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!


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 $c_1=c_2=c$.

  • 1
    $\begingroup$ +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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.