Exercise 8:[Hard] Let f be a function differentiable on (c-p, c+p) for some p > 0. Suppose f has a maximum at c, ie., f(c)2 f(x) for all x € (c - p,c+ p) Last modified: October 29, 2019, Due: November 6, 2019. 1 ΜΑΤΗ 15100, SECTION 13 2 (a) Prove that f'(c) < 0. f(c+h)-f(c) Hint: Consider the limit: lim0+ What can be said about the numerator? (b) Prove that f'(c) > 0. f(c+h)-f(c) Hint: Consider the limit: lim/0- (c) Conclude that f'(c) = 0. Remark: The same result is true for minima as well, with basically the same proof. Note this is not an if and only if statement: let g(x) = x3. Then g'(0) = 0, but g has neither a maximum nor minimum at x = 0

Question
Exercise 8:[Hard] Let f be a function differentiable on (c-p, c+p) for some p > 0. Suppose f has a maximum
at c, ie., f(c)2 f(x) for all x € (c - p,c+ p)
Last modified: October 29, 2019, Due: November 6, 2019.
1
ΜΑΤΗ 15100, SECTION 13
2
(a) Prove that f'(c) < 0.
f(c+h)-f(c)
Hint: Consider the limit: lim0+
What can be said about the numerator?
(b) Prove that f'(c) > 0.
f(c+h)-f(c)
Hint: Consider the limit: lim/0-
(c) Conclude that f'(c) = 0.
Remark: The same result is true for minima as well, with basically the same proof.
Note this is not an if and only if statement: let g(x) = x3. Then g'(0) = 0, but g has neither a
maximum nor minimum at x = 0

Image Transcription

Exercise 8:[Hard] Let f be a function differentiable on (c-p, c+p) for some p > 0. Suppose f has a maximum at c, ie., f(c)2 f(x) for all x € (c - p,c+ p) Last modified: October 29, 2019, Due: November 6, 2019. 1 ΜΑΤΗ 15100, SECTION 13 2 (a) Prove that f'(c) < 0. f(c+h)-f(c) Hint: Consider the limit: lim0+ What can be said about the numerator? (b) Prove that f'(c) > 0. f(c+h)-f(c) Hint: Consider the limit: lim/0- (c) Conclude that f'(c) = 0. Remark: The same result is true for minima as well, with basically the same proof. Note this is not an if and only if statement: let g(x) = x3. Then g'(0) = 0, but g has neither a maximum nor minimum at x = 0

Expert Answer

Want to see the step-by-step answer?

See Answer

Check out a sample Q&A here.

Want to see this answer and more?

Experts are waiting 24/7 to provide step-by-step solutions in as fast as 30 minutes!*

See Answer
*Response times vary by subject and question complexity. Median response time is 34 minutes and may be longer for new subjects.

Related Advanced Math Q&A

Find answers to questions asked by student like you
Show more Q&A