Chapter 1.2, Problem 52E

Solution

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

It is neither even nor odd.

Given information: 

The function $f\left(x\right)=x^3+0.04x^2+3$ .

Calculation:

Graph of

  $f\left(x\right)=x^3+0.04x^2+3$

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

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

And $x=1$ , $y=4.04$

And $x=-1$ , $y=2.04$

And $x=2$ , $y=11.16$

And $x=-2$ , $y=-4.84$

Plot the graph:

  

The graph of $g\left(x\right)=x^3+0.04x^2+3$ is not symmetric about the $y-$ axis or the origin.

Hence, it is neither even nor odd.

To show that a function is even, then show that $g\left(-x\right)=g\left(x\right)$ and to show that a function is odd and show that $g\left(-x\right)=-g\left(x\right)$ .

  $\begin{array}{c}g(x)=x^3+0.04x^2+3\\ $ g(-x)={(-x)}^3+0.04{(-x)}^2+3$g(-x)=-x^3+0.04x^2+3\end{array}$

Then

  $g\left(-x\right)\ne g\left(x\right)$ and $g\left(-x\right)\ne -g\left(x\right)$

Neither is true for this function.