Question

In Problems 73, determine where the graph of $f$is below the graph of $g$by solving the inequality $f\left(x\right)\le g\left(x\right)$. Graph $f$and $g$together.

$\begin{array}{rcl}f\left(x\right)& =& x^4-1\\ g\left(x\right)& =& -2x^2+2\end{array}$

Short answer

The graph is shown below:

Step-by-step solution

Given information

The given functions are$f\left(x\right)=x^4-1$and$g\left(x\right)=-2x^2+2$.

 Write the inequality so that the function fx and gx  is on the left side and zero is on the right side.

Substitute $f\left(x\right)=x^4-1$and $g\left(x\right)=-2x^2+2$into the inequality $f\left(x\right)\le g\left(x\right)$.

$\begin{array}{rcl}{x}^4-1& \le & -2x^2+2\\ x^4& \le & -2x^2+2+1\\ x^4+2x^2& \le & 3\\ x^4+2x^2-3& \le & 0\\ x^4+3x^2-x^2-3& \le & 0\\ x^2\left(x^2+3\right)-1\left(x^2+3\right)& \le & 0\\ \left(x^2-1\right)\left(x^2+3\right)& \le & 0\\ \left(x-1\right)\left(x+1\right)\left(x^2+3\right)& \le & 0\end{array}$

Determine the zeros of the function.

Now, find the zeros of the function.

$\begin{array}{rcl}\left(x-1\right)\left(x+1\right)\left(x^2+3\right)& =& 0\\ x& =& 1,-1\end{array}$

$x^2+3$gives the imaginary solution.

 Use the zeros to divide the real number line into three intervals.

$(-\infty ,-1),(-1,1),(1,+\infty )$

Determine the solution of the inequality fx≤gx

Select a test number in each interval found in above step and evaluate $\left(x-1\right)\left(x+1\right)\left(x^2+3\right)\le 0$at each number to determine if $\left(x-1\right)\left(x+1\right)\left(x^2+3\right)$is positive or negative.

Interval	$\left(-\infty ,-1\right)$	$\left(-1,1\right)$	$\left(1,\infty \right)$
Number Chosen	$-5$	$0$	$5$
Value of $\left(x-1\right)\left(x+1\right)\left(x^2+3\right)$	localid="1646800368047" $\left(-5-1\right)\left(-5+1\right)\left({\left(-5\right)}^2+3\right)=672$	$\left(0-1\right)\left(0+1\right)\left(0^2+3\right)=-3$	$\left(5-1\right)\left(5+1\right)\left(5^2+3\right)=672$
Conclusion	Positive	Negative	Positive

Notice, the function $\left(x-1\right)\left(x+1\right)\left(x^2+3\right)$is negative in the interval $(-1,1)$

Since we are looking where the function $\left(x-1\right)\left(x+1\right)\left(x^2+3\right)$is less than or equal to zero, the solution is:

$\left[-1,1\right]$

Graph the function

The solution of the inequality is $\left[-1,1\right]$

Use the graphing calculator, to graph the given function.

The graph is shown below:

Write the conclusion

The function $f\left(x\right)$is below $g\left(x\right)$in the interval $[-1,1]$.