- Identify the exponent on \(x\). - Recall how the exponent changes when the power rule is applied. - The derivative of a nonconstant monomial has degree one less than the original monomial.

1. Apply the power rule to \(x^6\). 2. The derivative is \(f'(x)=6x^5\).

\(f'(x)=6x^5\)

- Separate the constant coefficient from the power of \(x\). - Check what happens to both the coefficient and the exponent under the power rule. - Compare Adaeze’s coefficient with the original exponent.

1. The coefficient \(5\) remains a factor, but the exponent \(3\) must also multiply the coefficient when the power rule is applied. 2. Therefore, \(g'(x)=15x^2\).

Adaeze forgot to multiply by the original exponent \(3\). The correct derivative is \(g'(x)=15x^2\).

- Compare each original exponent with the coefficient in the derivative. - Compare each original exponent with the exponent that remains after differentiation. - Apply the same two changes to \(x^7\).

1. In each row, the original exponent becomes the coefficient and the exponent decreases by \(1\). 2. Therefore, the derivative of \(x^7\) is \(7x^6\). 3. In general, the pattern is \(\frac{d}{dx}(x^n)=nx^{n-1}\).

The derivative of \(x^7\) is \(7x^6\). In general, \(\frac{d}{dx}(x^n)=nx^{n-1}\).

- Identify the numerical coefficient and the power of \(x\). - How does the power rule change these two parts of a monomial?

1. Multiply the coefficient by the exponent: \(4\cdot3=12\). 2. Reduce the exponent by one: \(f'(x)=12x^2\).

\(f'(x)=12x^2\)

- Work out how the \(x^2\) term changes. - Does adding the same number to every output change any slope?

1. The power rule gives the derivative of \(x^2\) as \(2x\). 2. The constant \(7\) has rate of change zero, so \(g'(x)=2x\).

\(g'(x)=2x\). The constant contributes \(0\).

- The units of a derivative are the output units divided by the input units. - What physical quantity is distance traveled per unit of time? - Use the power rule to differentiate the function. - Include units after substituting \(t = 4\).

1. Since \(s\) is measured in meters and \(t\) in seconds, the derivative \(s'(t)\) has units \(\frac{\text{m}}{\text{s}}\). 2. The derivative of distance traveled with respect to time is the vehicle’s instantaneous speed at time \(t\). 3. By the power rule, \(s'(t) = t+2\). Therefore, \(s'(4) = 4+2 = 6\,\frac{\text{m}}{\text{s}}\).

1. \(\frac{\text{m}}{\text{s}}\) 2. \(s'(t)\) is the vehicle’s instantaneous speed at time \(t\). 3. \(s'(4) = 6\,\frac{\text{m}}{\text{s}}\)

- The derivative at a point gives the tangent-line slope there. - Differentiate each term of the function. - Substitute \(x = 3\) into the derivative. - Interpret the sign of the result.

1. Differentiate using the power and sum rules: \(f'(x) = x-4\). 2. Evaluate at \(x = 3\): \(f'(3) = 3-4 = -1\). 3. Because \(f'(3) < 0\), the tangent line has negative slope.

The tangent line has negative slope because \(f'(3) = -1\).

- Instantaneous rate of change means the derivative at the point. - First find the derivative function. - Substitute \(x = 4\). - Use the sign of the result, not only its magnitude.

1. Differentiate using the power rule: \(g'(x) = 6x^2 - 18x\). 2. Evaluate at \(x = 4\): \(g'(4) = 6(4)^2-18(4) = 96-72 = 24\). 3. Since \(24 > 0\), the instantaneous rate of change is positive.

The instantaneous rate of change is positive because \(g'(4) = 24\).

- Instantaneous rate of change is the derivative at the given point. - Find a function that gives the slope at any x-value. - Substitute \(x = 2\) after differentiating.

1. Differentiate using the power and sum rules: \(f'(x) = 6x^2-18x+12\). 2. Evaluate at \(x = 2\): \(f'(2) = 6(2)^2-18(2)+12 = 24-36+12 = 0\).

