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The question

If $f:[0,4] \rightarrow \mathbb{R}$ is differentiable, then prove that $$ [f(4)]^2-[f(0)]^2=8f'(a)f(b) \text{ for } a,b \in (0,4) $$

My Solution

Let's choose $b$ such that $$ f(b)=\frac{f(4)+f(0)}{2} \tag{1} $$

Since $f$ is differentiable in it domain, $$ \exists \text{ } c \in (0,4) \text{ | } f'(c)=\frac{f(4)-f(0)}{4-0} \tag{2} $$

If we consider $a=c$ and multiply (1) and (2), we get $$ f'(a)f(b)=\frac{[f(4)]^2-[f(0)]^2}{8} $$

and we're done.

My doubt

This solution has $b$ locked to one particular value and $a$ could assume a couple of values. I'm not entirely sure if the question is asking us to prove all values of $a$ and $b$ in $(0,4)$ satisfies the given assertion. I'm pretty sure that all values in $(0,4)$ would not satisfy the given assertion.

Any comments is highly appreciated.

Abhishek A Udupa
  • 1,156
  • $c$ indeed exists because $f$ is differentiable in $(0,4)$ by the virtue of Mean Value Theorem. I'm using that to my advantage by choosing $a$ to be equal to $c$ and not the other way round. How about that? – Abhishek A Udupa Sep 05 '21 at 07:19
  • The question sim badly worded. It should say for some $a, b$ or for all $a,b$. – Kavi Rama Murthy Sep 05 '21 at 07:20
  • @mathcounterexamples.net: But $c$ satisfies the conditions placed on $a$, so it is a valid choice for $a$. A more elegant way would have been to just start out with calling the number from the MVT $a$ instead of $c$. – Vercassivelaunos Sep 05 '21 at 07:21
  • @Vercassivelaunos Yep. That'd have been more neat. – Abhishek A Udupa Sep 05 '21 at 07:22

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