Estimate a derivative at a point

Learning goals

Estimates tangent slope from a graph.
Uses nearby secant slopes or a numerical table to estimate the derivative.
Compares derivative estimates obtained from different input spacings.
Interprets the sign and magnitude of the estimated derivative.

Difficulty: 1 – 5

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Estimate a derivative at a point 1

At \(x=4\), the tangent line to the graph of a function rises from left to right. Is \(f'(4)\) positive, negative, or zero?

Hints

- Interpret the derivative as a tangent-line slope. - Think about the sign of the slope of a line that rises as \(x\) increases.

Solution

1. The derivative at a point is the slope of the tangent line there. 2. A line that rises from left to right has positive slope. 3. Therefore, \(f'(4)>0\).

Answer

Positive; \(f'(4)>0\).

Estimate a derivative at a point 2

Use the tangent line \(t\) to estimate \(f'(2)\) from the grid.

Hints

- The derivative at the marked point is represented by the slope of the tangent line. - Choose two clear grid intersections on the tangent line rather than estimating from the curved graph. - Compare the tangent line's vertical change with its horizontal change.

Solution

1. Read two convenient grid points on the tangent line, such as \((1,0)\) and \((3,4)\). 2. The tangent slope is \(\frac{4-0}{3-1}=2\). 3. From the graph, the estimated derivative at the marked input is \(f'(2)\approx2\).

Answer

\(f'(2)\approx2\)

Estimate a derivative at a point 3

At \(x=0\), is the derivative of the graphed function positive, negative, or zero?

Hints

- Imagine a tangent line at the requested input. - What would its rise and run tell you about the slope's sign?

Solution

1. The graph falls from left to right, so its tangent slope at \(x=0\) is negative.

Answer

Negative.

Estimate a derivative at a point 4

Use the straight tangent line \(t\) to estimate \(f'(1)\) from the graph.

Hints

- You need the slope of the tangent, not the height of the curve. - Pick two clear grid points on the straight line. - Use rise divided by run.

Solution

1. The tangent line passes through \((0,-1)\) and \((2,3)\). 2. Its slope is \(\frac{3-(-1)}{2-0}=2\), so \(f'(1)\approx2\).

Answer

\(f'(1)\approx2\)

Estimate a derivative at a point 5

Estimate \(f'(1)\) from the graph. Give one reason for your answer.

Hints

- Compare the graph's heights just to the left and right of \(x=1\). - If you drew a line touching the curve right at the top, would it tilt upward or downward?

Solution

1. The curve has a smooth maximum at \(x=1\). 2. The tangent there is horizontal, so \(f'(1)\approx0\).

Answer

\(f'(1)\approx0\); the tangent is horizontal.

Estimate a derivative at a point 6

Which graph has the larger derivative at \(x=0\), a) or b)? Use rise over run for each line to explain.

Hints

- A straight line's derivative is its slope at every point. - Both lines rise. Does that mean they have equal slopes?

Solution

1. In a), the line rises \(1\) unit for each unit to the right, so its slope is \(1\). 2. In b), the line rises \(2\) units for each unit to the right, so its slope is \(2\). 3. The derivative is larger in b), since \(2>1\).

Answer

b), because its slope is \(2\), compared with \(1\) in a).

Estimate a derivative at a point 7

The graph of \(f\) is shown with tangent lines \(t_1\) at \(A(0,2)\) and \(t_2\) at \(B(4,2)\). a) Use the tangent lines to find \(f'(0)\) and \(f'(4)\). Briefly explain your method. b) Decide whether the derivative is positive, negative, or zero at each input. Briefly justify each answer. 1) \(x=2\) 2) \(x=5\) 3) \(x=-2\)

Hints

- A derivative value equals the slope of the tangent line. - Use two exact points on each tangent to calculate its slope. - Increasing, decreasing, and horizontal behavior correspond to positive, negative, and zero derivative values.

Solution

1. The line \(t_1\) passes through \(A(0,2)\) and \((-2,0)\). Its slope is \(\frac{2-0}{0-(-2)}=1\), so \(f'(0)=1\). 2. The line \(t_2\) passes through \(B(4,2)\) and \((6,0)\). Its slope is \(\frac{0-2}{6-4}=-1\), so \(f'(4)=-1\). 3. At \(x=2\), the graph has a local maximum and a horizontal tangent, so \(f'(2)=0\). 4. At \(x=5\), the graph is decreasing, so \(f'(5)<0\). 5. At \(x=-2\), the graph is increasing, so \(f'(-2)>0\).

Answer

a) \(f'(0)=1\) and \(f'(4)=-1\) b) 1) \(f'(2)=0\) 2) \(f'(5)<0\) 3) \(f'(-2)>0\)

Estimate a derivative at a point 8

The graph of a function \(f\) and its tangent line \(t\) at the origin are shown. Use the graph to determine: a) the zeros of \(f^{\prime}\), b) the value of \(f^{\prime}(0)\), and c) the interval on which \(f^{\prime}(x)<0\).