\(f'(2) = 0\)

- A secant slope gives the average rate between two times. - A derivative gives the tangent slope at one time. - Interpret each rate as volume added per second.

1. Evaluate the volumes: \(V(2) = 6\,\text{L}\) and \(V(5) = 22.5\,\text{L}\). 2. The average inflow rate is \(\frac{V(5)-V(2)}{5-2} = \frac{22.5-6}{3} = 5.5\,\frac{\text{L}}{\text{s}}\). 3. Differentiate: \(V'(t) = t+2\). Thus, \(V'(4) = 6\,\frac{\text{L}}{\text{s}}\). 4. The average rate describes the mean volume increase per second throughout \([2, 5]\). The instantaneous rate describes the inflow at the single time \(t = 4\).

a) \(5.5\,\frac{\text{L}}{\text{s}}\) b) \(6\,\frac{\text{L}}{\text{s}}\) c) The first is an average over an interval; the second is the rate at one exact time.

- The slope of a position-time graph represents velocity. - Differentiate the position function. - The derivative’s units are meters per second.

1. Instantaneous velocity is the derivative of the position function. 2. Differentiate: \(s'(t) = 0.8t+3\). 3. Evaluate at \(t = 5\): \(s'(5) = 0.8(5)+3 = 7\,\frac{\text{m}}{\text{s}}\). 4. In context, \(s'(t)\) gives the car’s instantaneous velocity at time \(t\).

a) \(7\,\frac{\text{m}}{\text{s}}\) b) \(s'(t)\) is the car’s instantaneous velocity at time \(t\).

- Average growth uses the change over a time interval. - Instantaneous growth uses the derivative at one time. - Keep track of the population unit, which is thousands of bacteria.

1. Evaluate the population values: \(B(2) = 122\) and \(B(4) = 168\), in thousands of bacteria. 2. The average growth rate is \(\frac{B(4)-B(2)}{4-2} = \frac{168-122}{2} = 23\) thousand bacteria per hour. 3. Differentiate: \(B'(t) = 6t+5\). Then \(B'(3) = 6(3)+5 = 23\) thousand bacteria per hour.

a) \(23\) thousand bacteria per hour b) \(23\) thousand bacteria per hour

- Use the slope formula for the secant line. - The derivative gives the tangent-line slope. - Compare the two numerical slopes.

1. Evaluate the function: \(f(-2) = \frac{1}{2}(-2)^2-2 = 0\), and \(f(2) = \frac{1}{2}(2)^2+2 = 4\). 2. The secant slope is \(\frac{f(2)-f(-2)}{2-(-2)} = \frac{4-0}{4} = 1\). 3. Differentiate: \(f'(x) = x+1\). The tangent slope at \(x = 0\) is \(f'(0) = 1\). 4. Both slopes are \(1\), so the secant and tangent lines are parallel.

1. The secant slope is \(1\). 2. The tangent slope is \(f'(0) = 1\), so the slopes are equal.

- Instantaneous rate of change is the derivative at a point. - Apply the power rule. - Substitute each x-value into the derivative.

1. Differentiate using the power rule: \(f'(x) = -2x\). 2. Evaluate at \(x = 2\): \(f'(2) = -2(2) = -4\). 3. Evaluate at \(x = -1\): \(f'(-1) = -2(-1) = 2\).

At \(x = 2\): \(-4\) At \(x = -1\): \(2\)

- The derivative gives tangent-line slope. - Expanding the square first may make differentiation easier. - Evaluate the derivative at each requested point.

1. Expand the function: \(g(x) = x^2+2x+1\). 2. Differentiate: \(g'(x) = 2x+2\). 3. At \(x = 1\), \(g'(1) = 2(1)+2 = 4\). 4. At \(x = -3\), \(g'(-3) = 2(-3)+2 = -4\).

At \(x = 1\): \(4\) At \(x = -3\): \(-4\)

- The derivative gives the slope of the graph at each x-value. - Set the derivative equal to \(5\). - Solve the resulting linear equation.

