# Help proving existence of an specific $x_0$ for which $f''(x_0)=0$

I have a function $f: [0,1]\rightarrow \mathbb{R}$ that is twice differentiable. For this function is stated that there is a line segment with endpoints $A=(0,f(0))$ and $B=(1,f(1))$ that cuts function $f$ at $(a,f(a))$ where $0<a<1$. I have to prove that there exists such $x_0$ that $f''(x_0)=0$.

Should I use some theorem for solving this? Like Rolle or Lagrange?

Somehow it seems like I should formulate a function between these two points or something, but I don't know whether it's linear or not.

Any help would be appreciated. Thank you in advance.

• By mean value theorem, there exists points $c\in(0,a), d\in(a,1)$ such that $f'(c) = f'(d) = \frac {f(a) - f(0)}{a} = \frac {f(1) - f(a)}{1-a} = f(1) - f(0)$ And by Rolle's there exists a point $e\in(c,d)$ such that $f''(e) = 0$ – Doug M Jun 19 '17 at 16:02

Hint. For $t\in [0,1]$ consider the auxiliary function $$F(t):=f(t)-(1-t)f(0)-tf(1).$$ Then $F(0)=F(a)=F(1)$. Now use Rolle's theorem two times: one with $F$ and the other with $F'$.

• How did come to auxiliary function F(t)? i mean, i see that it will work, but i wonder how and why is it exactly this one? – MathIsTheWayOfLife Jun 19 '17 at 16:02
• @MathIsTheWayOfLife It is the difference between $f$ and the line segment with extreme points $A$ and $B$ (similarly to the proof of Lagrange's Theorem). – Robert Z Jun 19 '17 at 16:05
• Oh now I see, thank you. – MathIsTheWayOfLife Jun 19 '17 at 16:12

Apply the mean value theorem on $[0,a]$ and $[a,1]$. This gives points $b,c$ with $0<b<a<c<1$ and $$f'(b) = \frac{f(a)-f(0)}{a-0} = \frac{f(1)-f(a)}{1-a} = f'(c),$$ because the gradients of the line segments are equal. Now use Rolle on $f'$ on $[b,c]$.

Apply Rolle's Theorem for the function $$g(x)=f(x)-x$$

For the points $(0,f(0))$ and $(a,f(a)$ to get that $$\exists c_1 , c_1\in (0,a) \, g'(c_1)=0$$

Similarly, apply Rolle Theorem for the points $(a,f(a))$ and $(1,f(1)$ to get that $$\exists c_1 , c_2 \in (a,1) \,g'(c_2)=0$$

Now apply Rolle's Theorem to $$h(x)=g'(x)$$

For the points $(c_1,g'(c_1))$ and $(c_2,g'(c_3))$ To get $$\exists x_0 \in (c_1,c_2) \, h'(x_0)=0 \implies f''(x_0)=0$$