Section 6.1 – Antiderivatives Graphically and Numerically

Home , Antiderivative, Graph of a function, Slope

1. For each of the following functions f, sketch two antiderivatives of f, one that satisﬁes F (0) = 0 and one that satisﬁes F (0) = 1. f(x) f(x) f(x) y y y 1 1 1

0.5 0.5 0.5

x x x 0.5 1 1.5 2 0.5 1 1.5 2 0.5 1 1.5 2 -0.5 -0.5 -0.5

-1 -1 -1

Antiderivatives: Antiderivatives: Antiderivatives: y y y 1 1 1

0.5 0.5 0.5

x x x 0.5 1 1.5 2 0.5 1 1.5 2 0.5 1 1.5 2 -0.5 -0.5 -0.5

-1 -1 -1

2. Given below is the graph of a function f 0(x). Given that f(0) = 1, sketch an accurate graph of f(x) on the blank axes to the right. Label all critical points and inﬂection points of f with their x and y coordinates. f 0(x) f(x) y y 2 2 H1,2L 1 1 H2,1L H3,0L x x 1 2 3 4 1 2 3 4 -1 -1

-2 -2

Since f 0(1) = f 0(3) = 0, we see that f will have critical points at x = 1 and x = 3; we have labeled the critical points (1, 2) and (3, 0) on our sketch of f above.

Since f 0 is decreasing on 0 ≤ x ≤ 2 and increasing on 2 ≤ x ≤ 4, it follows that f 00 is negative on 0 ≤ x ≤ 2 and positive on 2 ≤ x ≤ 4. Therefore, f has an inflection point at x = 2; we have labeled the inflection point (2, 1) on our sketch of f above.

Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2017 1 3. Given below is the graph of a function f(x). Sketch a graph of an antiderivative F (x) of f(x) that satisﬁes F (0) = 5. Label all critical points and inﬂection points of f with their approximate x and y coordinates. f(x) Area=30 Area=20 x 1 2 3 4 5 6 Area = 4

20 CP F(x) 10 IP x IP 1 2 3IP 4 5 6 −10 −20 CP

Since F 0(x) = f(x), we see that F 0(x) = f(x) = 0 when x = 1.9 and x = 5.1. Therefore, F has critical points at x = 1.9 and x = 5.1; we have estimated the coordinates of these critical points as (1.9, −15) and (5.1, 15) (see above sketch).

Also, we can see that f(x) changes from decreasing to increasing (or vice versa) at each of the points x = 0.5, x = 3.3, and x = 5.5. Since F 00(x) = f 0(x), this means that f 0(x) (and hence F 00(x)) changes sign at these three values of x; therefore, F has inflection points at these points. We have estimated the coordinates of these inflection points as (5, 0), (3.3, −2), and (5.5, 13) (see above sketch).

2 Developed by Jerry Morris Section 6.2 – Constructing Antiderivatives Analytically

1. Evaluate each of the of the following exactly. Z 3 10 − 5x (a) √ dx 1 x

Z 3 10 − 5x Z 3 Z 3 √ dx = (10 − 5x) x−1/2 dx = (10x−1/2 − 5x1/2) dx 1 x 1 1

1/2 3/2 3 x x = 10 · − 5 · 1/2 3/2 1 3 1/2 10 3/2 = 20x − x 3 1 √ √ 10 = (20 3 − 10 3) − (20 − ) 3 √ 50 = 10 3 − . 3

Z 0 (b) (2ex − 4 cos x) dx −5

Z 0 x x 0 0 −5 (2e − 4 cos x) dx = (2e − 4 sin x)|−5 = (2e − 4 sin 0) − (2e − 4 sin(−5)) −5 = 2 − 2e−5 + 4 sin(−5) 2 = 2 − − 4 sin 5. e5

Z (c) x(x2 + 1)2 dx

Z Z x6 x4 x2 x(x2 + 1)2 dx = (x5 + 2x3 + x) dx = + 2 · + + C 6 4 2 1 1 1 = x6 + x4 + x2 + C. 6 2 2

Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2017 3 1 2. Find the exact area under y = between x = 1 and x = 4. 4x The exact area is given by

Z 4 Z 4 4 1 1 −1 1 dx = x dx = ln x 1 4x 4 1 4 1 1 1 = ln 4 − ln 1 4 4 ln 4 = . 4

3. At Bill’s Gas ’n’ Snax, the most happening gas station in Dudleyville, Arizona, the demand for gasoline is changing at a rate of r(t) = −90t − 100 gallons per day, where t = 0 represents the number of days after a R 4 major jump in gas prices. Calculate 1 r(t) dt and give a practical interpretation of this integral, including the appropriate units.