Hints

- Look for points where the graph has horizontal tangent lines. - The derivative at a point is the slope of the tangent line there. - Use two points on line \(t\) to calculate its slope. - The derivative is negative where the function is decreasing.

Solution

1. The zeros of \(f^{\prime}\) occur where the graph of \(f\) has horizontal tangent lines. The graph has a local maximum at \(x=-3\) and a local minimum at \(x=3\), so the zeros are \(x=-3\) and \(x=3\). 2. The value \(f^{\prime}(0)\) is the slope of tangent line \(t\). The line passes through \((0,0)\) and \((2,-3)\), so its slope is \(\frac{-3-0}{2-0}=-1.5\). Therefore, \(f^{\prime}(0)=-1.5\). 3. The graph of \(f\) decreases between the local maximum and local minimum. Therefore, \(f^{\prime}(x)<0\) for \(x\in(-3,3)\).

Answer

a) \(x=-3\) and \(x=3\) b) \(f^{\prime}(0)=-1.5\) c) \(x\in(-3,3)\)

Estimate a derivative at a point 9

Use the graph of \(f\) to decide whether the tangent slope is positive, negative, or zero at \(x=-2\), \(x=0\), and \(x=2\). Explain briefly.

Hints

- Look at the direction of the curve near each input, not just the height of the point. - Imagine a tangent drawn at each input and compare its slope with zero.

Solution

1. The graph has a smooth high point at \(x=-2\), so its tangent slope there is \(0\). 2. The graph is falling at \(x=0\), so its tangent slope is negative. 3. The graph has a smooth low point at \(x=2\), so its tangent slope there is \(0\).

Answer

\(f'(-2)=0\): the graph has a smooth high point there. \(f'(0)<0\): the curve falls at \(0\). \(f'(2)=0\): the graph has a smooth low point there.

Estimate a derivative at a point 10

The graph of a quadratic function \(g\) is shown with tangent lines at \(x=0\) and \(x=4\). Point \(S\) marks the vertex. a) Use the tangent lines to find \(g^{\prime}(0)\) and \(g^{\prime}(4)\). b) At what input is \(g^{\prime}(x)=0\)? Give the correct mathematical name for point \(S\). c) Compare the slopes at \(x=1\) and \(x=3\). Which is greater?

Hints

- Calculate each line slope as change in \(y\) divided by change in \(x\). - The tangent line at the vertex of a smooth parabola is horizontal. - Compare a negative slope with a positive slope.

Solution

1. The tangent line at \(x=0\) passes through \((0,-1)\) and \((1,-2)\). Its slope is \(\frac{-2-(-1)}{1-0}=-1\), so \(g^{\prime}(0)=-1\). 2. The tangent line at \(x=4\) passes through \((4,-1)\) and \((5,0)\). Its slope is \(\frac{0-(-1)}{5-4}=1\), so \(g^{\prime}(4)=1\). 3. At the vertex \(S(2,-2)\), the tangent line is horizontal. Thus, \(g^{\prime}(2)=0\). Point \(S\) is the vertex and a local minimum. 4. At \(x=1\), the graph is decreasing, so the slope is negative. At \(x=3\), it is increasing, so the slope is positive. Therefore, \(g^{\prime}(3)>g^{\prime}(1)\).

Answer

a) \(g^{\prime}(0)=-1\) and \(g^{\prime}(4)=1\) b) \(g^{\prime}(2)=0\); \(S\) is the vertex and a local minimum. c) The slope at \(x=3\) is greater.

Estimate a derivative at a point 11

The graph of a parabola \(f\) and its tangent line \(t\) at \(A(2,0)\) are shown. 1. Use the tangent line to determine \(f'(2)\). 2. Use that derivative value to find the equation of \(t\). 3. Find the angle of inclination \(\alpha\) of \(t\) to the nearest hundredth of a degree.

Hints

- The derivative value at \(x=2\) is the slope of the displayed tangent. - Use two exact grid points on the tangent to determine its slope. - Use the derivative value and point \(A\) in point-slope form. - Relate slope and inclination angle with \(m=\tan(\alpha)\).

Solution

1. The tangent passes through \((2,0)\) and \((3,2)\), so its slope is \(\frac{2-0}{3-2}=2\). Therefore, \(f'(2)=2\). 2. Using \(A(2,0)\), the tangent is \(y-0=2(x-2)\), or \(y=2x-4\). 3. Since \(\tan(\alpha)=2\), \(\alpha=\arctan(2)\approx63.43^\circ\).

