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.
 A: If $x_1$ or $x_2$ is a multiple root then we can take $\xi = x_1$ or $\xi = x_2$, respectively. Otherwise, if $f'(x_1) < 0$ then $f'(x_2) > 0$, and vice versa. Try drawing it to understand why. Now we can use the intermediate value theorem for derivatives.
A: 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 $f(\xi)+f^{'}(\xi)=0$
A: I suppose you want to say that $f$ is defined on $\left[ x_{1},x_{2}\right] $
(at least) such that $f(x_{1})=f(x_{2})=0$ and $f(x)>0$ on the open $%
(x_{1},x_{2}).$ Moreover, $f$ is continously differentiable over (at least) $%
\left[ x_{1},x_{2}\right] .$ Show that there exists $\xi \in \left[
x_{1},x_{2}\right] $ such that $f(\xi )+f^{\prime }(\xi )=0.$
It suffices to show that $f^{\prime }(x_{1})\geq 0$ and $f^{\prime
}(x_{2})\leq 0.$ Then the function $g(x)=f(x)+f^{\prime }(x)$ is continous
on $\left[ x_{1},x_{2}\right] $ and $g(x_{1})=0+f^{\prime }(x_{1})\geq 0$
and $g(x_{2})=0+f^{\prime }(x_{2})\leq 0.$ Then by the intermediate value
theorem, there exists $\xi \in \left[ x_{1},x_{2}\right] $ such that $g(\xi
)=f(\xi )+f^{\prime }(\xi )=0.$
Now let us give a proof for $f^{\prime }(x_{1})\geq 0.$ Since $%
f(x)>0=f(x_{1})$ for $x>x_{1}$ then $f(x)-f(x_{1})>0$ and $x-x_{1}>0$ for
all $x>x_{1},$ then $\frac{f(x)-f(x_{1})}{x-x_{1}}>0$ for all $x>x_{1}$ and
by passing to the limit as $x$ tends to $x_{1}^{+}$ it follows that  $%
f^{\prime }(x_{1})=\lim_{x\rightarrow x_{1}^{+}}\frac{f(x)-f(x_{1})}{x-x_{1}}%
\geq 0.$ Same proof for $f^{\prime }(x_{2})\leq 0.$
A: 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 $f(x_3)>0.$
If $f'(x_2)=0,$ that $x_2$ is required point. Let $r_0=x_2.$ If $f'(x_2)>0,$ then we have $f(x_2-\varepsilon)<0$ for sufficiently small epsilon and a root $r_1\in(x_3, x_2-\varepsilon)$ etc. If $f'(r_k)<0,$ then let $x_r=r_k,$ wherein $x_r\leq x_2$
$f(x)+f'(x)$ is continuos in $(x_3,x_r),$ wherein 
$$f(x_3)+f'(x_3)>0,\quad f(x_r)+f'(x_r)<0.$$
So there is a point $X\in(x_3, x_2)$ with $f(X)+f'(X)=0.$
Case $f(x_3)<0$ is proved.
Case $f(x_3)<0$ can be proved similarly.
