Question

If$f\left(x\right)={\int }_0^x{e}^{t^2}(t-2)(t-3)dt$, $\forall x\in (0,\infty )$, then

• A. $f$ has a local maximum at $x=2$
• B. $f$ is decreasing on $(2,3)$
• C. there exists some $c\in (0,\infty )$ such that $f\text{'}\left(c\right)=0$
• D. $f$ has a local minimum at $x=3$

Answer 1

The correct option is D

$f$ has a local minimum at $x=3$

Explanation for the correct options:-

Step 1: Apply Leibnitz rule

Given:-

$f\left(x\right)={\int }_0^x{e}^{t^2}(t-2)(t-3)dt$, $\forall x\in (0,\infty )$

Differentiating both sides with respect to $x$

$f\text{'}\left(x\right)=\frac{d}{dx}\left({\int }_0^x{e}^{t^2}(t-2)(t-3)dt\right)$

We know that if the functions $\mathrm{\psi }\left(\mathrm{x}\right)$ and $\mathrm{\varphi }\left(\mathrm{x}\right)$ are defined on $\left[\mathrm{a},\mathrm{b}\right]$ and differentiable at a point $\mathrm{x}\in \left(\mathrm{a},\mathrm{b}\right)$, and $f\left(t\right)$ is continuous on $\left[\mathrm{\varphi }\left(\mathrm{a}\right),\mathrm{\varphi }\left(\mathrm{b}\right)\right]$, then

$\frac{\mathrm{d}}{\mathrm{dx}}\left({\int }_{\varphi \left(x\right)}^{\psi \left(x\right)}f\left(t\right)dt\right)=f\left(\psi \left(x\right)\right)\frac{d}{dx}\left\{\psi \left(x\right)\right\}-f\left(\varphi \left(x\right)\right)\frac{d}{dx}\left\{\varphi \left(x\right)\right\}$

By applying the above Leibnitz rule, we get

$\begin{array}{rcl}\therefore f\text{'}\left(x\right)& =& \left(e^{x^2}(x-2)(x-3)\right)\frac{d}{dx}\left(x\right)-\left(e^0(0-2)(0-3)\right)\frac{d}{dx}\left(0\right)\\ & \Rightarrow & f\text{'}\left(x\right)=\left(e^{x^2}(x-2)(x-3)\right)\end{array}$

Step 2: Find the critical points

For critical point $f\text{'}\left(x\right)=0$

$$\begin{gathered} \Rightarrow \begin{array}{l}{e}^{x^2}\left(x-2\right)\left(x-3\right)\end{array}=0 \\ \begin{array}{rcl}& \Rightarrow & \left(x-2\right)\left(x-3\right)\end{array}\begin{array}{rcl}& =& \end{array}0\left[\because e^{x^2}\ne 0,\forall x\in \mathbb{\text{ℝ}}\right] \end{gathered}$$

$\Rightarrow x=2,3$ are the critical points of the function $f\left(x\right)$

Step 3: Check the interval of monotonicity

If $x<2,f\text{'}\left(x\right)=e^{x^2}\left(x-2\right)\left(x-3\right)>0$, because $e^{x^2}$ is always positive and for $x<2,\left(x-2\right)<0$ and $\left(x-3\right)<0\Rightarrow \left(x-2\right)\left(x-3\right)>0$

So, $f\left(x\right)$ is increasing if $x<2$.

If $2<x<3$$,f\text{'}\left(x\right)=e^{x^2}\left(x-2\right)\left(x-3\right)<0$, because $e^{x^2}$ is always positive and for $2<x<3,\left(x-2\right)>0$ and $\left(x-3\right)<0\Rightarrow \left(x-2\right)\left(x-3\right)<0$.

So, $f\left(x\right)$ is decreasing if $2<x<3$.

If $x>3,f\text{'}\left(x\right)=e^{x^2}\left(x-2\right)\left(x-3\right)>0$, because $e^{x^2}$ is always positive and for $x>3,\left(x-2\right)>0$ and $\left(x-3\right)>0\Rightarrow \left(x-2\right)\left(x-3\right)>0$

So, $f\left(x\right)$ is increasing if $x>3$.

Thus, the function has a local maximum at $x=2$ and has a local minimum at $x=3$.

Step 4: Apply Rolle's theorem

Now,

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \left(e^{x^2}(x-2)(x-3)\right)\end{array}$

So, for $x=2$ and $x=3$

$\begin{array}{rcl}f\text{'}\left(2\right)& =& \left(e^4(2-2)(2-3)\right)=0\end{array}$

$\begin{array}{rcl}f\text{'}\left(3\right)& =& \left(e^9(3-2)(3-3)\right)=0\end{array}$

$\Rightarrow f\text{'}\left(2\right)=f\text{'}\left(3\right)$

So, by Rolle's theorem, there exists some $c\in (2,3)$ such that $f"\left(c\right)=0$

As $(2,3)\subset \left(0,\infty \right)$

Therefore, there exists some $c\in (0,\infty )$ such that $f"\left(c\right)=0$

But in option C it is given that there exists some $c\in (0,\infty )$ such that $f\text{'}\left(c\right)=0$, which is not correct because there is no interval which is subset of $(0,\infty )$ at which the value of the function $f$ is equal.

Hence, the correct options are A, B, and D.