Answer

1. \(f'(2)=2\) 2. \(t:y=2x-4\) 3. \(\alpha\approx63.43^\circ\)

Estimate a derivative at a point 12

The graph of a rational function \(f\) and its tangent line \(t\) at \(P(1,0)\) are shown. 1. Use the tangent line to determine \(f'(1)\). 2. Use that derivative value to find the equation of \(t\). 3. Find the angle of inclination \(\alpha\) of \(t\) to the nearest hundredth of a degree.

Hints

- The derivative at the marked input is the slope of the displayed tangent. - Use two exact points on the tangent to determine its slope. - A line with negative slope has an inclination angle between \(90^\circ\) and \(180^\circ\).

Solution

1. The tangent passes through \((1,0)\) and \((3,-1)\), so its slope is \(\frac{-1-0}{3-1}=-\frac12\). Therefore, \(f'(1)=-\frac12\). 2. Using \(P(1,0)\), the tangent is \(y=-\frac12(x-1)\), or \(y=-\frac12x+\frac12\). 3. Since \(\arctan(-\frac12)\approx-26.57^\circ\), the inclination angle is \(180^\circ-26.57^\circ=153.43^\circ\).

Answer

1. \(f'(1)=-\frac12\) 2. \(t:y=-\frac12x+\frac12\) 3. \(\alpha\approx153.43^\circ\)

Estimate a derivative at a point 13

The graph of \(f\) and its tangent line \(t\) at \(W(0,0)\) are shown. 1. Use the tangent line to determine \(f'(0)\). 2. Use that derivative value to find the equation of \(t\). 3. Find the angle of inclination \(\alpha\) of \(t\) to the nearest hundredth of a degree.

Hints

- The derivative at \(x=0\) is the slope of the displayed tangent. - Use the origin and the second marked point to determine rise over run. - For a negative slope, convert the calculator's negative reference angle to an inclination between \(90^\circ\) and \(180^\circ\).

Solution

1. The tangent passes through \((0,0)\) and \((2,-1.5)\), so its slope is \(\frac{-1.5-0}{2-0}=-0.75\). Therefore, \(f'(0)=-0.75\). 2. Because the tangent passes through the origin, its equation is \(y=-0.75x\). 3. Since \(\arctan(-0.75)\approx-36.87^\circ\), its inclination angle is \(180^\circ-36.87^\circ=143.13^\circ\).

Answer

1. \(f'(0)=-0.75\) 2. \(t:y=-0.75x\) 3. \(\alpha\approx143.13^\circ\)

Estimate a derivative at a point 14

The graph of \(f\) and its tangent line at \(P(0,2)\) are shown. Find the slope of the tangent line and therefore the value of \(f'(0)\).

Hints

- Choose two points on the tangent line. - The tangent slope equals the derivative value at the point of tangency. - Use vertical change divided by horizontal change.

Solution

1. The derivative value \(f'(0)\) equals the slope of the tangent line at \(P\). 2. The tangent passes through \(P(0,2)\) and the grid point \((4,3)\). 3. Its slope is \(m=\frac{3-2}{4-0}=\frac{1}{4}=0.25\). 4. Therefore, \(f'(0)=0.25\).

Answer

The tangent slope is \(0.25\), so \(f'(0)=0.25\).

Estimate a derivative at a point 15

Use the graph of \(f\) to decide whether \(f'(-3)\), \(f'(0)\), and \(f'(3)\) are positive, negative, or zero. Explain your choices using the graph.

Hints

- Examine the curve close to each of the three inputs. - A derivative describes the direction and steepness of the tangent at that input.

Solution

1. The graph rises at \(x=-3\), so \(f'(-3)>0\). 2. The graph falls at \(x=0\), so \(f'(0)<0\). 3. The graph rises at \(x=3\), so \(f'(3)>0\).

Answer

\(f'(-3)>0\) because the curve rises near \(-3\); \(f'(0)<0\) because it falls near \(0\); and \(f'(3)>0\) because it rises near \(3\).

Estimate a derivative at a point 16

A test cart moves along a straight track. Its position after \(t\) seconds is \(s(t)=t^2+3t\) feet. The average velocities from \(t=4\) to nearby later times are shown below. <table><tr><th>Interval</th><th>Average velocity</th></tr><tr><td>\([4, 4.5]\)</td><td>\(11.5\,\text{ft/s}\)</td></tr><tr><td>\([4, 4.1]\)</td><td>\(11.1\,\text{ft/s}\)</td></tr><tr><td>\([4, 4.01]\)</td><td>\(11.01\,\text{ft/s}\)</td></tr></table> Use the pattern to estimate the cart's velocity at \(t=4\) seconds. Explain what the estimate means in terms of average velocity over shorter intervals.

