Proving that there exists $c$ that $f'(c)= \frac{f(c)}{c}$

Question

So i have a function $f$ that is continuous on $[1,2]$ and that is differentable on $(1,2)$. Also, let $f(2)=2f(1)$

Now I have to prove that for some $a \in (1,2)$ is $f'(a)=\frac{f(a)}{a}$.

Since the function works for Rolle's theorem I thought this should be used.

So I thought I would define function $g(x)=\frac{f(x)}{x}$.

And since it's continuity and differentiability is the same as $f$ I might use it for Rolle's theorem.

So I got $g(1)=f(1)$ and $g(2)=f(1)$ hence there exists $a\in (1,2)$ such that $g'(a)=0$.

But somehow that didn't help me.

Did use the wrong concept, or am I missing something?

Any help would be appreciated.

How did that not help you? What is $ g'(x) $ according to the quotient rule? — Bananach, Jun 19 '17 at 15:41
Well, $g'(x)=\frac {f'(x)-g(x)}x$ so $g'(a)=0\implies 0=f'(a)-g(a)$ . That looks like what you want. — lulu, Jun 19 '17 at 15:42

score 1 · Answer 1 · edited Jun 19 '17 at 15:44

1

You are on the right track! Note that $1<a<2$ and $$g'(a)=\frac{af'(a)-f(a)}{a^2}=0\implies f'(a)=\frac{f(a)}{a}$$

edited Jun 19 '17 at 15:44

Jaideep Khare

answered Jun 19 '17 at 15:42

Robert Z

score 1 · Answer 2 · answered Jun 19 '17 at 15:42

$\exists c \in (1,2)$ such that $g'(c)=0$

Now how didn't it help?

You simply need to put $$\left(\frac{f(c)}{c}\right)'=0 \tag1$$

$$\frac{cf'(c)-f(c)}{c^2}=0 \tag 2$$

$$\implies f'(c)=\frac{f'(c)}{c}$$

You will get $(2)$ by applying quotient rule to $(1)$

2 Answers2