We have Z 4 Z 4 2 4 r(t) dt = (−90t − 100) dt = −45t − 100t 1 1 1 = (−45 · 42 − 400) − (−45 − 100) = −975.

Since r(t) is measured in gallons per day, and t is measured in days, we conclude that the units of the above integral are gallons. Therefore, the evaluation of the above integral indicates that the demand for gasoline will drop by 975 gallons between the first day and the fourth day after the jump in gas prices.

4 Developed by Jerry Morris Section 6.2 – Derivatives and Antiderivatives

0 1. Let f (x) = 3 − |2x|. Sketch the following graphs clearly y and accurately and label all important points. 3 f ’’ (a) On one set of axes, sketch a graph of f 0(x) and f 00(x). 2 f ’ 1 (See figure to the right.) x -3 -2 -1 1 2 3 -1 -2 -3 -4

(b) On another set of axes, sketch possible graphs of y f(x), one having f(0) = −2 and one having f(0) = 0. 3 2 (See figure to the right.) 1 x -3 -2 -1 1 2 3 -1 -2 -3 -4 2. Find derivatives and antiderivatives of the following. (a) ex ex is both the derivative and an antiderivative.

(b) sin x cos x is the derivative and − cos x is an antiderivative.

(d)5 x4 20x3 is the derivative and x5 is an antiderivative.

(e)3 e2x 6e2x is the derivative and (3/2)e2x is an antiderivative.

(f) cos 3x + sin 4x −3 sin(3x) + 4 cos(4x) is the derivative and (1/3) sin(3x) − (1/4) cos(4x) is an antiderivative.

−x x+3 5 (g) f(x) = e − x2 + cos2 2x First, we rewrite the function as 5 e−x − x−1 − 3x−2 + . cos2(2x) Differentiating term by term, we see that the derivative is given by d f 0(x) = −e−x + x−2 + 6x−3 + 5 (cos(2x))−2 dx 1 6 = −e−x + + − 10(cos(2x))−3 · (− sin(2x)) · 2 x2 x3 1 6 20 sin(2x) = −e−x + + + x2 x3 cos3(2x)

Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2017 5 To find an antiderivative for f, we first recall that d 1 2 tan(2x) = · 2 = , dx cos2(2x) cos2(2x)

which means that (1/2) tan(2x) is an antiderivative of 1/ cos2(2x). Therefore, an antiderivative of f is given by

1 F (x) = −e−x − ln x + 3x−1 + 5 · tan(2x) 2 3 5 = −e−x − ln x + + tan(2x). x 2

3. Explain the diﬀerence between an antiderivative and the antiderivative.

When we say an antiderivative of a function f, we are simply referring to one of many possible functions whose derivatives equal f. For example, x3 is an antiderivative of 3x2 even though 3x2 has many other possible antiderivatives, such as x3 + 1, x3 + 2, etc. When we say the antiderivative of a function f, we are actually referring to the so-called "general" antiderivative of f, a family of functions that describes all possible antiderivatives of f at once. For example, if C is any constant, then x3 + C is the (general) antiderivative of 3x2.

6 Developed by Jerry Morris Section 6.4 – The Second Fundamental Theorem of Calculus

Given to the right is a graph of the function Example. y = sin(x2) y = sin(x2). Deﬁne a new function y Z x 1.5 F (x) = sin(t2) dt. 0 1 F

0.5

x 0.5 1 1.5 2 2.5 3 -0.5

-1

1. Use your calculator to help you ﬁll in the following table entries. x 0 0.5 1 1.5 2 2.5 3 F (x) 0 0.041 0.310 0.778 0.805 0.431 0.774

Z 0 F (0) = sin(t2) dt = 0 0 Z 0.5 F (0.5) = sin(t2) dt = 0.041 0 Z 1 F (1) = sin(t2) dt = 0.310 0 Z 1.5 F (1.5) = sin(t2) dt = 0.778 0 Z 2 F (2) = sin(t2) dt = 0.805 0 Z 2.5 F (2.5) = sin(t2) dt = 0.431 0 Z 3 F (3) = sin(t2) dt = 0.774 0

2. On the same set of axes as the above graph, sketch the graph of F (x). Do you notice a relationship between f and F ? How are these two functions related?

