Chapter 4, Problem 3P

To determine

To examine: Whether the absolute maximum and absolute minimum of the function $f\left(x\right)=e^{10\left|x-2\right|-x^2}$ exist.

Answer to Problem 3P

The absolute maximum of $f\left(x\right)=e^{10\left|x-2\right|-x^2}$ is $f\left(-5\right)=e^{45}$.

The absolute minimum of $f\left(x\right)=e^{10\left|x-2\right|-x^2}$ does not exist.

Explanation of Solution

The given function can be represented in the form $f\left(x\right)=e^{g\left(x\right)}$, where $g\left(x\right)=10\left|x-2\right|-x^2$.

Thus, $f\left(x\right)$ will have absolute maximum (minimum) when $g\left(x\right)$ have absolute maximum (minimum).

Obtain the absolute maximum and minimum for $g\left(x\right)$.

Express $g\left(x\right)$ as follows.

$g\left(x\right)=\{\begin{array}{ll}10\left(x-2\right)-x^2\hfill & \text{if}x-2>0\hfill \\ 10\left[-\left(x-2\right)\right]-x^2\hfill & \text{if}x-2<0\hfill \end{array}$

That is, $g\left(x\right)=\{\begin{array}{ll}-x^2+10x-20\hfill & \text{if}x>2\hfill \\ -x^2-10x+20\hfill & \text{if}x<2\hfill \end{array}$

Obtain the first derivative of $g\left(x\right)$.

$g^{\prime }\left(x\right)=\{\begin{array}{ll}-2x+10\hfill & \text{if}x>2\hfill \\ -2x-10\hfill & \text{if}x<2\hfill \end{array}$

Set $g^{\prime }\left(x\right)=0$ and obtain the critical numbers as follows.

$\begin{array}{l}-2x+10=0\\ x=5\\ -2x-10=0\\ x=-5\end{array}$

Since the derivative does not exist at $x=2$, it is also a critical point.

Therefore, the critical points are $x=-5,x=5\text{and}x=2$.

Apply the critical points in $f\left(x\right)$.

Substitute $x=2$ in $f\left(x\right)$,

$\begin{array}{c}f\left(2\right)=e^{10\left|2-2\right|-2^2}\\ =e^{-4}\end{array}$

Substitute $x=5$ in $f\left(x\right)$,

$\begin{array}{c}f\left(5\right)=e^{10\left|5-2\right|-5^2}\\ =e^5\end{array}$

Substitute $x=-5$ in $f\left(x\right)$,

$\begin{array}{c}f\left(-5\right)=e^{10\left|-5-2\right|-{\left(-5\right)}^2}\\ =e^{45}\end{array}$

Since the largest functional value is the absolute maximum, the absolute maximum of $f\left(x\right)$ is $e^{45}$.

To find absolute minimum consider the following limit.

$\begin{array}{c}\underset{x\to \infty }{\lim }g\left(x\right)=\underset{x\to \infty }{\lim }\left(-x^2+10x-20\right)\\ =-\infty \end{array}$

Or

$\begin{array}{c}\underset{x\to \infty }{\lim }g\left(x\right)=\underset{x\to \infty }{\lim }\left(-x^2-10x+20\right)\\ =-\infty \end{array}$

Therefore, $\underset{x\to \infty }{\lim }g\left(x\right)=-\infty$.

Using the above value of limit, find $\underset{x\to \infty }{\lim }f\left(x\right)$.

$\begin{array}{c}\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }{e}^{g\left(x\right)}\\ =e^{-\infty }\\ =0\end{array}$

Since $f\left(x\right)>0$ as the exponential function is always positive for all values of x, so the absolute minimum of $f\left(x\right)$ does not exist.

Therefore, the absolute maximum value of $f\left(x\right)$ is $f\left(-5\right)=e^{45}$ and the absolute minimum of $f\left(x\right)$ does not exist.