Hints

- Look for the number the listed average velocities are approaching. - Think about what happens to the time interval as its right endpoint moves closer to \(4\). - Connect the limiting average velocity to the idea of velocity at a single instant.

Solution

1. The average velocities \(11.5\), \(11.1\), and \(11.01\) approach \(11\) as the right endpoint moves toward \(4\). 2. The instantaneous velocity is modeled by the limiting value of these average velocities. 3. Therefore the velocity at \(t=4\) is \(11\,\text{ft/s}\).

Answer

\(11\,\text{ft/s}\). It is the value approached by the average velocity over intervals whose endpoints get closer to \(t=4\).

Estimate a derivative at a point 17

A robotic arm's angle is \(\theta(t)=1.5t^2+0.2t\) degrees, where \(t\) is measured in seconds. Average angular velocities were computed on intervals immediately to the left and right of \(t=2\). <table><tr><th>Distance from \(2\)</th><th>Left average</th><th>Right average</th></tr><tr><td>\(0.5\)</td><td>\(5.45^\circ/\text{s}\)</td><td>\(6.95^\circ/\text{s}\)</td></tr><tr><td>\(0.1\)</td><td>\(6.05^\circ/\text{s}\)</td><td>\(6.35^\circ/\text{s}\)</td></tr><tr><td>\(0.01\)</td><td>\(6.185^\circ/\text{s}\)</td><td>\(6.215^\circ/\text{s}\)</td></tr></table> Estimate the angular velocity at \(t=2\), and explain why using values from both sides strengthens the estimate.

Hints

- Compare the left and right entries in the row closest to \(t=2\). - Look for a common value between the two sequences. - An instantaneous rate is supported when shrinking-interval averages from both sides agree.

Solution

1. The left averages increase toward \(6.2^\circ/\text{s}\). 2. The right averages decrease toward \(6.2^\circ/\text{s}\). 3. Because the average rates from both sides approach the same value, the instantaneous angular velocity is \(6.2^\circ/\text{s}\).

Answer

\(6.2^\circ/\text{s}\). The left- and right-side average angular velocities converge to the same value.

Estimate a derivative at a point 18

For \(f(x)=x^3\), define the average rate from \(x=2\) to \(x=2+h\) by \(Q(h)=\frac{f(2+h)-f(2)}{h}\), where \(h\ne0\). <table><tr><th>\(h\)</th><th>\(Q(h)\)</th></tr><tr><td>\(-0.5\)</td><td>\(9.25\)</td></tr><tr><td>\(-0.1\)</td><td>\(11.41\)</td></tr><tr><td>\(0.1\)</td><td>\(12.61\)</td></tr><tr><td>\(0.5\)</td><td>\(15.25\)</td></tr></table> Estimate the instantaneous rate of change of \(f\) at \(x=2\) without using a derivative rule.

Hints

- Compare the values for small positive and negative \(h\). - Look for a number between \(11.41\) and \(12.61\) that both sides appear to approach. - The quotient represents an average rate over an interval whose width is \(|h|\).

Solution

1. For negative \(h\) close to \(0\), \(Q(h)\) rises toward \(12\). 2. For positive \(h\) close to \(0\), \(Q(h)\) falls toward \(12\). 3. Therefore the instantaneous rate of change at \(x=2\) is \(12\).

Answer

\(12\).

Estimate a derivative at a point 19

An engineer estimates an instantaneous flow rate from pairs of average rates on opposite sides of \(t=5\) minutes. The closest pair is \(7.998\,\text{L/min}\) and \(8.002\,\text{L/min}\). A report lists the instantaneous flow rate as \(8.000\,\text{L/min}\). Is that estimate consistent with the data? Explain why reporting \(8.000\) is more reasonable than reporting \(8.002\) as the instantaneous rate.

Hints

- Use both sides of the instant, not just one measurement. - Ask which number the two nearby averages appear to bracket. - Distinguish a finite-interval average from the limiting instantaneous estimate.

Solution

1. The two closest averages lie nearly symmetrically around \(8.000\,\text{L/min}\). 2. Their behavior supports a common limiting value of \(8.000\,\text{L/min}\). 3. The value \(8.002\,\text{L/min}\) is one nearby average rate, not the common value suggested by both sides.

Answer

Yes. \(8.000\,\text{L/min}\) is consistent with the two-sided data; \(8.002\,\text{L/min}\) is only one finite-interval average.

Estimate a derivative at a point 20

Four data sets show average rates over intervals shrinking toward \(t=a\). Which set gives the strongest evidence that the instantaneous rate at \(t=a\) is \(2\)? A. Left: \(1.8,1.98,1.998\); right: \(2.2,2.02,2.002\) B. Left: \(1,1,1\); right: \(3,3,3\) C. \(1,-1,1,-1,1,-1\) D. \(2,20,200,2000\) Explain your choice.

