,\frac{{}^\circ\text{F}}{\text{h}}\) b) 8:00–10:00 a.m.; the temperature rises by \(10\,{}^\circ\text{F}\), more than during any other marked two-hour interval.

- Instantaneous speed is the tangent slope on a distance-time graph. - Use two readable points on the drawn tangent line. - Convert feet per second using \(1\,\text{mi}=5280\,\text{ft}\) and \(1\,\text{h}=3600\,\text{s}\). - A graph that becomes steeper represents increasing speed.

1. The drawn tangent at \(t = 5\) passes through \((6, 100)\) and \((10, 250)\). 2. Its slope is \(\frac{250-100}{10-6} = 37.5\,\frac{\text{ft}}{\text{s}}\). 3. Convert to miles per hour: \(37.5\cdot\frac{3600}{5280}\approx 25.6\,\frac{\text{mi}}{\text{h}}\). 4. The graph becomes steeper as time increases, so the instantaneous speed increases. Its concave-up shape indicates positive acceleration.

1. \(\approx 37.5\,\frac{\text{ft}}{\text{s}}\) 2. \(\approx 25.6\,\frac{\text{mi}}{\text{h}}\) 3. The speed increases because the graph becomes progressively steeper.

- Recall the difference-quotient formula. - What happens to an interval’s midpoint when both endpoints are moved outward by the same amount \(k\)? - Factor the difference of the quadratic expressions so that \(b - a\) cancels. - How are two lines related when they have the same slope?

1. Over \(I_1\), \(\frac{f(5) - f(1)}{5 - 1} = \frac{5 - (-3)}{4} = 2\). 2. Over \(I_2\), \(\frac{f(6) - f(0)}{6 - 0} = \frac{12 - 0}{6} = 2\). 3. Over \([a, b]\), \(\frac{f(b) - f(a)}{b - a} = \frac{b^2 - 4b - (a^2 - 4a)}{b - a} = \frac{(b - a)(a + b - 4)}{b - a} = a + b - 4\). 4. Over \([a - k, b + k]\), the sum of the endpoints is \((a - k) + (b + k) = a + b\). Applying the same simplification, the average rate of change is \((a - k) + (b + k) - 4 = a + b - 4\). 5. The corresponding secant lines have the same slope, so they are parallel. The two intervals have the same midpoint, and for this quadratic the secant slope depends only on the sum, or equivalently the midpoint, of the endpoints.

a) \(I_1\): \(2\) \(I_2\): \(2\) b) Both average rates simplify to \(a + b - 4\). c) The corresponding secant lines are parallel. For this quadratic, intervals with the same midpoint produce the same secant slope.

- What factor remains after a quantity decreases by \(15\%\)? - Use the units of the numerator and denominator to interpret each expression. - Average rate of change is the slope of a secant line through two graph points. - Make sure the hourly factor \(a\) matches the fact that the stated percent decrease occurs every \(4\) hours.

1. The initial value is \(c = 200\). A \(15\%\) decrease every \(4\) hours gives \(f(4) = 200 \cdot 0.85\). From \(200a^4 = 200 \cdot 0.85\), \(a = \sqrt[4]{0.85} \approx 0.9603\). Thus, \(f(t) = 200(\sqrt[4]{0.85})^t\), or approximately \(f(t) = 200 \cdot 0.9603^t\). 2. The expression \(f(t) - f(t_0)\) is the net change in the amount, in milligrams, from time \(t_0\) to time \(t\). The quotient \(\frac{f(t) - f(t_0)}{t - t_0}\) is the average rate of change over that interval, in milligrams per hour. 3. Over \([0, 4]\), \(f(0) = 200\) and \(f(4) = 170\), so the average rate is \(\frac{170 - 200}{4} = -7.5\,\frac{\text{mg}}{\text{h}}\). 4. Over \([12, 16]\), \(f(12) = 200 \cdot 0.85^3 = 122.825\) and \(f(16) = 200 \cdot 0.85^4 = 104.40125\). The average rate is \(\frac{104.40125 - 122.825}{4} \approx -4.6059\,\frac{\text{mg}}{\text{h}}\).

a) \(c = 200\), \(a = \sqrt[4]{0.85} \approx 0.9603\), and \(f(t) = 200(\sqrt[4]{0.85})^t\) b) \(f(t) - f(t_0)\) is the net change in milligrams; \(\frac{f(t) - f(t_0)}{t - t_0}\) is the average change in milligrams per hour. c) \([0, 4]\): \(-7.5\,\frac{\text{mg}}{\text{h}}\) \([12, 16]\): approximately \(-4.6059\,\frac{\text{mg}}{\text{h}}\)