Question

Let $f$ be a twice differentiable function $[0,2]$ show that if $f\left(0\right)=0$, $f\left(1\right)=2$ and $f\left(2\right)=4$ then there is some point $x_0\in [0,2]$ such that $f^{\prime \prime }\left(x_0\right)=0$

Answer 1

$f\left(0\right)=0$ $\left(1\right)=2$ $f\left(2\right)=4$

Let $C_1\in (0,2)$ and $C_2\in (2,4)$

then according to $LMVT$ there exist some $C_1$ and $C_2$ such that

$f\left(c_1\right)=\frac{2-0}{1-0}=2$

$f\left(c_2\right)=\frac{4-2}{2-1}=2$

Now, let $g\left(x\right)=f\left(x\right)$

then according to $LMVT$ there exists $C_3\in (C_1,C_2)$ such that

$g\left(c_3\right)=g\frac{\left(c_2\right)-g\left(c_1\right)}{c_2-c_1}=0$

$\Rightarrow g\left(c_3\right)=0$

$\Rightarrow f\left(c_3\right)=0$ and $c_3\in (c_1,c_2)$

$\Rightarrow C_3\in (0,2)$

Hence proved