It appears that F 0(x) = sin(x2), meaning that the function F appears to be an antiderivative of the function y = sin(x2) whose graph was given. The Second Fundamental Theorem of Calculus confirms this conjecture.

Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2017 7 Exercises. 1. For each of the following, ﬁnd F 0(x). R x (a) F (x) = 3 (2t + 1) dt By the Second Fundamental Theorem of Calculus, we have d Z x F 0(x) = (2t + 1) dt dx 3 = 2x + 1.

R ln x 1 (b) F (x) = 1 1+t4 dt By the Chain Rule and the Second Fundamental Theorem of Calculus, we have Z ln x ! 0 d 1 F (x) = 4 dt dx 1 1 + t 1 d = · ln x 1 + (ln x)4 dx 1 = . x + x(ln x)4

R x sin t (c) F (x) = x Si(x), where Si(x) = 0 t dt By the Product Rule and the Second Fundamental Theorem of Calculus, we have d F 0(x) = x Si(x) + Si(x) dx d Z x sin t Z x sin t = x dt + dt dx 0 t 0 t sin x Z x sin t = x · + dt x 0 t Z x sin t = sin x + dt, 0 t 0 R x sin t 0 so F (x) = sin x + 0 t dt, which can also be written as F (x) = sin x + Si(x).

R x 2. Let F (x) = 0 f(t) dt, where f(x) is the function whose graph is given below. 2 f(x) A2 A3 A5 −4 −2 2 4 6 A1 A4 −2

(a) What are the critical points of F (x)? By the Second Fundamental Theorem of Calculus, we have F 0(x) = f(x), so the critical points of F will occur when f(x) = 0. Therefore, using the graph of f, we see that x = −5, x = 0, x = 3, and x = 5 are all critical points of F.

8 Developed by Jerry Morris (b) Where is F (x) increasing? decreasing? Again, since F 0(x) = f(x), the function F is increasing where f(x) > 0 and decreasing where f(x) < 0. Using the provided graph of f, we therefore see that F is increasing on −5 ≤ x ≤ 3 and 5 ≤ x ≤ 6, and decreasing on −6 ≤ x ≤ −5 and on 3 ≤ x ≤ 5.

(c) Locate all places where F (x) has a local maximum or a local minimum, and make it clear which are which. F will have a local maximum where F 0 = f changes sign from positive to negative, and a local minimum where F 0 = f changes sign from negative to positive. Therefore, F has a local maximum at x = 3 and local minima at x = −5 and x = 5.

(d) Where is F (x) concave up? concave down? F will be concave up when F 00(x) > 0, which in turn will happen when f, the function whose graph we are given, is increasing. Therefore, F is concave up when −6 < x < −2.8, when 0 < x < 2, and when 4 < x < 6. On the other hand, F is concave down when −2.8 < x < 0 and when 2 < x < 4.

(e) Where does F attain its maximum value? Use a method of your choice to estimate this maximum value. Since the longest interval on which F increases is −5 ≤ x ≤ 3, a likely location for where the maximum value of F occurs is at x = 3. To help confirm this, we have labeled various "pockets of area" between f and the x-axis as shown on the graph above. Note that since the sum A2 + A3 is significantly greater than A1, the function F increases significantly more than it decreases between −6 and 3, and from this we can conclude that F (3) is greater than any value of F to the left of x = 3. Similarly, since A4 is greater than A5,F decreases more than it increases to the right of x = 3, so F (3) is also greater than any value of F to the right of x = 3. This confirms that F attains its maximum value at x = 3. R 3 To estimate this maximum value, we note that F (3) = 0 f(t) dt, so using a middlesum with n = 3, the approximate value of F (3) is given by

0.2(1) + 1(1) + 1(1) = 2.2,

so we estimate that the maximum value of F is about 2.2.

R x 1 3. Let F (x) = 2 ln t dt for x ≥ 2. (a) Find F 0(x). By the Second Fundamental Theorem of Calculus, we have d Z x 1 1 F 0(x) = dt = . dx 2 ln t ln x

(b) Is F increasing or decreasing? What can you say about the concavity of F ? Since ln x > 0 for all x ≥ 2, we see that F 0(x) = 1/(ln x) > 0 for all x ≥ 2; therefore, F is increasing for all x ≥ 2. To discuss the concavity of F, we need to calculate F 00; we have d 1 −1 F 00(x) = = . dx ln x x(ln x)2 Since x(ln x)2 is positive for all x ≥ 2, it follows that F 00(x) is negative for all x ≥ 2, so we conclude that F is concave down.

Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2017 9 4. Let f(x) be the function whose graph is given to the right. f(x) Deﬁne 2 Z x F (x) = f(t) dt. 0 1 Fill in the entries in the table that follow: 1 2 3 45 6 −1 x 1 2 3 4 5 −2 F (x) 1.5 2 1.5 0.5 −0.25 F 0(x) 1 0 −1 −1 −0.5 F 00(x) −1 −1 DNE DNE 0.5

Since F is defined as an integral, we use the area under the provided graph as a guide to finding values of F and filling in the first row of the table; for example, we have

Z 1 F (1) = f(t) dt = 1.5 0 Z 2 F (2) = f(t) dt = 1.5 + 0.5 = 2 0 Z 3 F (3) = f(t) dt = 2 − 0.5 = 1.5 0 . .

To find the entries in the 2nd row of the table, note that, by the Second Fundamental Theorem of Calculus, we have d Z x F 0(x) = f(t) dt = f(x), dx 0 so F 0(1) = f(1) = 1,F 0(2) = f(2) = 0,F 0(3) = f(3) = −1, etc. In other words, to fill in the 2nd row of the table, we are simply reading off the output values of the function f whose graph we are given.

Finally, to find the entries in the 3rd row of the table, note that since F 0(x) = f(x), we have F 00(x) = f 0(x). Therefore, F 00(1) = f 0(1) = −1 since the slope of f at x = 1 is −1. Similarly, F 00(2) = f 0(2) = −1, and F 00(3) does not exist because f has a sharp point at x = 3. In summary, to fill in the 3rd row of the table, we are calculating the slope of the graph of f at the relevant point.

10 Developed by Jerry Morris The Calculus Workout Sheet

1. A motorcyclist traveling along a straight east/west road begins travel at her home at t = 0. Below you are given a graph of her velocity, v, as a function of time, where positive velocities indicate easterly travel, and negative velocities indicate westerly travel. v (ft/sec)

f(t) 80 60 40 20 t(sec) 10 20 30 40 50 60 70 80 90 −20 −40

(a) What are the location, velocity, and acceleration of the rider after 20 seconds?

Let s(t) represent the position of the motorcyclist, in feet, relative to her home, where positive values of s(t) represent easterly displacements and negative values of s(t) represent westerly displacements. Using the Total Change Principle, we have

Z 20 s(20) = s(0) + f(t) dt = 0 + 600 = 600, 0 so the rider’s position is 600 feet east of her home after 20 seconds. Since f(20) = 40, we conclude that her velocity at this same time is 40 feet per second toward the east. Finally, since f 0(20) = 0, we conclude that her acceleration after 20 seconds is 0 ft/s2.

(b) Estimate the time(s) when her speed is the greatest, the time(s) when her acceleration is the greatest, and the time(s) when her distance from home is the greatest.

The rider’s speed will be the greatest when the given velocity graph is the farthest away from the t-axis; that is, her maximum speed of 80 feet per second is reached when t = 90 seconds. On the other hand, since f 0 gives the acceleration of the rider, her acceleration will be the greatest when the graph of f is the steepest; that is, her maximum acceleration appears to occur at about t = 82.5 seconds. Finally, estimating the rider’s distance from home amounts to analyzing various areas between the graph of f and the t-axis. We can see that during the first 35 seconds, the rider drives east to a maximum distance of Z 35 f(t) dt = 1000 feet. 0 Between t = 35 and t = 75, she travels west a distance of 1200 feet, putting her 200 feet west of her home. Since her remaining travel is east, but the area under f is small enough to indicate that she never reaches 1000 feet east of her home again, we conclude that her distance from home is the greatest after t = 35 seconds.

(c) At what time does she pass her house going west? R 35 R 65 By part (b), 0 f(t) dt = 1000, and a similar calculation reveals that 35 f(t) dt = −1000. Since the first integral indicates that the motorcyclist ended up 1000 feet east of her starting position, and the second integral indicates that she ended up 1000 feet west of her position at t = 35 seconds, we conclude that she has returned to her house at t = 65 seconds. Also, since f(65) < 0, we see that the rider is traveling west at this same time, which confirms that she passes her house traveling west at t = 65 seconds.

Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2017 11 2. Between noon and 6 p.m., the electrical power in use by a family, in kilowatts, is given by the function P = 0.6t2 − 0.1t3 + 0.2, where t is the number of hours after 12:00 noon.

