Q 73. - Precalculus Enhanced with Graphing Utilities

Question

$F\left(x\right)=-x^4+8x^2+8$

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

(b) There is a local maximum value of $24$ at $x = 2$. Determine a second local maximum value. 

(c) Suppose the area under the graph of F between$x = 0$ and $x = 3$ that is bounded below by the $x$-axis is $47.4$ square units. Using the result from part (a), determine the area under the graph of F between $x=-3$ and $x=0$ bounded below by the $x$-axis.

Step-by-Step Solution

Answer

Part (a). The given function $F\left(x\right)$ is even.

Part (b). The second local maximum value is $24$ at $x=-2$.

Part (c). The area under the graph of F between $x=-2$ and $x=0$ bounded below by the $x$-axis is $47.4$ square units.

Part (a) Step 1. Given information

The given function is $F\left(x\right)=-x^4+8x^2+8$ and the local maximum is $24$ at $x=2$.

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 $F\left(x\right)$.

$\begin{array}{rcl}F(-x)& =& -{(-x)}^4+8{(-x)}^2+8 \\ & =& -{\left(x\right)}^4+8\left(x^2\right)+8\\ & =& -x^4+8x^2+8\\ & =& F\left(x\right)\end{array}$

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

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

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

There is local maximum $24$ at $x=2$, so there is second local maximum at $x=-2$.

Part (b) Step 2. Determine the second local maximum 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 $F\left(2\right)=24$ into $\left(1\right)$.

$\begin{array}{rcl}F(-2)& =& F\left(2\right)\\ & =& 24\end{array}$

Hence, the second local maximum value is $24$ at $x=-2$.

Part (c) Step 1. Find the area under the graph of F between x = - 3 and x = 0 .

It is given that the area under the graph of $F$ between $x=0$ and $x=3$ is bounded below by the $x$-axis is $47.4$ 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 $F$ between $x=-3$ and $x=0$ bounded below by the $x$-axis is $47.4$ square units.