Question

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

• A. $f$ has a local maximum at $x=2$.
• B. $f$ is decreasing on $\left(2,3\right)$.
• C. There exists some $c\in \left(0,\infty \right)$ such that $f"\left(c\right)=0$.
• D. $f$ has a local minimum at $x=3$.
• E. All

Answer 1

The correct option is E

All

Explanation for the correct options:

Option(A): The given equation is $f\left(x\right)={\int }_0^x{e}^{x^2}\left(x-2\right)\left(x-3\right)dx$.

Differentiate both sides of the equation with respect to $x$.

$$\begin{gathered} \frac{d}{dx}\left[f\left(x\right)\right]=\frac{d}{dx}\left[{\int }_0^x{e}^{x^2}\left(x-2\right)\left(x-3\right)dx\right] \\ \Rightarrow f\text{'}\left(x\right)=\left[e^{x^2}\left(x-2\right)\left(x-3\right)\right]\frac{d}{dx}\left(x\right)-\left[e^{0^2}\left(0-2\right)\left(0-3\right)\right]\frac{d}{dx}\left(0\right)\left[\because {\mathbf{\int }}_{\mathbf{a}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}^{\mathbf{b}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}\mathit{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{d}\mathit{x}\mathbf{=}\mathit{f}\mathbf{\left[}\mathbf{b}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\right]}\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{\left[}\mathbf{b}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\right]}\mathbf{-}\mathit{f}\mathbf{\left[}\mathbf{a}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\right]}\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{\left[}\mathbf{a}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\right]}\right] \\ \Rightarrow f\text{'}\left(x\right)=e^{x^2}\left(x-2\right)\left(x-3\right) \end{gathered}$$ Differentiate both sides of the equation with respect to $x$.

$$\begin{gathered} \frac{d}{dx}\left[f\text{'}\left(x\right)\right]=\frac{d}{dx}\left[e^{x^2}\left(x-2\right)\left(x-3\right)\right] \\ \Rightarrow f"\left(x\right)=\left(x-2\right)\left(x-3\right)\frac{d}{dx}\left(e^{x^2}\right)+e^{x^2}\left(x-3\right)\frac{d}{dx}\left(x-2\right)+e^{x^2}\left(x-2\right)\frac{d}{dx}\left(x-3\right)\left[\because \frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{\left(}\mathbf{u}\mathbf{v}\mathbf{\right)}\mathbf{=}\mathit{v}\frac{\mathbf{d}\mathbf{u}}{\mathbf{d}\mathbf{x}}\mathbf{+}\mathit{u}\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{x}}\right] \\ =\left(x-2\right)\left(x-3\right)e^{x^2}\frac{d}{dx}\left(x^2\right)+e^{x^2}\left(x-3\right)+e^{x^2}\left(x-2\right) \\ =2x{e}^{x^2}\left(x-2\right)\left(x-3\right)+e^{x^2}\left(x-3+x-2\right) \\ =e^{x^2}\left(2x^3-10x^2+12x\right)+e^{x^2}\left(2x-5\right) \\ =e^{x^2}\left(2x^3-10x^2+14x-5\right) \end{gathered}$$

To calculate the maxima, $f\text{'}\left(x\right)=0$.

$\Rightarrow e^{x^2}\left(x-2\right)\left(x-3\right)=0$

Thus, either $x=2$ or $x=3$.

So, for $x=2$.

$$\begin{gathered} \Rightarrow f"\left(2\right)=e^{{\left(2\right)}^2}\left(2{\left(2\right)}^3-10{\left(2\right)}^2+14\left(2\right)-5\right) \\ =e^4\left(2\left(8\right)-10\left(4\right)+28-5\right) \\ =e^4\left(16-40+23\right) \\ =e^4\left(-1\right) \\ =-e^4 \end{gathered}$$

Thus, $f"\left(2\right)<0$.

Therefore, $f$ has a local maximum at $x=2$.

Thus option(A) is correct.

Option(B): For decreasing function, $f\text{'}\left(x\right)<0$.

$$\begin{gathered} \Rightarrow e^{x^2}\left(x-2\right)\left(x-3\right)<0 \\ \Rightarrow \left(x-2\right)\left(x-3\right)<0 \\ \Rightarrow x\in \left(2,3\right) \end{gathered}$$

Therefore, $f$ is decreasing on $\left(2,3\right)$.

Thus option(B) is correct.

Option(C): It is determined that $f"\left(x\right)=e^{x^2}\left(2x^3-10x^2+14x-5\right)$, so $f"\left(x\right)$ is a polynomial of degree $3$.

Thus, $f"\left(x\right)=0$ has at least one real solution.

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

Hence, option(C) is correct.

Option(D): It is previously determined that for $x=3$, $f\text{'}\left(x\right)=0$.

So, for $x=3$.

$$\begin{gathered} \Rightarrow f"\left(3\right)=e^{{\left(3\right)}^2}\left(2{\left(3\right)}^3-10{\left(3\right)}^2+14\left(3\right)-5\right) \\ =e^9\left(2\left(27\right)-10\left(9\right)+42-5\right) \\ =e^9\left(54-90+37\right) \\ =e^9\left(1\right) \\ =e^9 \end{gathered}$$

Thus, $f"\left(3\right)>0$.

Therefore, $f$ has a local minimum at $x=3$.

Thus option(D) is correct.

So, all options are correct.

Hence, option(E) is the correct option.