1. Differentiate using the power and sum rules: \(f'(x) = 4x-7\). 2. Set the derivative equal to the required slope: \(4x-7 = 5\). 3. Solve: \(4x = 12\), so \(x = 3\).

\(x = 3\)

- Apply the power rule to the quadratic term. - A marginal rate is an instantaneous rate of change. - Convert the decimal rate to a percentage for interpretation.

1. Differentiate using the power and sum rules: \(S'(x) = 0.000004x+0.15\). 2. Evaluate at \(x = 30{,}000\): \(S'(30{,}000) = 0.000004(30{,}000)+0.15 = 0.27\). 3. The value \(0.27\) is equivalent to \(27\%\). Near an income of \(\$30{,}000\), the model predicts that each additional dollar of taxable income increases the tax owed by approximately \(\$0.27\).

1. \(S'(x) = 0.000004x+0.15\) 2. \(S'(30{,}000) = 0.27 = 27\%\) 3. Near \(\$30{,}000\), one additional dollar of taxable income increases the tax owed by approximately \(\$0.27\).

- Which terms depend on \(x\), and what powers do they contain? - Consider the contribution of each term before combining the results.

1. Differentiate \(x^4\) to get \(4x^3\). 2. Differentiate \(3x^2\) to get \(6x\). 3. Add the results: \(h'(x)=4x^3+6x\).

\(h'(x)=4x^3+6x\)

- First differentiate the squared input. - Compare the derivative's sign at each input with the graph's direction there.

1. The power rule gives \(f'(x)=2x\). 2. At \(x=-1\), \(f'(-1)=-2<0\); the graph falls as x increases there. 3. At \(x=1\), \(f'(1)=2>0\); the graph rises as x increases there.

\(f'(x)=2x\); \(f'(-1)=-2\) (negative, falling in the graph) and \(f'(1)=2\) (positive, rising in the graph).

- Calculate the power-rule result first for a value to check. - In the difference quotient, simplify before letting h approach zero.

1. The power rule gives \(f'(x)=2x\), so \(f'(3)=6\). 2. Expand the difference quotient: \(\frac{(3+h)^2-9}{h}=\frac{6h+h^2}{h}\). 3. For nonzero h this equals \(6+h\), whose limit is \(6\). Both methods agree.

For \(h\ne0\), the difference quotient simplifies to \(6+h\), whose limit is \(6\). The power rule also gives \(f'(3)=6\).

- What physical quantity is the rate of change of distance traveled with respect to time? - Differentiate with respect to \(t\), treating the angle factor as a constant. - Substitute \(t = 2\) after differentiating. - Make sure your calculator is in degree mode.

1. The instantaneous rate of change \(s'(t)\) is the sled’s instantaneous speed, measured in meters per second. 2. For a fixed angle \(\theta\), the factor \(4.9\sin(\theta)\) is constant. By the power rule, \(s'(t) = 9.8\sin(\theta)t\). 3. For \(\theta = 30^\circ\) and \(t = 2\), \(s'(2) = 9.8\sin(30^\circ)(2) = 9.8\,\frac{\text{m}}{\text{s}}\). 4. For \(\theta = 50^\circ\) and \(t = 2\), \(s'(2) = 9.8\sin(50^\circ)(2) \approx 15.01\,\frac{\text{m}}{\text{s}}\).

The instantaneous rate of change is the sled’s instantaneous speed. For \(\theta = 30^\circ\): \(9.8\,\frac{\text{m}}{\text{s}}\) For \(\theta = 50^\circ\): \(\approx 15.01\,\frac{\text{m}}{\text{s}}\)

- What does the derivative of an area function represent? - Differentiate with respect to time, treating \(v\) as a constant. - Apply the power rule to \(t^2\). - The output unit is area divided by time.

1. The derivative \(A'(t)\) gives the instantaneous rate at which the oil slick’s area is growing. Its units are square meters per hour. 2. Treating \(v\) as a constant and applying the power rule gives \(A'(t) = \sqrt{v}\,t\). 3. For \(v = 10\) and \(t = 3\), \(A'(3) = 3\sqrt{10} \approx 9.49\,\frac{\text{m}^2}{\text{h}}\). 4. For \(v = 25\) and \(t = 3\), \(A'(3) = 3\sqrt{25} = 15\,\frac{\text{m}^2}{\text{h}}\).