Hints

- Check whether each set settles toward a finite value. - When two sides are listed, compare their limiting behavior. - Eliminate patterns that disagree, oscillate, or grow without bound.

Solution

1. In A, the left-side rates approach \(2\) from below and the right-side rates approach \(2\) from above. 2. In B, the two sides approach different values. 3. C oscillates rather than settling, and D grows without bound. 4. Therefore A gives the strongest evidence for an instantaneous rate of \(2\).

Answer

A. Both sides approach \(2\).

Estimate a derivative at a point 21

The graph of \(f\) and its tangent line at \(P\) are shown. Use the grid to estimate \(f'(1)\) to the nearest tenth.

Hints

- The derivative at a point is the slope of the tangent line there. - Choose two well-separated x-values on the tangent and estimate the corresponding y-values from the grid. - Use rise divided by run; small differences in your visual readings are expected.

Solution

1. The derivative value \(f'(1)\) is the slope of the tangent line at \(P\). 2. From the grid, the tangent is approximately at \((0,0.3)\) and \((2,3.7)\). These readings are estimates because the line does not pass through exact grid intersections there. 3. The estimated slope is \(\frac{3.7-0.3}{2-0}\approx1.7\). Therefore, \(f'(1)\approx1.7\).

Answer

\(f'(1)\approx1.7\)

Estimate a derivative at a point 22

The table gives values near \(x=2\). <table><tr><th>\(x\)</th><td>\(1.8\)</td><td>\(1.95\)</td><td>\(2\)</td><td>\(2.1\)</td></tr><tr><th>\(f(x)\)</th><td>\(4.04\)</td><td>\(4.0025\)</td><td>\(4\)</td><td>\(4.01\)</td></tr></table> Find the secant slopes on \([1.8,2]\), \([1.95,2]\), and \([2,2.1]\). Estimate \(f'(2)\).

Hints

- Use the output change divided by the input change in each interval. - Compare the signs and sizes of the left and right secant slopes. - Decide what number the nearby rates appear to approach.

Solution

1. The left secants have slopes \(\frac{4-4.04}{0.2}=-0.2\) and \(\frac{4-4.0025}{0.05}=-0.05\). 2. The right secant has slope \(\frac{4.01-4}{0.1}=0.1\). 3. These small slopes lie on opposite sides of zero, so \(f'(2)\) is estimated as \(0\).

Answer

Secant slopes: \(-0.2\), \(-0.05\), \(0.1\). Estimated derivative: \(f'(2)\approx0\).

Estimate a derivative at a point 23

For a function \(f\), \(f(1.9)=6.859\), \(f(2)=8\), and \(f(2.1)=9.261\). Estimate \(f'(2)\) using (a) the forward secant from \(2\) to \(2.1\) and (b) the symmetric secant from \(1.9\) to \(2.1\). An exact calculation gives \(f'(2)=12\). Which estimate is closer?

Hints

- The two estimates use different pairs of table values. - A symmetric estimate uses inputs equally far from the target input. - Compare the absolute errors from the stated target.

Solution

1. The forward estimate is \(\frac{9.261-8}{0.1}=12.61\). 2. The symmetric estimate is \(\frac{9.261-6.859}{0.2}=12.01\). 3. The symmetric estimate is closer to \(12\).

Answer

(a) \(12.61\) (b) \(12.01\); the symmetric secant gives the closer estimate.

Estimate a derivative at a point 24

Estimates of the average rate of change of \(f\) on four intervals are shown: <table><tr><th>Interval</th><th>Average rate</th></tr><tr><td>\([3.5,4]\)</td><td>\(2.8\)</td></tr><tr><td>\([4,4.5]\)</td><td>\(3.2\)</td></tr><tr><td>\([3.9,4]\)</td><td>\(2.95\)</td></tr><tr><td>\([4,4.1]\)</td><td>\(3.05\)</td></tr></table> Estimate \(f'(4)\) from the wider and narrower interval pairs. What do the estimates suggest? Do these data show which estimate is more accurate?

Hints

- What number lies between the two rates in each pair? - Does each pair suggest the same pointwise slope? - Without knowing the exact derivative, how could you measure either estimate's error?

Solution

1. The longer intervals give rates on either side of \(x=4\): \(2.8\) and \(3.2\). Their average is \(3\). 2. The shorter intervals give \(2.95\) and \(3.05\), whose average is also \(3\). 3. Both pairs suggest \(f'(4)\approx3\). The data do not give the exact derivative, so they do not establish that one numerical estimate is more accurate than the other.

Answer

Both interval pairs suggest \(f'(4)\approx3\). Neither pair's accuracy can be established from these data alone.

