Prove that if f is diff. on $(-\infty,\infty)$ such that $f(0)=0$ and $f(1)=1$,then there is a $c\in(0,1)$ such that $f'(c)=2c$

Question

I have been solving last exam questions of calculus course and encountered with a problem which I couldn't solve completely.The question is following;

Determine whether the statement is true or false, and prove it;

Suppose that f is differentiable function on $(-\infty,\infty)$ such that $f(0)=0$ and >$f(1)=1$,then there is a $c\in(0,1)$ such that $f'(c)=2c$

My incomplete proof;

Since it is diff. and cont. on $(-\infty,\infty)$, by Mean Value Theorem;

$f'(r)=\frac{f(1)-f(0)}{1-0}=1$, where $r\in(0,1)$

Since for every f satisfying the conditions, there is $\exists r\in(0,1)$, and we want to prove that $\forall f$, there is $\exists c\in(0,1)$, so if I could establish a connection between r and c, I prove that the statement is true.

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Everything up to now seems quite correct, but after that I continued like that;

So, $f'(r)=f'(c)=1=2c$

$c=1/2$

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But it is obviously a wrong a answer since the functions $x,x^3,x^5..$ don't satisfy this property.In other words;

$\frac{d(x^3)}{dx}|_{x=c=0.5}=3(0.5)^2\neq1$

So what I am missing, how can I improve this proof and is there any alternative proof other than what I have done?

Edit:Guys, I'm mainly interested in what I am missing in my proof, the other questions are side questions.

Answer 1

Hint: Consider the function $g(x)=f(x)-x^2$...

Additional consideration: For a continuous function $h$, finding a point $c\in(a,b)$ on an interval such that $f'(c)=h(c)$ equates to saying that $$g(x)=\int_a^x f'(t)-h(t)\,dt$$ has a stationary point in $(a,b)$. This gives you a trick to backwards-engineer a good candidate function for Rolle (though it can still be uninformative, since you have no guarantee that $g(a)=0\stackrel?=f(b)-f(a)+\int_a^b h(t)\,dt=g(b)$ ).

Answer 2

Starting reverse engineering from the quotient $\frac{f'(c)}{2c}$ one could also get the idea to employ the extended mean value theorem with $g(x)=x^2$ leading to $$ \frac{f(b)-f(a)}{b^2-a^2}=\frac{f'(c)}{2c} $$ and using $a=0$, $b=1$ and some $c\in(a,b)$.