Q 71. - Precalculus Enhanced with Graphing Utilities

Question

$g\left(x\right)=x^3-27x$

(a) Determine whether the given function $g$ is even, odd, or neither.

(b) There is a local minimum value of $-54$ at $3$. Determine the local maximum value.

Step-by-Step Solution

Answer

Part (a). The given function $g\left(x\right)$ is odd.

Part (b). The local maximum value is $54$ at $x=-3$.

Part (a) Step 1. Given information

The given function is $g\left(x\right)=x^3-27x$ and the local miniumum is $-54$ at $3$.

Part (a) Step 2. Find whether the given function g is even, odd, or neither.

• A function is even, if $g(-x)=g\left(x\right)$.
• A function is odd, if $g(-x)=-g\left(x\right)$.

Replace $x$ by $-x$ in $g\left(x\right)$.

$\begin{array}{rcl}g(-x)& =& {(-x)}^3-27(-x)\\ & =& -x^3+27x\\ & =& -(x^3-27x)\\ & =& -g\left(x\right)\end{array}$

Since $g(-x)=-g\left(x\right)$, the given function is odd. 

Part (b) Step 1. Determine the point of local maximum value.

Since  is an odd function, conclude that the graph is symmetric with respect to the origin.

There is local minimum $-54$ at $x=3$, so there is local maximum at $x=-3$.

Part (b) Step 2. Determine the local maximum value at the point x = - 3 .

The function is odd, so $g(-x)=-g\left(x\right)$.

This implies that $g(-3)=-g\left(3\right)\dots \left(1\right)$.

Substitute the value of $g\left(3\right)=-54$ into $\left(1\right)$.

$\begin{array}{rcl}g(-3)& =& -g\left(3\right)\\ & =& -(-54)\\ & =& 54\end{array}$

Hence, the local maximum value is $54$ at $x=-3$.