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Calculus Level 4

Let $f(x)$ be a double differentiable function such that $\left| f''(x) \right | \le 5$ , $\forall x \in \left[ 0,4 \right]$ .

And $f$ takes its largest value at an interior point of this interval.

Find the maximum possible value of $\large\ \left| f'(0) \right| + \left| f'(4) \right|$ .

The answer is 20.

1 solution

Mark Hennings
Aug 15, 2018

If $f$ achieves its maximum at $0 < c < 4$ , then $f'(c) = 0$ , and the MVT tells us that $|f'(0)| \; = \; |f'(0) - f'(c)| \;\le \; 5c \hspace{2cm} |f'(4)| \; = \; |f'(4)-f'(c)| \; \le \; 5(4-c)$ so that $|f'(0)| + |f'(4)| \; \le \; 20$ The particular function $f(x) = -\tfrac52(x-2)^2$ shows that the maximum value of $\boxed{20}$ can be achieved.

@Mark Hennings , In the first inequality why you have maximised it to (5c)? Why not (c) only?

How 5 came in your term?

Priyanshu Mishra - 2 years, 9 months ago

The MVT says that $f'(c) - f'(0) \; = \; cf''(d)$ for some $0 < d < c$ . Thus $|f'(c) - f'(0)| \le 5c$ , since $|f''(x) \le 5$ for all $x$ .

Mark Hennings - 2 years, 9 months ago

Derivable on all Tracks!

The answer is 20.

1 solution

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