# Let $f:[-1,1]\to \mathbb R$ be $C^2((-1,1))$. If $f(-1)=1, f(0)=1, f(1)=3$, then show $\exists c \in (-1,1)$ such that $f''(c)=2$.

Let $f:[-1,1]\to \mathbb R$ be a continuous function that is twice differentiable on the open interval $(-1,1)$. Prove that if $f(-1)=1$, $f(0)=1$, and $f(1)=3$, then there exists a point $c \in (-1,1)$ such that $f''(c)=2$.

By MVT, there exists $a \in (-1,0)$ such that $f'(a)=\frac{f(0)-f(-1)}{0-(-1)}=0$.

Also, there exists $b\in (0,1)$ such that $f'(b)=\frac{f(1)-f(0)}{1-0}=2$.

We can also say there exists $d \in (-1,1)$ such that $f'(d)=\frac{f(1)-f(-1)}{1-(-1)}=1$.

How can I proceed in finding $c \in (-1,1)$ such that $f''(c)=2$.

• Try with $g(x)=f(x)-x^2$ and find $a$ and $b$ again. – Nosrati Jun 25 '18 at 1:19
• @user108128 How did you think of this? – Al Jebr Jun 25 '18 at 1:28
• @user108128 Your suggestion almost worked. If we write instead $g(x)=f(x)-x^2-x$, then we can use MVT on $g$ and find that it works. – Al Jebr Jun 25 '18 at 1:43
• And for $g=f-x^2-kx$? – Nosrati Jun 25 '18 at 1:47
• @user108128 Not working for $k\ne 1$. – Al Jebr Jun 25 '18 at 2:16

The key is Taylor's theorem.

We have $$f(1)=f(0)+f'(0)+\frac{1}{2}f''(c_1),f(-1)=f(0)-f'(0)+\frac{1}{2}f''(c_2)$$ for some $c_1\in(0,1),c_2\in(-1,0)$. Adding these we get $$f(1)+f(-1)=2f(0)+\frac{f''(c_1)+f''(c_2)}{2}$$ or $$\frac{f''(c_1)+f''(c_2)}{2}=2$$ and since derivatives satisfy intermediate value property it follows that there is a $c\in[c_2,c_1]\subseteq (-1,1)$ with $f''(c) =2$.

• Where does the intermediate value property come into play? – Al Jebr Jun 25 '18 at 2:13
• @AlJebr: The value $(f''(c_1)+f''(c_2))/2=2$ lies between $f''(c_1)$ and $f''(c_2)$ and hence it is taken by $f''$ somewhere in interval $[c_2,c_1]$ via intermediate value property. – Paramanand Singh Jun 25 '18 at 2:26
• How do you know $2$ lies between $f''(c_1)$ and $f''(c_2)$? – Al Jebr Jun 25 '18 at 2:42
• @AlJebr: read my comment again and note that $(a+b) /2$ always lies between $a$ and $b$ (exactly halfway between). – Paramanand Singh Jun 25 '18 at 2:44

By looking at $g(x)=f(x+1)-f(x)$ and applying the MVT twice:

$g$ is defined on $[-1,0]$ and has boundary values $0$ and $2$, so there is a $c\in (-1,0)$ for which $g'(c)=f'(c+1)-f'(c)=2$ and therefore there is a $d\in (c,c+1)$ for which $f''(d)=2$ .

• If the Question is worth answering, surely it deserves at least one complete sentence. In general one should beware an Answer that is substantially shorter than a concisely worded Question is responds to. – hardmath Jun 25 '18 at 0:49
• +1 for simpler (compared to mine) argument. – Paramanand Singh Jun 25 '18 at 15:39