Chapter 1, Problem 36RE

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 $x=2$ .

Given:

  $y=\frac{4x}{x^2+4}$

Calculation:

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

  

In the above graph it is visible that the function is relative maximum at $x=2$ .

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

  $\begin{array}{c}y=\frac{4\left(2\right)}{{\left(2\right)}^2+4}\\ y=\frac{8}{4+4}\\ y=\frac{8}{8}\\ y=1\end{array}$

Thus, the relative maximum value is $1$ and it occurs at $x=2$ .

b.

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

The relative minimum value is $-1$ and it occurs at $x=-2$ .

Given:

  $y=\frac{4x}{x^2+4}$

Calculation:

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

  

In the above graph it is visible that the function is relative minimum at $x=-2$ .

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

  $\begin{array}{c}y=\frac{4\left(-2\right)}{{\left(-2\right)}^2+4}\\ y=\frac{-8}{4+4}\\ y=\frac{-8}{8}\\ y=-1\end{array}$

The relative minimum value is $-1$ and it occurs at $x=-2$ .