Discussion Section #9 – Selected Answers1. a. b. c. d.

f ' > 0 and f '' > 0 f ' > 0 and f '' < 0 f ' < 0 and f '' > 0 f ' < 0 and f '' < 0

2. Many correct solutions, basic shape should resemble:

3. a. f '(c) = 0 or f '(c) does not existb. If f '(c) = 0 , we know the function has a horizontal tangent at x = c . Then we can check

the sign of f ' to the left and right of x = c to determine whether the graph has a localmax/min or inflection point at x = c . If f '(c) does not exist, we know f is discontinuous,has a sharp corner, or a vertical tangent at x = c .

4. a. Set the function equal to zero and solve for x.b. Evaluate the function at x = 0 .

5. a. To find discontinuities of a fraction, set the denominator equal to zero.To find discontinuities of a piecewise function, check the endpoints of each “piece”.A function containing a root is continuous for all points in its domain where the innerfunction is continuous.

b. If the one-sided limits of a function at its discontinuity are infinite, then the discontinuitycorresponds to a vertical asymptote. If the two-sided limit of a function at its discontinuityexists and is finite, then the discontinuity corresponds to a hole. (In the case of a rationalfunction, a factor that cancels from the numerator and denominator indicates a hole. Afactor that doesn’t cancel from the denominator indicates a vertical asymptote.)

6. The domain of a rational function includes all points where the denominator is non-zero.The domain of an even root includes all points where the inner function is non-negative.The domain of a log includes all points where the inner function is positive.

7. a. Evaluate the limits of the function as x→∞ and x→ −∞ .b. If the limit of the function as x→ ±∞ exists and is finite, then the function has a HA.c. If the end of a graph doesn’t go to a HA, then it may go to ±∞ or it may oscillate.

8. x-intercepts: x = ±1y-intercepts: y = 1

4

vertical asymptotes: x = ±2horizontal asymptotes: y = 1local max value: 14 (occurs at x = 0 )local min: noneinflection points: none

2

-2

2

-2

2

- 2

f x( ) = -e-x

2

-2

4

2

-2

- 5 5

9. a. x-intercepts: none y-intercepts: (0, 1)absolute minimum value: 1, occurs at x = 1 local/absolute max: noneinflection points: π

2 .π2( ) and 3π

2 .3π2( )

b. x-intercepts: − 6,0( ) , 0,0( ) , and 6,0( ) y-intercepts: 0,0( )local minimum value: -9, occurs at x = − 3 and x = 3absolute maximum value: 0, occurs at x = 0inflection points: −1,−5( ) and 1,−5( )

c. x-intercepts: none y-intercepts: 0, 1e( )local minimum: none absolute maximum value: 1, occurs at x = 1inflection points: 1− 1

2 ,1e( ) and 1+ 1

2 ,1e( )

10. a. x = 0 and x = 2 b. increasing on (0,2) , decreasing on (−∞,0) and (2,∞)c. concave up on −∞,2 − 2( ) and 2 + 2,∞( )

concave down on 2 − 2,2 + 2( )d. absolute minimum value of 0 occurs at x = 0

local maximum value of 4e2 occurs at x = 211. Determine whether the following statements are true. Give an explanation or counterexample.

a. False. f is increasing on (0,∞) . Never reaches a maximum.b. True. f is continuous on the closed interval [0, 1]. By the Extreme Value Theorem, f must

have an absolute max and an absolute min. (These occur at critical points or endpoints.)c. False. Local extrema must occur at interior points of an interval. Furthermore, if a

function is differentiable on an interval, then local extrema can only occur at critical points.f is differentiable, but f '(x) ≠ 0 on (0, 1). Therefore, no local extrema.

12. Determine whether the following statements are true. Give an explanation or counterexample.

a. False. Counterexample: f (x) =x, x < 02 − x, x ≥ 0

⎧⎨⎩

f has an absolute max value of 2 at x = 0 even though f is discontinuousb. False. Counterexample: f (x) = x − 2( )3 , f '(2) = 0 , but no max/min at x = 2c. True. Follows from the Extreme Value Theorem and the Local Extreme Point Theorem.d. False. We must also know that f (3) exists. (Critical points must be in the domain of f.)

13. Derivative at x = 1 does not exist because limh→0−

f (x + h) − f (x)h

= −∞ and limh→0+

f (x + h) − f (x)h

= 1 .

3

2

1

-2 2 4

4

2

- 2

- 5 5

f x( ) = 5

ex+1

14. Note that ddx f (x)[ ] = f '(x) and

ddx f (x) + c[ ] = f '(x) . So if we only know thederivative of a function, there are manypossibilities for the function itself. All ofwhich will be translations y a constant.f '(x) tells us the shape of f.

Specifying f (0) = 0 tells us the location of f.If we were told f (0) = 1 then the graph wouldbe shifted up one unit.

15. a. Many possible answers. One example: f (x) = x3 . f '(0) = 0 , but f '(x) ≥ 0 for all x.b. Many possible answers. One example: f (x) = x3 . f '(0) = 0 and f ''(0) = 0 .c. Many possible answers. One example: f (x) = x3 . f ''(0) = 0 and f '' changes from

negative to positive at x = 0 , so f changes from concave down to concave up.d. Many possible answers. One example: f (x) = x4 . f ''(0) = 0 but f ''(x) ≥ 0 for all x, so f

is always concave up.16. a. False. If f ' is constant on an interval, then f '' must be zero on the entire interval.

b. True. Example: f (x) = x2

c. True. Example: f (x) = x

d. True. Example: f (x) = x3

e. True. Example: f (x) = sin x (Changes concavity at x = 0 , but f '(0) = 1.)f. True. Anytime the critical point is a local extremum the concavity doesn’t change.

17. a. Many possible graphing windows, one good one is: [-10,25] by [-50,1800]b. Many possible graphing windows, one good one is: [-15,15] by [-10,10]c. Many possible graphing windows, one good one is: [0,30] by [-10,30]

18. a. Suppose f is continuous on an interval. Find the critical points of f within this interval.For each critical point c, determine the sign of f ' to the left of c and to the right of c.If the sign of f ' changes from positive to negative, then f has a local max at x = c .If the sign of f ' changes from negative to positive, then f has a local min at x = c .If f ' doesn’t change sign, then f does not have a local extremum at x = c .

b. Suppose f '' is continuous on an interval. Find the critical points where f '(c) = 0 withinthis interval. For each c, evaluate f ''(c) . If f ''(c) < 0 , then f has a local max at x = c . Iff ''(c) > 0 , then f has a local min at x = c . If f ''(c) = 0 , the test is inconclusive.

19.Horizontal Asymptotes: y = 0 and y = 5Inflection Point: 0, 52( )No local/absolute extrema.

20. absolute maximum value: 23 ≈ 1.155 occurs at x = 1

2 ≈ 0.707 and x = − 12 ≈ −0.707

local minimum value: 1 occurs at x = 0inflection points: (-0.295, 1.068), (0.295, 1.068), (-1.458, 0.986), and (1.458, 0.986)

21. Maximum monthly revenue of $81,000 occurs when rent is set at $900 (90 units are occupied).22. Boats are closest together at 2:21:36 PM.