We have, $f$ is continuous and differentiably function on $[a, b]$ Also, $f(a)=f(b)=0$
By Rolle's theorem, there exists $c \in(a, b)$ such the $f^{\prime}(c)=0$
Thus, there exists $x \in(a, b)$ such that $f^{\prime}(x)=0$ Let at $x=c \in(a, b), f^{\prime}(c)=0$
Now, $f$ is continuously differentiable on $|a, b|$ $\Rightarrow \quad f^{\prime}$ is continuous on $\left.\mid a, b\right]$ Also, $f$ is twice differentiable on $(a, b)$ $\therefore f^{\prime}$ is differentiable on $(a, b)$
and $f^{\prime}(a)=0=f^{\prime}(c)$
By Rolle's theorem, there exists $k \in(a, c)$ such that. $f^{\prime \prime}(k)=0$
Thus, there exists $x \in(a, c)$ such that $f^{\prime \prime}(x)=0$ So, there exists $x \in(a, b)$ such that $f^{\prime \prime}(x)=0$ Let us consider, $f(x)=(x-a)^{2}(x-b)$
where $f(a)=f(b)=f^{\prime}(a)=0$ but
$f^{\prime \prime}(a) \neq 0$ and $f^{m}(x) \neq 0$ for any $x \in(a, b)$