To help us answer the questions that follow, note that P has units of kilowatts, dP/dt has R b units of kilowatts per hour, and a P dt has units of kilowatt hours.

(a) How fast is the family’s power usage increasing, in kilowatts per hour, at 2:30 p.m.?

First, note that 2:30 p.m. corresponds to t = 2.5 hours according to our model. The rate at which the family’s power usage is changing is given by d 2 3 2 dP/dt = dt (0.6t − 0.1t + 0.2) = 1.2t − 0.3t , so

dP 2 = 1.2(2.5) − 0.3(2.5) = 1.125 kilowatts per hour. dt t=2.5 Therefore, we conclude that the family’s power usage is increasing at a rate of 1.125 kilowatts per hour at 2:30 p.m.

(b) What is the family’s power usage, in kilowatts, at 2:30 p.m.?

When t = 2.5, we have P = 0.6(2.5)2 − 0.1(2.5)3 + 0.2 = 2.3875, so we conclude that the family’s power usage is 2.3875 kilowatts at 2:30 p.m.

We have

Z 6 Z 6 3 4 6 2 3 t t P dt = (0.6t − 0.1t + 0.2) dt = 0.6 − 0.1 + 0.2t 0 0 3 4 0 = (43.2 − 32.4 + 1.2) − 0 = 12,

so the family consumes 12 kilowatt hours of energy between noon and 6:00 p.m.

3. The volume of water remaining in a leaking tank is given by the function V = 50e−0.1t, where V is measured in gallons and t in minutes.

(a) How fast is the tank leaking after 10 minutes? Include appropriate units.

The rate at which the tank leaks has units of gallons per minute, so we are interested in dV/dt, the rate of change of volume with respect to time. Since dV/dt = 50e−0.1t · (−0.1) = −5e−0.1t, we have

dV −0.1(10) 5 = −5e = − , dt t=10 e so we conclude that, at t = 10 minutes, water is leaking out of the tank at a rate of 5/e, or about 1.84 gallons per minute.

(b) How much water remains in the tank after 10 minutes? Include appropriate units.

When t = 10, we have V = 50e−0.1(10) = 50/e, so after 10 minutes, there are 50/e, or about 18.4 gallons of water in the tank.

4. The rate at which water is leaking from an initially full 100-gallon tank is given by the function r = −7e−0.1t, where r is measured in gallons per minute and t is measured in minutes. (The negative sign indicates a decrease as opposed to an increase in volume.)

12 Developed by Jerry Morris (a) How fast is the tank leaking after 10 minutes? Include appropriate units. When t = 10, we have r = −7e−0.1(10) = −7/e, so after 10 minutes, the tank is leaking at a rate of 7/e, or about 2.58 gallons per minute. (b) How much water remains in the tank after 10 minutes? Include appropriate units. Let V (t) represent the number of gallons of water in the tank after t minutes, and note that r measures the rate of change of V with respect to time; that is, V 0(t) = r. Therefore, by the Total Change Principle, we have Z 10 Z 10 V (10) = V (0) + V 0(t) dt = 100 + (−7e−0.1t) dt 0 0 10 = 100 + 70e−0.1t 0 70 = 30 + , e so we conclude that there are 30 + (70/e), or about 55.8 gallons of water in the tank after 10 minutes.

5. Let f be a diﬀerentiable function on the interval 0 ≤ x ≤ 4, and suppose that the derivative of f is also continuous. Suppose that the table below is the only information about f that we have.

x 0 0.5 1 1.5 2 2.5 3 3.5 4 f(x) 1.1 2.1 2.9 3.6 4.2 4.6 4.9 5.1 5.2

Estimate each of the following as accurately as you can based on the data given. (a) f 0(1) (b) f 0(3) 0 0 Graphically, f (1) and f (3) represent the slopes of the f(x) tangent lines to the graph of f at x = 1 and x = 3. To estimate y 0 f (1), we use the slope of the secant line L1 that goes through L the points (0.5, 2.1) and (1.5, 3.6) (see diagram to the right). 5 1 We have 3.6 − 2.1 4 L2 Slope of L1 = = 1.5, 1.5 − 0.5 3 so f 0(1) is approximately equal to 1.5. Similarly, to 0 2 approximate f (3), we use the slope of the secant line L2 that goes through (2.5, 4.6) and (3.5, 5.1); we have 1

