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Let $f:[a,b] \rightarrow \mathbb R$ be a differentiable function. Show that $\exists c_1,c_2 \in (a,b)$ such that $2f(c_1)f'(c_1)=f'(c_2)[f(a)+f(b)]$

I have no idea how to do this sum. Please help!

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Use that $$ f(b)^2-f(a)^2 = 2\int_a^b f(x)f'(x)\, dx $$ and use the mean value theorem (s of integration and differentiation) in conjunction with the binomial theorems.


Or use the intermediate value theorem to find a $c\in(a,b)$ such that $2f(x)=f(a)+f(b)$ and multiply with $f'(c)$. Then $c_1=c_2=c$.

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    $\begingroup$ +1 The second approach is nice, but if $f(a)=f(b)$ it might not be possible to choose a $c\in(a,b)$. In that case, you can pick a single $c$ with $f'(c)=0$. The first approach it is a little weird to bring in integrals. You are really just saying that if $g(x)=(f(x))^2$ then $g'(x)=2f(x)f'(x)$. $\endgroup$ – Thomas Andrews Oct 11 '15 at 14:49
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Let $g(x)=(f(x))^2 $..now apply the MVT for the function $g$ ..we have $(g(b)-g(a))/(b-a)=g'(c) $ for some $c$ in $(a,b)$ which implies that $(f(b)^2 -f(a)^2)/(b-a) = 2.f(c).f'(c)$ ......$(i)$

Now apply the MVT for $f$ and you will get some $d$ in $(a,b)$ such that $(f(b)-f(a))/(b-a) = f'(d) $ so from $(i)$ we have the result

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