The instantaneous rate is the rate at which the slick’s area is growing, in square meters per hour. For \(v = 10\,\text{mph}\): \(\approx 9.49\,\frac{\text{m}^2}{\text{h}}\) For \(v = 25\,\text{mph}\): \(15\,\frac{\text{m}^2}{\text{h}}\)

- Distinguish between average velocity over an interval and velocity at one instant. - Use a difference quotient for part a. - Differentiate the height function to obtain the velocity function. - Positive and negative vertical velocity indicate opposite directions.

1. Since \(s(0) = 4\) and \(s(2) = 4\), the average velocity is \(\frac{s(2)-s(0)}{2-0} = 0\,\frac{\text{m}}{\text{s}}\). 2. Differentiate to obtain velocity: \(v(t) = s'(t) = -10t+10\). Then \(v(1.5) = -10(1.5)+10 = -5\,\frac{\text{m}}{\text{s}}\). The negative sign means the ball is moving downward. 3. Set the velocity equal to \(2\): \(-10t+10 = 2\). Solving gives \(t = 0.8\,\text{s}\). Because the velocity is positive, the ball is rising at that time.

a) \(0\,\frac{\text{m}}{\text{s}}\) b) \(-5\,\frac{\text{m}}{\text{s}}\); the ball is falling. c) \(t = 0.8\,\text{s}\); the ball is rising.

- Position comes from substituting the time into \(s(t)\). - Velocity is the rate of change of position. - After finding the velocity rule, set it equal to the requested speed.

1. The position is \(s(2.5)=0.4(2.5)^3=6.25\,\text{m}\). 2. By the power rule, \(v(t)=s'(t)=1.2t^2\). Therefore, \(v(2.5)=1.2(2.5)^2=7.5\,\frac{\text{m}}{\text{s}}\). 3. Solve \(1.2t^2=30\), giving \(t^2=25\). In the modeled interval, \(t=5\,\text{s}\).

a) Position \(6.25\,\text{m}\); instantaneous velocity \(7.5\,\frac{\text{m}}{\text{s}}\) b) \(t=5\,\text{s}\)

- A function value gives the covered area at a specific time. - The derivative gives the instantaneous growth rate. - Apply the power rule to the cubic term. - To find a time from a given rate, set the derivative equal to that rate and solve.

1. Evaluate the area: \(A(4) = 0.5(4)^3+10 = 42\,\text{mm}^2\). 2. Differentiate to find the growth-rate function: \(A'(t) = 1.5t^2\). 3. At \(t = 4\), \(A'(4) = 1.5(4)^2 = 24\,\frac{\text{mm}^2}{\text{h}}\). 4. Set the growth rate equal to \(54\): \(1.5t^2 = 54\), so \(t^2 = 36\). Because \(t \ge 0\), \(t = 6\,\text{h}\).

a) Area: \(42\,\text{mm}^2\); instantaneous growth rate: \(24\,\frac{\text{mm}^2}{\text{h}}\) b) \(t = 6\,\text{h}\)

- Simplify the function rule by expanding the square first. - Identify the differentiation variable and treat the parameter as a constant. - Apply the sum rule and power rule. - After finding the derivative, substitute the specified input.

1. Expand the square: \(f(x)=x^2+4kx+4k^2-k^2x\). 2. Combine the linear terms: \(f(x)=x^2+(4k-k^2)x+4k^2\). 3. Differentiate with respect to \(x\), treating \(k\) as a constant: \(f'(x)=2x+4k-k^2\). 4. Substitute \(x=0\): \(f'(0)=4k-k^2\).

\(f'(x)=2x+4k-k^2\) \(f'(0)=4k-k^2\)

- Expand the product before differentiating. - Combine only terms that have the same power of \(t\). - Remember that \(a\) is constant when differentiating with respect to \(t\). - Substitute \(t=0\) after finding the derivative.