Estimate a derivative at a point 25

For an exponential function \(f(x)=b^x\), where \(b>0\), the slope at \(x=0\) can be approximated by \(\frac{f(h)-f(0)}{h}=\frac{b^h-1}{h}\) for a small positive value of \(h\). a) Use \(h=0.0001\) to approximate \(f'(0)\) for \(b=2\) and \(b=4\). b) A special base has slope \(1\) at \(x=0\). Solve \(\frac{b^h-1}{h}=1\) for \(b\), and use \(h=0.0001\) to approximate this base.

Hints

- Substitute each base into the given difference quotient. - Keep enough calculator precision before rounding. - Isolate \(b^h\) before taking the \(h\)th root. - Rewrite an \(h\)th root as a power of \(1/h\).

Solution

1. For \(b=2\), \(\frac{2^{0.0001}-1}{0.0001}\approx 0.6932\). For \(b=4\), \(\frac{4^{0.0001}-1}{0.0001}\approx 1.3864\). 2. Solve the equation algebraically: \(\frac{b^h-1}{h}=1\) implies \(b^h=1+h\). Therefore, \(b=(1+h)^{1/h}\). With \(h=0.0001\), \(b=(1.0001)^{10000}\approx 2.7181\).

Answer

a) For \(b=2\), \(f'(0)\approx 0.6932\). For \(b=4\), \(f'(0)\approx 1.3864\). b) \(b\approx 2.7181\)

Estimate a derivative at a point 26

The graph of \(f\) is shown with points \(A(0,2)\), \(B(2,3)\), and \(C(4,2)\) and their tangent lines. a) Use the tangent slopes to determine \(f'(0)\), \(f'(2)\), and \(f'(4)\). b) Use those derivative values to write the equation of the tangent line at each point.

Hints

- At a marked point, the derivative value is the slope of the displayed tangent line. - Read two convenient grid points on each tangent to determine its slope. - After finding the derivative value, use the marked point in point-slope form.

Solution

1. At \(A\), the tangent rises \(1\) unit for a run of \(1\), so \(f'(0)=1\). 2. At \(B\), the tangent is horizontal, so \(f'(2)=0\). 3. At \(C\), the tangent falls \(1\) unit for a run of \(1\), so \(f'(4)=-1\). 4. Through \(A(0,2)\) with slope \(1\), the tangent is \(y=x+2\). 5. Through \(B(2,3)\) with slope \(0\), the tangent is \(y=3\). 6. Through \(C(4,2)\) with slope \(-1\), the tangent is \(y=-x+6\).

Answer

a) \(f'(0)=1\), \(f'(2)=0\), \(f'(4)=-1\) b) At \(A\): \(y=x+2\); at \(B\): \(y=3\); at \(C\): \(y=-x+6\)

Estimate a derivative at a point 27

The graphs of \(f\) and \(g\) are shown. The graph of \(g\) is a translation of the graph of \(f\). The tangent line \(t\) to the graph of \(f\) at \(x=2\) is also shown. a) Use the marked vertices \(S_f\) and \(S_g\) to describe the horizontal and vertical translation. Write \(g\) in the form \(g(x)=f(x-d)+c\). b) Find the slope of \(t\) and therefore the value of \(f'(2)\). c) What is the slope of the graph of \(g\) at \(x=5\)? Justify your answer geometrically, and find the equation of the tangent line to \(g\) at that input.

Hints

- Compare the coordinates of the two vertices. - Use two readable points on \(t\) to calculate its slope. - A translation does not change the slope at corresponding points. - Use the translated point and the slope in point-slope form.

Solution

1. The vertices are \(S_f(1,0.5)\) and \(S_g(4,2.5)\). Therefore, the graph is translated \(3\) units right and \(2\) units up, so \(g(x)=f(x-3)+2\). 2. The line \(t\) passes through \((1,0)\) and \((2,1)\), so its slope is \(\frac{1-0}{2-1}=1\). Therefore, \(f'(2)=1\). 3. A translation preserves slopes at corresponding points. Since the graph moved \(3\) units right, \(x=5\) on \(g\) corresponds to \(x=2\) on \(f\). Thus, \(g'(5)=f'(2)=1\). 4. The translated point is \((5,3)\). A line of slope \(1\) through that point is \(y-3=x-5\), or \(y=x-2\).

Answer

a) The graph is translated \(3\) units right and \(2\) units up, so \(g(x)=f(x-3)+2\). b) The slope is \(1\), so \(f'(2)=1\). c) The slope is \(1\), and the tangent line is \(y=x-2\).

Estimate a derivative at a point 28

The graph of a quadratic function \(f\) and its tangent lines \(t_1\) and \(t_2\) at \(A(0,-1)\) and \(B(2,3)\) are shown. a) Use the tangent lines to determine \(f'(0)\) and \(f'(2)\). b) Use those derivative values to find the equations of \(t_1\) and \(t_2\). c) Find the angle of inclination of each tangent line to the nearest tenth of a degree.

