Prove if $f(x)$ has 3 distinct solutions on $[a,b]$ and $f(x)$ has a continuous second derivative on $[a,b]$, then $f''(x)+2f(x)$ has solutions on $[a,b]$.
I'm stuck for ideas on this problem. Where can I start? Is it possible to generalize with random c (instead of 2)?
We solve for the general case $f''(x) +c\cdot f(x)$.
(for example, $c = 2$ as in the question)
In this case, the question is wrong, that is: $f''(x)+2f(x)$ may have no solution. Here is a counter-example:
Let us define $$f(x) := \sin(\sqrt{c} \cdot x) -1$$ The function $f(x)$ has $3$ solutions $x \in \left\{ \frac{\pi}{2\sqrt{c}}, \frac{5\pi}{2\sqrt{c}}, \frac{9\pi}{2\sqrt{c}} \right\}$ over the interval $\left[\frac{\pi}{2\sqrt{c}},\frac{9\pi}{2\sqrt{c}} \right]$.
However, $$f''(x) + c\cdot f(x) = \underbrace{-c\sin(\sqrt{c} \cdot x)+c\sin(\sqrt{c} \cdot x)}_{=0}- c \ne 0 \hspace{1cm} \forall x \in \mathbb{R}$$
(for example, $c = -2$)
In this case, the question is correct. Here is the proof:
Denote $g(x):=e^{ \sqrt{|c|}\cdot x}\cdot f(x)$ .
The function $g(x)$ has $3$ distinct solutions in $[a,b]$, then $g'(x)$ has $2$ distinct solutions in $[a,b]$, we have $$g'(x) =e^{\sqrt{|c|}\cdot x}\cdot \left(\sqrt{|c|}\cdot f(x)+f'(x) \right) $$
The function $\left(\sqrt{|c|}\cdot f(x)+f'(x)\right)$ must have $2$ distinct solutions in $[a,b]$.
Denote $h(x) = e^{\color{red}{-}\sqrt{|c|}\cdot x}\cdot \left(\sqrt{c}\cdot f(x)+f'(x)\right)$, its derivative $h'(x)$ must have $1$ solution in $[a,b]$:
$$h'(x) =e^{- \sqrt{|c|}\cdot x}\cdot \left( -|c|\cdot f(x)+f''(x) \right)=e^{- \sqrt{|c|}\cdot x}\cdot \left(\color{red}{ c\cdot f(x)+f''(x)} \right) $$
The function $\left(c\cdot f(x)+f''(x)\right)$ must have a solution in $[a,b]$.