1. Distribute \(t\): \(g(t)=at^3-2t^2+a^2t-\frac{1}{2}at^3\). 2. Combine the cubic terms: \(g(t)=\frac{1}{2}at^3-2t^2+a^2t\). 3. Differentiate with respect to \(t\), treating \(a\) as a constant: \(g'(t)=\frac{3}{2}at^2-4t+a^2\). 4. Substitute \(t=0\): \(g'(0)=a^2\).

\(g'(t)=\frac{3}{2}at^2-4t+a^2\) \(g'(0)=a^2\)

- Use a secant slope for the average rate over the first five days. - Use the derivative for the instantaneous rate. - Apply the power rule term by term. - For part c, set the derivative equal to the target rate and solve the quadratic equation.

1. Since \(h(0) = 2\) and \(h(5) = -0.008(5)^3+0.24(5)^2+2 = 7\), the average growth rate on \([0, 5]\) is \(\frac{7-2}{5} = 1\,\frac{\text{in.}}{\text{day}}\). 2. Differentiate: \(h'(t) = -0.024t^2+0.48t\). Then \(h'(10) = -0.024(10)^2+0.48(10) = 2.4\,\frac{\text{in.}}{\text{day}}\). 3. Set the derivative equal to \(1.8\): \(-0.024t^2+0.48t = 1.8\). This is equivalent to \(t^2-20t+75 = 0\), which factors as \((t-5)(t-15)=0\). 4. Therefore, the growth rate is \(1.8\,\frac{\text{in.}}{\text{day}}\) on days \(5\) and \(15\).

a) \(1\,\frac{\text{in.}}{\text{day}}\) b) \(2.4\,\frac{\text{in.}}{\text{day}}\) c) \(t = 5\) days and \(t = 15\) days

- Average velocity uses a difference quotient over an interval. - Instantaneous velocity is the derivative of height with respect to time. - For part c, set the derivative equal to the target velocity and solve the quadratic equation. - Check that each solution lies in the stated time interval.

1. Evaluate the endpoint heights: \(H(2) = \frac{124}{3}\) and \(H(5) = \frac{475}{3}\). 2. The average upward velocity is \(\frac{H(5)-H(2)}{5-2} = \frac{117}{3} = 39\,\frac{\text{m}}{\text{s}}\). 3. Differentiate: \(H'(t) = -t^2+12t+10\). Then \(H'(4) = -16+48+10 = 42\,\frac{\text{m}}{\text{s}}\). 4. Set the derivative equal to \(43\): \(-t^2+12t+10 = 43\), or \(t^2-12t+33 = 0\). 5. The solutions are \(t = 6\pm\sqrt{3}\), approximately \(4.27\,\text{s}\) and \(7.73\,\text{s}\). Both lie in the model’s domain.

a) \(39\,\frac{\text{m}}{\text{s}}\) b) \(42\,\frac{\text{m}}{\text{s}}\) c) \(t = 6-\sqrt{3} \approx 4.27\,\text{s}\) and \(t = 6+\sqrt{3} \approx 7.73\,\text{s}\)

- Treat the parameter as a constant when differentiating with respect to \(x\). - Write the derivative in terms of \(k\). - Use the given rate at \(x = 3\) to form an equation.

1. Differentiate with respect to \(x\), treating \(k\) as a constant: \(h_k'(x) = 3x^2-2kx\). 2. Apply the condition \(h_k'(3) = 15\): \(3(3)^2-2k(3) = 15\). 3. Solve \(27-6k = 15\), which gives \(k = 2\).

\(k = 2\)

- Average change over an interval is found with a secant slope. - Instantaneous change at one time is found with the derivative. - Differentiate the quadratic function. - Set the derivative equal to the target rate and solve for \(t\).

