Chapter 1, Problem 35RE

Solution

a.

To Determine: The relative maximum. And express the value of $x$ at which the relative maximum occurs.

The relative maximum value is $-1$ and it occurs at $x=0$ .

Given:

  $y=\frac{x^2+4}{x^2-4}$

Calculation:

Graph the function $y=\frac{x^2+4}{x^2-4}$:

  

In the above graph it is seen that the function is relative maximum at $x=0$ .

To find the relative maximum value, substitute $x=0$ in the function $y=\frac{x^2+4}{x^2-4}$ :

  $\begin{array}{l}y=\frac{{\left(0\right)}^2+4}{{\left(0\right)}^2-4}\\ y=\frac{4}{-4}\\ y=-1\end{array}$

Thus, the relative maximum value is $-1$ and it occurs at $x=0$ .

b.

To Determine: The relative minimum. And express the value of $x$ at which the relative minimum occurs.

The relative minimum is none.

Given:

  $y=\frac{x^2+4}{x^2-4}$

Calculation:

Graph the function $y=\frac{x^2+4}{x^2-4}$:

  

In the above graph it is visible that there is no relative minimum occurs and thus there is no relative minimum value.

Thus, the relative minimum is none.