1
$\begingroup$

This task is quite simple and I am ashamed that I can't realize how to prove it. I know that I must use the mean value thoerem but I have no idea right now. So if everyone can just help me with a clue I will be very grateful. And sorry for my bad English. Here we go.

A function $f$ is differentiable on the interval $[a,b]$ and $f(a) = f(b)$. I need to prove that there exists a point $c$ in $(a,b)$ such that $f(a) - f(c) = \frac{c}{2}*f'(c)$.

  • 0
    Are you sure this is the correct statement? It seems no such $c$ exists in the case of the function $f(x)=x$ on $[1,2]$. – 2017-01-29
  • 0
    But $f(1)=f(2)$ is not true in your example, Jonathan. – 2017-01-29

1 Answers 1

1

Let $g(x)=f(a) - f(x) -\dfrac{x}{2}f'(x)$ which is continuous on $(a,b)$. $$g(a)=f(a) - f(a) -\dfrac{a}{2}f'(a)=-\dfrac{a}{2}f'(a)$$ $$g(b)=f(a) - f(b) -\dfrac{b}{2}f'(b)=-\dfrac{b}{2}f'(b)$$

Case $f'(a) so $g(a)>g(b)$ by Rolle's Theorem, there exist a $c\in(a,b)$ such that $g(c)=0$.

Use this and complete the proof.

  • 0
    I think you'd need continuity of $\;f'(x)\;$ to use Rolle... – 2017-01-29
  • 0
    $f$ is differentiable on the interval so $f'$ exist in interval. – 2017-01-29
  • 0
    @My I know that, yet Rolle's Theorem *requires* continuity of the function under study. – 2017-01-29
  • 0
    @DonAntonio Sorry. I haven't any way, write right solution, please. – 2017-01-29
  • 0
    @MyGlasses It doesn't matter any more: the OP is happy. ;) – 2017-01-29
  • 0
    @DonAntonio Oh yes, I understood. Seems like the task is incopmtlete or has another solution. Anyway thanks everyone for your answers – 2017-01-29