Hints

- A derivative value at a marked point is represented by the slope of its tangent. - Use exact grid points to determine each tangent slope. - Once the slope is known, use the marked point to write the line equation. - Relate a line's slope to its inclination angle with \(m=\tan(\alpha)\).

Solution

1. Line \(t_1\) passes through \((0,-1)\) and \((1,0)\), so its slope is \(1\). Hence \(f'(0)=1\). 2. Line \(t_2\) passes through \((1,0)\) and \((2,3)\), so its slope is \(3\). Hence \(f'(2)=3\). 3. Using the marked points, \(t_1\) is \(y=x-1\), and \(t_2\) is \(y-3=3(x-2)\), or \(y=3x-3\). 4. Since \(m=\tan(\alpha)\), \(\alpha_1=\arctan(1)=45.0^\circ\) and \(\alpha_2=\arctan(3)\approx71.6^\circ\).

Answer

a) \(f'(0)=1\), \(f'(2)=3\) b) \(t_1:y=x-1\); \(t_2:y=3x-3\) c) \(45.0^\circ\) and \(71.6^\circ\)

Estimate a derivative at a point 29

The graph of a function \(f\) and the tangent line \(t\) at \(x=0\) are shown. Within \([-5,5]\), answer the following questions about the derivative \(f'\). a) Find the zeros of \(f'\). Briefly justify your answer. b) Use tangent line \(t\) to find \(f'(0)\). c) For \(-5<x<5\), state the intervals on which \(f'(x)>0\). Justify your answer using the behavior of \(f\).

Hints

- Zeros of the derivative correspond to horizontal tangents on the function graph. - Use two readable points on tangent line \(t\) to calculate its slope. - The derivative is positive where the function graph is increasing.

Solution

1. The zeros of \(f'\) occur where the graph of \(f\) has horizontal tangent lines. The graph has a local maximum at \(x=-2\) and a local minimum at \(x=2\), so the zeros are \(x=-2\) and \(x=2\). 2. The tangent line passes through \((0,0)\) and \((1,-3)\). Its slope is \(\frac{-3-0}{1-0}=-3\), so \(f'(0)=-3\). 3. On the requested interval, the graph of \(f\) is increasing for \(-5<x<-2\) and for \(2<x<5\). Therefore, \(f'(x)>0\) on \((-5,-2)\) and \((2,5)\).

Answer

a) \(x=-2\) and \(x=2\) b) \(f'(0)=-3\) c) \((-5,-2)\) and \((2,5)\)

Estimate a derivative at a point 30

The graph of \(f\) is shown with tangent lines at \(A(1,4)\), \(B(2,2)\), and \(C(4,1)\). a) Determine \(f'(1)\), \(f'(2)\), and \(f'(4)\) from the displayed tangent slopes. b) Use those derivative values to find the equation of the tangent line at each point.

Hints

- The derivative at a marked input is the slope of the tangent shown there. - Use two exact grid points on each tangent, choosing points farther apart for the shallow line. - Use the derivative value and the marked point to write each tangent equation.

Solution

1. At \(A\), the tangent passes through \((0,8)\) and \((2,0)\), so its slope is \(-4\). Thus \(f'(1)=-4\). 2. At \(B\), the tangent passes through \((0,4)\) and \((4,0)\), so its slope is \(-1\). Thus \(f'(2)=-1\). 3. At \(C\), the tangent passes through \((0,2)\) and \((8,0)\), so its slope is \(-\frac14\). Thus \(f'(4)=-\frac14\). 4. Using the marked points gives the tangent equations \(y=-4x+8\), \(y=-x+4\), and \(y=-\frac14x+2\), respectively.

Answer

a) \(f'(1)=-4\), \(f'(2)=-1\), \(f'(4)=-\frac14\) b) At \(A\): \(y=-4x+8\); at \(B\): \(y=-x+4\); at \(C\): \(y=-\frac14x+2\)

Estimate a derivative at a point 31

The graph of a cubic function \(f\) is shown with tangent lines at \(A(-2,4)\), \(B(0,0)\), and \(C(2,-4)\). a) Determine \(f'(-2)\), \(f'(0)\), and \(f'(2)\) from the tangent lines. b) Use those derivative values to write the tangent equation at each point.

Hints

- A horizontal tangent represents derivative value \(0\). - At \(B\), use the grid to determine the tangent's rise over run. - After finding each derivative value, combine it with the marked point to write the tangent equation.

Solution

1. At \(A\), the tangent is horizontal, so \(f'(-2)=0\). 2. At \(B\), the tangent falls \(3\) units for each \(1\)-unit increase in \(x\), so \(f'(0)=-3\). 3. At \(C\), the tangent is horizontal, so \(f'(2)=0\). 4. Therefore, the tangent equations are \(y=4\) at \(A\), \(y=-3x\) at \(B\), and \(y=-4\) at \(C\).

