Chapter 3: Q. 19 (page 260)
Find the possibility graph of its derivative f'.

Short Answer
The possibility graph of its derivative f' is

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Chapter 3: Q. 19 (page 260)
Find the possibility graph of its derivative f'.

The possibility graph of its derivative f' is

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Prove that the lateral surface area of a right circular cone
is equal to πrl, where r is the radius of the cone and
l is the length of the diagonal of the cone, that is, the
distance from the vertex of the cone to a point on its
circumference.
For Exercises 15–20, sketch the graph of a function f that has the indicated characteristics. If a graph is not possible, explain why.
f negative, f' positive, and f'' positive on
Use optimization techniques to answer the questions in Exercises 25–30.
Find the area of the largest rectangle that fits inside a circle of radius .
Q. True/False: Determine whether each of the statements that follow is true or false. If a statement is true, explain why. If a statement is false, provide a counterexample.
(a) True or False: Every local maximum is a global maximum.
(b) True or False: Every global minimum is a local minimum.
(c) True or False: If f has a global maximum at x = 2 on the interval , then the global maximum of fon the interval [0, 4] must also be at x = 2.
(d) True or False: Iff has a global maximum at x = 2 on the interval [0, 4], then the global maximum of f on the interval must also be at x = 2.
(e) True or False: If f is continuous on an intervalI, then f has both a global maximum and a global minimum on I.
(f) True or False: Suppose f has two local minima on the interval [0, 10], one at x = 2 with a value of 4 and one at x = 7 with a value of 1. Then the global minimum of fon [0, 10] must be at x = 7.
(g) True or False: If f has no local maxima on , then it will have no global maximum on the interval [0, 5].
(h) True or False: Iff'(3) =0, then f has either a local minimum or a local maximum at x = 3.
Sketch the graph of a function f with the following properties:
f is continuous and defined on R;
f(0) = 5;
f(−2) = −3 and f '(−2) = 0;
f '(1) does not exist;
f' is positive only on (−2, 1).
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