Q 72. - Precalculus Enhanced with Graphing Utilities

Question

$f\left(x\right)=-x^3+12x$

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

(b) There is a local maximum value of $16$ at $2$. Determine the local minimum value.

Step-by-Step Solution

Answer

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

Part (b). The local minimum value is $-16$ at $x=-2$.

Part (a) Step 1. Given information

The given function is $f\left(x\right)=-x^3+12x$ and the local maximum is $16$ at $2$.

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

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

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

$\begin{array}{rcl}f(-x)& =& -{(-x)}^3+12(-x)\\ & =& -(-x^3)-12x\\ & =& -(-x^3+12x)\\ & =& -f\left(x\right)\end{array}$

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

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

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

There is local maximum $16$ at $x=2$, so there is local minimum at $x=-2$.

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

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

This implies that $f(-2)=-f\left(2\right)\dots \left(1\right)$.

Substitute the value of $f\left(2\right)=16$ into $\left(1\right)$.

$\begin{array}{rcl}f(-2)& =& -f\left(2\right)\\ & =& -\left(16\right)\\ & =& -16\end{array}$

Hence, the local minimum value is $-16$ at $x=-2$.