Chapter 1.2, Problem 50E

Solution

To verify: the function is odd, even, or neither.

The function is even.

Given information: 

The function $g\left(x\right)=\frac{3}{1+x^2}$ .

Calculation:

Graph of

  $g\left(x\right)=\frac{3}{1+x^2}$

Let $y=g\left(x\right)$ .

Putting the value of $x=0$ then $y=3$ .

And $x=1$ , $y=1.5$

And $x=-1$ , $y=1.5$

And $x=2$ , $y=0.6$

And $x=-2$ , $y=0.6$

And $x=3$ , $y=0.3$

And $x=-3$ , $y=0.3$

Plot the graph:

  

The graph of $g\left(x\right)=\frac{3}{1+x^2}$ is symmetric about the $y-$ axis

To show that a function is even, its need to show that

  $\begin{array}{c}g\left(-x\right)=g\left(x\right)\\ g\left(x\right)=\frac{3}{1+x^2}\\ g\left(-x\right)=\frac{3}{1+{\left(x\right)}^2}\\ g\left(-x\right)=\frac{3}{1+x^2}\\ g\left(-x\right)=g\left(x\right)\end{array}$

Therefore, the function is even.