Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 4: Applications of Derivatives - Section 4.2 - The Mean Value Theorem - Exercises 4.2 - Page 198: 17

Answer

Proof given above.

Work Step by Step

$f'' \gt 0 \Rightarrow f'$ is differentiable on $(a,b) $ (and continuous on $[a,b].$) Assume that the opposite of the statement is true (that $f'$ has more than one zero in $[a,b]$. ) Let $c_{1},c_{2}\in[a,b]$ be two such zeros of $f'.$ Then, by Rolle's theorem (which applies for $f'$ on $[c_{1},c_{2}]\subset[a,b]$), since $f'(c_{1})=f'(c_{2})=0$, there must be a $c_{3}\in(c_{1},c_{2})$ such that $f''(c_{3})=0.$ This is a contradiction with ($f''$(x) $ \gt 0$ for all $x\in[a,b]$ ), because $c_{3}\in[a,b].$ Our assumption is wrong, which means that the statement is true: $f^\prime$ has at most one zero on the interval. --- The same logic applies for the case when $f'' \ \lt \ 0$ on $[a,b].$ (Make the same assumption, arrive at the same contradiction.)
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.