Let $f$ be a continuous function in $[0,5]$ and twice differentiable function in $(0,5)$ such that $f(4)=f(5)=0$. Prove the following:
- There exists some $a$ in $[0,5]$ such that $nf(a)+af'(a)=0, n\in N$
- There exists distinct $a_1,a_2 \in [0,5]$ such that $4f'(a_1)+f'(a_2)+f(0)=0$
- There exists some $a \in [0,5]$ such that $nf(a)=af'(a), n \in N$
By Rolles theorem, there will be $a \in [4,5]$ such that $f'(a)=0$. But $f(a)$ is not necessarily zero at that point, so it doesn't lead us anywhere.
If we assume $f(0)=0,$ there will be $a_1 \in [0,4]$ and $a_2 \in [4,5]$ such that $f'(a_1)=f'(a_2)=0$. Now I don't know how to prove this if $f(0) \neq 0$
Since $n,a>0$ either $f(a),f'(a)>0$ or $f(a),f'(a)<0$. I haven't been able to find a single function which violates this rule in $[4,5]$ but I can't prove this rigorously. Please help, thanks :)
Let $$F(x)=a^nf(x)$$ then $F$ is twice differentiable just as $f$ is, and $F(4)=F(5)=0$. Apply Rolle's theorem on $F$ on $[4,5]$, and you get $a$ in $[4,5]$ such that $$na^{n-1}f(a)+a^nf^{'}(a)=0$$ which by the positivity of $a$ leads to the wanted expression.
To ensure $a_1$ and $a_2$ are different, it's natural to choose them respectively on $(0,4)$ and $(4,5)$. On $(4,5)$ one can apply Rolle's theorem to get $a_2$ such that $f^{'}(a_2)=0$. Look at the rest of wanted expression, and the coefficient of $f(a_1)$ implies using Lagrange's theorem:
$$f(4)-f(0)=4f^{'}(a1)$$
plus that $f(4)=0$ and we are done.
(Now, it seems to me that the twice differentiability is excessive. I would welcome anyone to tell me how/whether that is necessary.)