Q 74. - Precalculus Enhanced with Graphing Utilities

Question

$G\left(x\right)=-x^4+32x^2+144$

(a) Determine whether $G$ is even, odd, or neither.

(b) There is a local maximum value of $400$ at $x=4$. Determine a second local maximum value.

(c) Suppose the area under the graph of $G$ between$x = 0$ and $x = 6$ that is bounded below by the $x$-axis is $1612.8$ square units. Using the result from part (a), determine the area under the graph of $G$ between$x = -6$ and $x = 0$ bounded below by the $x$-axis.

Step-by-Step Solution

Answer

Part (a). The given function $G$ is even.

Part (b). The second local maximum value is $400$ at $x=-4$.

Part (c). The area under the graph of $G$ between $x=-6$ and $x=0$ bounded below by the $x$-axis is $1612.8$ square units.

Part (a) Step 1. Given information

The given function is $G\left(x\right)=-x^4+32x^2+144$ and the local maximum is $400$ at $x=4$.

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 $G\left(x\right)$.

$\begin{array}{rcl}G(-x)& =& -{(-x)}^4+32{(-x)}^2+144 \\ & =& -{\left(x\right)}^4+32{\left(x\right)}^2+144\\ & =& -x^4+32x^2+144\\ & =& G\left(x\right)\end{array}$

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

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

Since $G$ is an even function, conclude that the graph is symmetric with respect to the $x$-axis.

There is local maximum $400$ at $x=4$, so there is second local maximum at $x=-4$.

Part (b) Step 2. Determine the second local maximum value at the point x = - 4 .

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

This implies that $G(-4)=G\left(4\right)\dots \left(1\right)$.

Substitute the value of $G\left(4\right)=400$ into $\left(1\right)$.

$\begin{array}{rcl}G(-4)& =& G\left(4\right)\\ & =& 400\end{array}$

Hence, the second local maximum value is $400$ at $x=-4$.

Part (c) Step 1. Find the area under the graph of G between x = - 6 and x = 0 .

It is given that the area under the graph of  between $x=0$ and $x=6$ is bounded below by the $x$-axis is $1612.8$ square units.

From part (a), the graph is even and symmetric about the $x$-axis.

This implies that the area bounded between the lines $x=a$ and $x=b$ will be the same as the area bounded between the lines $x=-a$ and $x=-b$.

Therefore, the area under the graph of $G$ between $x=-6$ and $x=0$ bounded below by the $x$-axis is $1612.8$ square units.