Answer

a) \(f'(-2)=0\), \(f'(0)=-3\), \(f'(2)=0\) b) At \(A\): \(y=4\); at \(B\): \(y=-3x\); at \(C\): \(y=-4\)

Estimate a derivative at a point 32

The graph of a quadratic function \(f\) is shown with tangent lines \(t_1\), \(t_2\), and \(t_3\) at \(x_1=-2\), \(x_2=0\), and \(x_3=2\). a) Read the slopes \(m_1\), \(m_2\), and \(m_3\) from the tangent lines. b) A point \(Q(x,y)\) on the derivative graph has the tangent slope of \(f\) at \(x\) as its y-coordinate. Give the coordinates of \(Q_1\), \(Q_2\), and \(Q_3\). c) What type of function do you expect \(f^{\prime}\) to be: constant, linear, quadratic, or another type? Justify your answer.

Hints

- Use rise over run to find each tangent-line slope. - The y-coordinate on the derivative graph equals the tangent slope at that input. - Plot the three derivative points mentally and identify their pattern. - Recall the degree of the derivative of a quadratic polynomial.

Solution

1. At \(x=-2\), tangent line \(t_1\) falls \(2\) units for each unit to the right, so \(m_1=-2\). At \(x=0\), tangent line \(t_2\) is horizontal, so \(m_2=0\). At \(x=2\), tangent line \(t_3\) rises \(2\) units for each unit to the right, so \(m_3=2\). 2. The corresponding derivative points are \(Q_1(-2,-2)\), \(Q_2(0,0)\), and \(Q_3(2,2)\). 3. These three points lie on a line through the origin. Also, the derivative of a quadratic function is linear. Therefore, \(f^{\prime}\) is a linear function.

Answer

a) \(m_1=-2\), \(m_2=0\), and \(m_3=2\) b) \(Q_1(-2,-2)\), \(Q_2(0,0)\), and \(Q_3(2,2)\) c) \(f^{\prime}\) is linear.

Estimate a derivative at a point 33

For \(p(t)=t^2\), the average rate of change over an interval \([a, b]\) containing \(t=1\) can be computed from \(\frac{p(b)-p(a)}{b-a}\). Three such rates are: <table><tr><th>Interval</th><th>Average rate</th></tr><tr><td>\([0.5, 1.2]\)</td><td>\(1.7\)</td></tr><tr><td>\([0.9, 1.05]\)</td><td>\(1.95\)</td></tr><tr><td>\([0.99, 1.001]\)</td><td>\(1.991\)</td></tr></table> The intervals are not centered at \(1\). Explain why the data can still be used to estimate the instantaneous rate at \(t=1\), and give the estimate.

Hints

- Track what happens to both interval endpoints from row to row. - Decide whether symmetry is essential to the idea of shrinking an interval. - Use the trend in the average rates to identify the estimate.

Solution

1. Each interval contains \(t=1\), and both endpoints move closer to \(1\) across the table. 2. The corresponding average rates \(1.7\), \(1.95\), and \(1.991\) approach \(2\). 3. Centering is not required; shrinking the interval around the instant is the key idea. The instantaneous rate is \(2\).

Answer

\(2\). The intervals need not be symmetric as long as their endpoints approach \(1\) and the average rates approach a common value.

Estimate a derivative at a point 34

A process has output \(E(t)=t^3-12t\). Near \(t=2\), average rates are: <table><tr><th>Distance from \(2\)</th><th>Left average</th><th>Right average</th></tr><tr><td>\(0.5\)</td><td>\(-2.75\)</td><td>\(3.25\)</td></tr><tr><td>\(0.1\)</td><td>\(-0.59\)</td><td>\(0.61\)</td></tr><tr><td>\(0.01\)</td><td>\(-0.0599\)</td><td>\(0.0601\)</td></tr></table> Estimate the instantaneous rate at \(t=2\). What do the opposite signs of nearby averages suggest about the behavior before and after that instant?

Hints

- Do not treat opposite signs as evidence that the limiting values are different. - Compare the magnitudes of the left and right averages as the distance shrinks. - Interpret what a negative rate before the instant and a positive rate after it mean.

Solution

1. Both columns approach \(0\) as the intervals shrink toward \(t=2\). 2. The instantaneous rate is \(0\). 3. Negative averages just before \(t=2\) and positive averages just after \(t=2\) suggest that the output is changing from decreasing to increasing near that time.

Answer

\(0\). The nearby averages suggest a change from decreasing before \(t=2\) to increasing after \(t=2\).

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Estimate a derivative at a point

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