5.1 − 4.6 x Slope of L = = 0.5, 1 2 3 4 2 3.5 − 2.5 so f 0(3) is approximately equal to 0.5. Z 4 (c) f(x) dx 0 Geometrically, this integral is the area under y y the graph of f from x = 0 to x = 4. We can get 5 5 one estimate of this integral by adding up the 4 4 areas of the leftsum rectangles in the diagram to the right, and we can get another estimate by 3 3 adding up the areas of the rightsum rectangles 2 2 in the diagram to the far right. We have 1 1 x x 1 2 3 4 1 2 3 4

L4 = 0.5(1.1 + 2.1 + 2.9 + 3.6 + 4.2 + 4.6 + 4.9 + 5.1) = 14.25

R4 = 0.5(2.1 + 2.9 + 3.6 + 4.2 + 4.6 + 4.9 + 5.1 + 5.2) = 16.3

Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2017 13 Therefore, we take the average of these two calculations to get our best estimate of R 4 0 f(x) dx, giving us a final answer of (14.25 + 16.3)/2 = 15.275. Z 3 (d) f 0(x) dx 1 Since f is an antiderivative of f 0, we have Z 3 f 0(x) dx = f(3) − f(1) = 4.9 − 2.9 = 2. 1 Note that this is an exact answer.

6. Let f be the function whose graph is given below, and deﬁne a new function F by the equation Z x F (x) = f(t) dt. 0 f(x) y

x 4 812 16 20

−4

Given below are several lists of numbers. Rank each list in order from smallest to largest.

(a)0 , f 0(3), f 0(4), f 0(9), f 0(14), f 0(15) Since we are given the graph of f, the above numbers represent slopes of tangent lines to the provided graph. Arranging these slopes in order from most negative to most positive gives f 0(14) < f 0(15) < 0 < f 0(9) < f 0(3) < f 0(4).

(b)0 ,F (3),F (4),F (6),F (13),F (14) First, note that since F is defined as an integral, F 0 = f, so we know that F is decreasing on 0 ≤ x ≤ 4 because f is below the x-axis on this entire interval. R 0 Therefore, since F (0) = 0 f(t) dt = 0, we have F (4) < F (3) < 0.

R 6 Next, note that, by comparing areas, we have 3 f(t) dt < 0; therefore, the change in F R 6 between 3 and 6 is negative, meaning that F (6) < F (3). Similarly, 4 f(t) dt > 0, so F (6) > F (4). Combining all of these facts gives

F (4) < F (6) < F (3) < 0.

R 13 Again, by comparing areas, 0 f(t) dt > 0, so the change in F between 0 and 13 is positive, meaning that F (13) > F (0) = 0. Therefore,

F (4) < F (6) < F (3) < 0 < F (13).

14 Developed by Jerry Morris Finally, note that F (14) > F (13) because f is above the x-axis between x = 13 and x = 14, meaning that F is increasing on this interval. Therefore, our final answer is

F (4) < F (6) < F (3) < 0 < F (13) < F (14).

Since F 0 = f, the above numbers are just the y-coordinates of points on the graph we were given. Therefore, we have

F 0(0) < F 0(15) < F 0(4) < F 0(18) < F 0(23) < 0 < F 0(7) < F 0(10).

7. Find the slope of the tangent line to each of the following functions at x = 2. Give the exact slope if possible; otherwise, round to the nearest 0.01. (a) f(x) = sin(cos(x))

By the Chain Rule, we have d f 0(x) = cos(cos x) (cos x) = − cos(cos x) · sin x, dx so the slope of the tangent line to f at x = 2 is given by f 0(2) = − cos(cos 2) · sin 2.

x Z 2 (b) F (x) = e−t dt 0 By the Second Fundamental Theorem of Calculus, we have

x d Z 2 2 F 0(x) = e−t dt = e−x , dx 0

2 so the slope of the tangent line to F at x = 2 is given by F 0(2) = e−2 = 1/e4.

Using the limit definition of the derivative, we have

f(2 + h) − f(2) (2 + h)2+h − 4 f 0(2) = lim = lim . h→0 h h→0 h To approximate the value of this limit, we make a table below. h 0.1 0.01 0.001 0.0001 0.00001 (2 + h)2+h − 4 7.496 6.840 6.779 6.773 6.773 h Therefore, to 2 decimal places of accuracy, the slope of the tangent line to f at x = 2 is f 0(2) = 6.77.

Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2017 15