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I tried the Question as so:

$$f^2(4)-f^2(0)=[f(4)+f(0)]\cdot[f(4)-f(0)]$$

Now $$\frac{f(4)-f(0)}{4-0}= f'(c) ~, ~~c \in (0,4)$$ $$f(4)-f(0)=4f'(c)$$

And

$$\frac{f(4)+f(0)}{2}= f(b) ~~, ~b \in ~ (0,4) $$

$$\therefore f^2(4)-f^2(0) = 4f'(c)\cdot2f(b) $$ $$= 8f'(c)f(b)$$

But the answer in the book says that b and c are the same points and the answer is given as $8f'(a)f(a)$

so basically I need to prove that c and b are the same points.

BTW -> what does this mean $f : [0,4] \rightarrow~~\mathbb{R}$

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It's kind of hard to understand what you're talking about if you don't give any details! –  Mercy Jul 4 '12 at 13:21

2 Answers 2

up vote 3 down vote accepted

I guess the problem asks you to use the MVT to find $f^2(4)-f^2(0)$.

Let $g(x)= (f(x))^2$.

By the MVT there exists a $a \in (0,4)$ so that

$$\frac{g(4)-g(0)}{4-0}=g'(a) \,.$$

Now, use that $g'(a)=2f'(a)f(a)$ and you are done...

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I guessed you meant let $f: [0,4] \rightarrow \mathbb{R}$ prove there is a $a$ such that $8 f'(a)f(a)= f^2 (4) - f^2(0)$. This is direct if you consider $ g(x) = f^2(x)$ because $g'(x)= 2 f(x)f'(x)$ so $$\frac{g(4)-g(0)}{4-0}= g'(a)$$ and get what you wanted. $f: [0,4] \rightarrow \mathbb{R}$ this means that the domain of $f$ is $[0,4]$ and it takes values in $\mathbb{R}$

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