which of the following statements are true regarding differentiability.

Question

which of the following statements are true?
(a)If $f$ is twice continuously differentiable in $(a,b)$ and if for all $x\in(a,b)$ , $$f''(x)+2f'(x)+3f(x)=0$$, then f is infinitely differentiable in($a,b$).
(b) Let $f\in C[a,b]$ be differentiable in $(a,b)$. If $f(a)=f(b)=0$, then, for any real number $\alpha$ , there exist $x\in(a,b)$ such that $f'(x)+\alpha f(x)=0$.

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I am totally clueless.seek your help.

Answer 1

for (a):

If $$f''(x)+2f'(x)+3f(x)=0$$ then $$f''(x)=-2f'(x)-3f(x)$$

Now, do you see why $f''$ is differentiable in $(a,b)$? Can you continue this process for higher order derivatives?

hint: Induction.

Now just an exercise for fun :) - if we rephrase question (b) to

(b*) Let $f\in C[a,b]$ be differentiable in $(a,b)$. If $f(a)=f(b)=0$, there exist $x\in(a,b)$ such that $f'(x)+f(x)=0$.

It turns out to be true (just did this last week in class).

hint: Look at the function $e^x\cdot f(x)$

Answer 2

For (a), $f''=-2f'-3f$, and hence $f''$ is differentiable, and thus $f\in C^3$, and recursively, if $f\in C^{k+2}$, then $-2f'-3f\in C^{k+1}$, and thus $f''\in C^{k+1}$, which means that $f\in C^{k+3}$.

For (b), $$ g(x)=\mathrm{e}^{\alpha x}f(x). $$ Then $g(a)=g(b)=0$, and hence there exists an $x_0\in(a,b)$, such that $0=g'(x_0)=\mathrm{e}^{\alpha x}\big(g'(x_0)+\alpha g(x_0)\big)$.

Answer 3

(a) It follows by induction that the $n$th derivative exists and is a linear combination of $f$ and $f'$. Indeed this is true for $f'=1\cdot f'+0\cdot f$ and if $f^{(n-1)}=af'+bf$ then $f^{(n)}=af''+bf = -2af'+(b-3a)f$

(b) Consider $g(x) = f(x)e^{\alpha x}$. Tnhen $g'(x)= (f'(x)+\alpha f(x))e^{\alpha x}$. So the question is whether $g'(x)=0$ for some $x\in (a,b)$. Rolle.