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

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.

• The problem statement is not clear. Please check it. Commented Nov 25, 2014 at 2:09
• Are you trying to say that $f'(x)$ exists for all $x$ and $f(x)$ has tw roots $x_1$ and $x_2$? Commented Nov 25, 2014 at 2:10

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

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.

• However, $f'(x)$ may be not continuous.
– Paul
Commented Nov 25, 2014 at 6:39
• Darboux Theorem ensures that a derivative that changes sign attains a zero, irrespective of whether or not it is continuous. Commented Nov 25, 2014 at 6:45

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 $f^{\prime }(x_{2})\leq 0.$

• Is it legal to make so many assumptions? :D Commented Nov 25, 2014 at 2:46
• I agree with you, but the problem statment is not clear, so i have tried to do my best! If he precise the set of assumptions I will arrange my proof accordingly! ''continously diff'' is too much, I beleive that it can be relaxed. BUt don't forget, he say no integration theory! That is we cannot use the Taylor thm with integral reminder for example! Commented Nov 25, 2014 at 2:52
• Second answer with no additional assumption. Put $h(x)=e^{x}f(x),$ then $h$ is as $f$ has a derivative over the interval $% \left[ x_{1},x_{2}\right] .$ Moreover, $h(x_{1})=e^{x_{1}}f(x_{1})=0,$ and $h(x_{2})=0.$ By Rolle's theorem, there exists $\xi \in \left( x_{1},x_{2}\right)$ such that $h^{\prime }(\xi )=0.$ However, $h^{\prime }(\xi )=e^{\xi }(f(\xi )+f^{\prime }(\xi )).$ So, $f(\xi )+f^{\prime }(\xi )=0$ since $e^{\xi }>0.$ Commented Nov 25, 2014 at 5:24

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.

• "More strict" than what?
– Did
Commented Aug 20, 2016 at 20:49
• @Did Which is founded at definitions, dear downvoter. Commented Aug 21, 2016 at 1:44