Chapter 6.5, Problem 3QQ

To determine

The value of x in which f(x) is minimum.

Answer to Problem 3QQ

The value of $x=\frac{3}{2}$ it is minimum point.

Explanation of Solution

Given information:

The function is $f\left(x\right)={\displaystyle \underset{-2}{\overset{x^2-3x}{\int }}{e}^{r^2}dt}$

Formula used:

Integration method is used.

Calculation:

Correct option is (C).

For check the point of minima, take the derivative

  $\begin{array}{l}f\text{'}\left(x\right)=\frac{d}{dx}{\displaystyle \underset{-2}{\overset{x^2-3x}{\int }}{e}^{r^2}dt}\\ =e^{{\left(x^2-3x\right)}^2}\frac{d}{dx}\left(x^2-3x\right)\\ f\text{'}\left(x\right)=\left(2x-3\right)e^{{\left(x^2-3x\right)}^2}\end{array}$

Now, equate it to zero, to find the point of minima,

  $\begin{array}{l}f\text{'}\left(x\right)=0\\ \left(2x-3\right)e^{{\left(x^2-3x\right)}^2}=0\\ x=\frac{3}{2}\end{array}$

Now, this is a point of minima or not, checks the double derivative.

  $\begin{array}{l}f"\left(x\right)=\frac{d}{dx}\left(\left(2x-3\right)e^{{\left(x^2-3x\right)}^2}\right)\\ f"\left(x\right)=\left(2x-3\right)e^{{\left(x^2-3x^2\right)}^2}2\left(x^2-3x\right)\left(2x\right)+e^{\left(x^2-3x\right)}\left(2\right)\end{array}$

At

  $\begin{array}{l}x=\frac{3}{2}\\ f"\left(\frac{3}{2}\right)>0\end{array}$

Hence, $x=\frac{3}{2}$ it is minimum point.

Conclusion:

The value of $x=\frac{3}{2}$ it is minimum point.