1. Evaluate the endpoint temperatures: \(T(0) = 80\,{}^\circ\text{F}\) and \(T(2) = 4-24+80 = 60\,{}^\circ\text{F}\). 2. The average rate of change on \([0, 2]\) is \(\frac{60-80}{2-0} = -10\,\frac{{}^\circ\text{F}}{\text{h}}\). 3. Differentiate: \(T'(t) = 2t-12\). 4. Set the derivative equal to \(-9\): \(2t-12 = -9\), so \(2t = 3\) and \(t = 1.5\,\text{h}\).

a) \(-10\,\frac{{}^\circ\text{F}}{\text{h}}\) b) \(t = 1.5\,\text{h}\)

- Average velocity is the slope through two position-time points. - Instantaneous velocity is the derivative of position. - Use the power rule to find the velocity function. - Set the velocity function equal to the given value and solve.

1. Evaluate the endpoint positions: \(s(1) = 6\) and \(s(4) = 36\). 2. The average velocity on \([1, 4]\) is \(\frac{s(4)-s(1)}{4-1} = \frac{36-6}{3} = 10\,\frac{\text{m}}{\text{s}}\). 3. Differentiate to obtain instantaneous velocity: \(s'(t) = 2t+5\). 4. Solve \(2t+5 = 12\), giving \(t = 3.5\,\text{s}\).

a) \(10\,\frac{\text{m}}{\text{s}}\) b) \(t = 3.5\,\text{s}\)

- The derivative gives the slope at a specific time. - What do positive and negative slopes mean for temperature? - If a quantity is momentarily neither increasing nor decreasing, what must its derivative equal?

1. Differentiate: \(T'(t) = -2t+12\). 2. Evaluate the derivative: \(T'(2) = -2(2)+12 = 8\,\frac{{}^\circ\text{C}}{\text{min}}\), and \(T'(8) = -2(8)+12 = -4\,\frac{{}^\circ\text{C}}{\text{min}}\). 3. The positive rate at \(t = 2\) means the solution is warming. The negative rate at \(t = 8\) means it is cooling. 4. Set the rate equal to zero: \(-2t+12 = 0\). Solving gives \(t = 6\,\text{min}\).

a) \(T'(2) = 8\,\frac{{}^\circ\text{C}}{\text{min}}\); \(T'(8) = -4\,\frac{{}^\circ\text{C}}{\text{min}}\) b) Positive means the temperature is rising; negative means it is falling. c) \(t = 6\,\text{min}\)

- Velocity is the derivative of position. - The sign of velocity indicates direction. - An object is momentarily at rest when its velocity is zero.

1. Differentiate the position function: \(v(t) = s'(t) = t^2-8t+12\). 2. Evaluate at \(t = 3\): \(v(3) = 9-24+12 = -3\,\frac{\text{m}}{\text{s}}\). The negative sign means the vehicle is moving in the negative coordinate direction. 3. The vehicle is at rest when \(v(t) = 0\). Solve \(t^2-8t+12 = 0\), which factors as \((t-2)(t-6)=0\). 4. Therefore, the vehicle is momentarily at rest at \(t = 2\,\text{s}\) and \(t = 6\,\text{s}\).

a) \(-3\,\frac{\text{m}}{\text{s}}\); the vehicle is moving in the negative coordinate direction. b) \(t = 2\,\text{s}\) and \(t = 6\,\text{s}\)

- Write the secant slope in terms of the unknown parameter \(a\). - Instantaneous rate of change means derivative. - After finding \(a\), set the derivative equal to the required rate and solve.

1. The secant slope is \(\frac{f_a(3)-f_a(1)}{3-1} = \frac{9a-a}{2} = 4a\). 2. Set the secant slope equal to \(4\): \(4a = 4\), so \(a = 1\). 3. For \(f_1(x) = x^2\), the derivative is \(f_1'(x) = 2x\). 4. Set the instantaneous rate equal to \(4\): \(2x = 4\), so \(x = 2\).

1. \(a = 1\) 2. \(x = 2\)

- Work backward from what the derivative of a monomial looks like under the power rule. - Compare the exponent in the given derivative with the exponent that would remain after differentiation. - After determining the exponent, compare coefficients.

1. By the power rule, \(\frac{d}{dx}(ax^n)=anx^{n-1}\). 2. Match exponents: \(n-1=4\), so \(n=5\). 3. Match coefficients: \(an=15\), so \(5a=15\) and \(a=3\). 4. Therefore, \(f(x)=3x^5\).

