Suppose that $f ' (x)$ exists and $f(x)$ has two roots $x_1$ and $x_2$. Try to prove that: Suppose that  $f'(x)$ exists and $f(x)$ has two roots $x_1$ and $x_2$. Try to prove that:


  $\exists \xi \in (x_1,x_2)$ such that $f(\xi)+f'(\xi)=0$.


We cannot use the knowedge of integration.
Consider $\phi(x)=e^xf(x)$ which is  continuous and differentiable since $f$ is so.

Also $\phi ^{'}(x)=e^x(f(x)+f^{'}(x))$

Now $\phi(x_1)=0,\phi(x_2)=0$

Since between any two roots of $\phi(x)$ lies one root of $\phi^{'}(x)$,
 so there exists $\xi \in(x_1,x_2)$ such that $\phi^{'}(\xi)=0$ 

Hence $e^\xi(f(\xi)+f^{'}(\xi))=0$

since $e^\xi\neq 0 $  so $f(\xi)+f^{'}(\xi)=0$
Real-valued differentiable function $f(x)$ attains equal zero values at two distinct points $x_1$ and $x_2.$ According with Rolle's theorem, so exists point $x_3$ with $f'(x_3)=0.$ 

If $f(x_3)=0$ that $x_3$ is required point. Consider the case $f(x_3)&gt;0.$

If $f'(x_2)=0,$ that $x_2$ is required point. Let $r_0=x_2.$ If $f'(x_2)&gt;0,$ then we have $f(x_2-\varepsilon)&lt;0$ for sufficiently small epsilon and a root $r_1\in(x_3, x_2-\varepsilon)$ etc. If $f'(r_k)&lt;0,$ then let $x_r=r_k,$ wherein $x_r\leq x_2$

$f(x)+f'(x)$ is continuos in $(x_3,x_r),$ wherein 
$$f(x_3)+f'(x_3)&gt;0,\quad f(x_r)+f'(x_r)&lt;0.$$
So there is a point $X\in(x_3, x_2)$ with $f(X)+f'(X)=0.$

Case $f(x_3)&lt;0$ is proved.

Case $f(x_3)&lt;0$ can be proved similarly.