If $f(x)=\int\limits_0^x e^{t^2}(t-2)(t-3) d t$ for all $x \in(0, \infty)$, then which of the following is incorrect?
Answer & explanation
Correct answer: option 4
We have,
$f(x)=\int\limits_0^x e^{t^2}(t-2)(t-3) d t$
$\Rightarrow f'(x)=e^{x^2}(x-2)(x-3)$
The changes in signs of f'(x) for different values of x are as given below:
Clearly, $f'(x)>0$ for $x \in(0,2) \cup(3, \infty)$ and $f'(x)<0$ for $x \in(2,3)$. So, $f(x)$ is increasing on $(0,2) \cup(3, \infty)$ and decreasing on $(2,3)$.
As $f'(x)$ changes sign from positive to negative as $x$ passes through 2 and negative to positive as $x$ passes through 3 . So, $x=2$ is a point of local maximum and $x=3$ is a point of local minimum of $f(x)$.
Clearly, $f'(x)=e^{x^2}(x-2)(x-3)$ is continuous on $[2,3]$ and differentiable on $(2,3)$. Also, $f'(2)=f'(3)=0$. Therefore, by Rolle's theorem there exists $c \in(2,3)$ such that $f''(c)=0$.