Let $f: [1,7]→ \mathbb R$ be a twice differentiable function such that $f(7) = 3f(3)−2f(1)$. Prove that there is $c∈(1,7)$ such that $f{′′}(c) = 0$

Question

Let $f: [1,7]→ \mathbb R$ be a twice differentiable function such that $f(7) = 3f(3)−2f(1)$. Prove that there is $c∈(1,7)$ such that $f{′′}(c) = 0$

Not certain what to do with this question. Am I supposed to come up with a function that holds this property?

Answer 1

$$f(7)=3f(3)-2f(1)\\\iff f(7)-f(3)=2(f(3)-f(1))\\\iff \frac{f(7)-f(3)}{7-3}=\frac{f(3)-f(1)}{3-1}$$

By LMVT and the above equality, there exists $c_1\in [1,3]$ and $c_2\in [3,7]$ such that $f'(c_1)=f'(c_2)$

Apply Rolle's theorem on $f'$ on $[c_1,c_2]$

Answer 2

If $f$ is twice differentiable on $[1,7]$, then it must be differentiable on $[1,3]$ and $[3,7]$. Applying the mean value theorem to each interval then implies that there exists an $x_1\in(1,3)$ and an $x_2\in (3,7)$ such that

$$f'(x_1)=\frac{f(3)-f(1)}{3-1}=\frac{f(3)-f(1)}{2}$$

and

$$f'(x_2)=\frac{f(7)-f(3)}{7-3}=\frac{f(7)-f(3)}{4}$$

But $f(7)=3f(3)-2f(1)$ implies that $f(7)-f(3)=2f(3)-2f(1)$, so

$$f'(x_2)=\frac{f(7)-f(3)}{4}=\frac{2f(3)-2f(1)}{4}=\frac{f(3)-f(1)}{2}=\frac{f(3)-f(1)}{3-1}=f'(x_1)$$

We thus have $f'(x_1)=f'(x_2)$, and since $f$ is twice differentiable on $[x_1,x_2]\subset[1,7]$, it must be the case that $f'$ is differentiable on $[x_1,x_2]$. We can now apply Rolle's theorem to $f'$ on the interval $[x_1,x_2]$ to deduce that there is a $c\in(x_1,x_2)\subset(1,7) $ where $f''(c)=0$. And that's what we wanted!