\(a=3\), \(n=5\), and \(f(x)=3x^5\).

- Start by finding the function's derivative using the power rule. - Check whether each candidate input agrees with a change in the graph's direction.

1. The derivative is \(f'(x)=3x^2-3\). 2. A horizontal tangent has slope zero, so \(3x^2-3=0\), giving \(x=-1\) or \(x=1\). 3. These are the two visible turning points.

\(f'(x)=3x^2-3\); its zeros are \(x=-1\) and \(x=1\), matching the two turning points in the graph.

- Read the input of the marked point and consider what its tangent tells you. - Find the derivative while keeping the unknown coefficient in the expression. - Check how the graph moves near \(x=1\) after you have found the coefficient.

1. The marked turning point has \(x=2\), so its tangent slope is \(g'(2)=0\). 2. The power rule gives \(g'(x)=3ax^2+4\). Thus \(12a+4=0\), giving \(a=-\frac{1}{3}\). 3. Now \(g'(1)=3\left(-\frac{1}{3}\right)+4=3>0\), matching the graph's upward direction near \(x=1\).

\(g'(x)=3ax^2+4\), so \(g'(2)=12a+4=0\) and \(a=-\frac{1}{3}\). The derivative at \(x=1\) is \(g'(1)=3\), so the curve is rising there.

- Use change in output over change in input for both pairs of endpoints. - What do the two pairs of secant endpoints have in common?

1. The first secant slope is \(\frac{9-1}{3-1}=4\). 2. The second slope is \(\frac{16-0}{4-0}=4\). 3. The power rule gives \(f'(2)=2\cdot2=4\). Both intervals are centered on \(x=2\), so their secant slopes agree with the midpoint tangent slope for this quadratic.

The drawn intervals are \([1,3]\) and \([0,4]\). Both secant slopes are \(4\), equal to \(f'(2)=4\). Both intervals have midpoint \(2\).

- Check the signs on each side of the input where the derivative vanishes. - A curve can move downward even while its output values are positive.

1. Expanding gives \(h(x)=x^2-4x+3\), so \(h'(x)=2x-4\). 2. For \(x<2\), the derivative is negative and the graph decreases. 3. For \(x>2\), the derivative is positive and the graph increases.

\(h'(x)=2x-4\), negative for \(x<2\) and positive for \(x>2\). The graph decreases for \(x<2\) and increases for \(x>2\).

- Differentiate the tax function to find the marginal tax rate. - Divide total tax by gross income to find the average tax rate. - Compare the fraction of an additional dollar that remains after tax with the current average fraction of income that remains after tax. - Use \(N(x) = x-T(x)\) to relate the two rates.

1. Differentiate: \(T'(x) = 0.15+0.000008x\). At \(x = 35{,}000\), \(T'(35{,}000) = 0.15+0.000008(35{,}000) = 0.43\), or \(43\%\). 2. Simplify the average tax rate: \(r(x) = \frac{0.15x+0.000004x^2}{x} = 0.15+0.000004x\). Thus, \(r(35{,}000) = 0.29\), or \(29\%\). The marginal rate is higher than the average rate. 3. The marginal net-income rate is \(N'(x) = 1-T'(x)\), so \(N'(35{,}000) = 0.57\). The current average net-income share is \(\frac{N(35{,}000)}{35{,}000} = 1-r(35{,}000) = 0.71\). 4. Because \(0.57<0.71\), each additional dollar contributes a smaller fraction to net income than the current average. Therefore, a small percentage increase in gross income produces a smaller percentage increase in net income.

1. \(T'(x) = 0.15+0.000008x\); \(T'(35{,}000) = 0.43 = 43\%\) 2. \(r(35{,}000) = 0.29 = 29\%\); the marginal tax rate is higher. 3. At \(\$35{,}000\), the marginal net-income rate is \(0.57\), while the average net-income share is \(0.71\). Therefore, net income grows proportionally more slowly than gross income.