# Given $f$ continuous on $[-1, 1]$ and twice differentiable on $(-1, 1)$. Prove $f''(c) = 1$ for some $c \in (-1, 1)$.

Got this question. Struggled with it for some time and recieved the hint to fit a polynomial through the given points and then use Rolle's Theorem. I think I have the proof for this question now. The full question is below. Please let me know if it makes sense, or if perhaps there is another simpler way for my own knowledge!

Given $f$ continuous on $[-1, 1]$ and twice differentiable on $(-1, 1)$ and $f(-1) = 0, f(0) = 0, f(1) = 1$. Prove $f''(c) = 1$ for some $c \in (-1, 1)$.

First, let us fit a function $g(x) = ax^2 + bx+c$ so that

$g(0) = c = 0$

$g(-1) = a - b = 0, a=b$

$g(1) = a + b = 1, a = \frac{1}{2}$.

Then we get $g(x) = \frac{x^2}{2} + \frac{x}{2}$. Then let us construct another function $h(x) = f(x) - g(x)$. Note that since $f$ is continuous on $[-1, 1]$ and twice differentiable on $(-1, 1)$ and since $g$ is a polynomial then $h(x)$ is also continuous on $[-1, 1]$ and twice differentiable on $(-1, 1)$.

Also note that $h''(x) = f''(x) - 1$.

Therefore, since $h(-1) = h(0) = 0$ and $h(0) = h(1) = 0$ then by Rolle's Theorem there exists a $c_1 \in (-1, 0), c_2 \in (0, 1)$ so that $h'(c_1) = h'(c_2) = 0$.

Now, since $h'(c_1) = h'(c_2) =0$ applying Rolle's Theorem again, there exists a $c \in (c_1, c_2) \subset (-1, 1)$ so that $h''(0) = 0.$

Hence, we have some $c \in (-1, 1)$ so that $h''(c) = f''(c) - 1 = 0$ and so $f''(c) = 1.$

• You have a typo., I think. You have $f(1) = 0, f(1) = 1$ above. – copper.hat Apr 5 '17 at 16:17
• Looks good ${}{}$ – user251257 Apr 5 '17 at 17:39

Just using Rolle's theorem between $-1$ and $0$

we have

$$f(-1)=0$$ $$f(0)=0$$

so there exist $c_1 \epsilon (-1,0)$ such that $$f'(c_1)=0$$ Applying Mean Value theorem between $0$ and $1$

we have

$$f(0)=0$$ $$f(1)=1$$ so there exist $c_2 \epsilon (0,1)$ such that

$$f'(c_2)={f(1)-f(0) \over1-0}=1$$ we have $(c_1,c_2)$ which is subset of $[-1,1]$ such that we can use MVT on $f'(x)$

Applying Mean Value Theorem on $f'(x)$

we have $$f'(c_1)=0$$ $$f'(c_2)=1$$

so there exist $c\epsilon(c_1,c_2) \subset [-1,1]$ such that

$$f''(c)={f(c_2)-f(c_1)\over c_2 - c_1}={1-0\over c_2-c_1}$$

we have lot have value for which $a-b =1$ where a and b are some value of $c_2$ and $c_1$ respectively so

$$f''(c)=1$$