Chapter 11.1, Problem 38E

To determine

the $x-\mathrm{int}ercepts$ , the local maximums and local minimums points and the intervals for which the function is increasing or decreasing.

Answer to Problem 38E

  $x-\mathrm{int}ercepts$ are $x\approx -2.8$ and $x\approx 3.2$ .

The local maximum of a function isat $\left(-0.25,-2.87\right)$ .

The local minimum of this function is at $\left(2.1,-9.06\right)$ .

The function is increasing when $-1.86<x<-0.25$ and $x>2.1$ and decreasing when $x<-1.86$ and $-0.25<x<2.1$

Explanation of Solution

Given:

  $g\left(x\right)=\frac{1}{4}{x}^4-2x^2-x-3$

Calculation:

To graph the function $g\left(x\right)=\frac{1}{4}{x}^4-2x^2-x-3$ , make a table of values and plot the corresponding points. Connect the points with a smooth curve and check the end behavior.

$x$	$-2$	$-1$	$0$	$1$	$2$
$g\left(x\right)$	$-5$	$-3.75$	$-3$	$-5.75$	$-9$

The degree of the polynomial is even and the leading coefficient is positive. So, $g\left(x\right)\to +\infty$ as $x\to +\infty$ and $g\left(x\right)\to +\infty$ as $x\to -\infty$

  

From the graph, it is clear that $x-\mathrm{int}ercepts$ are $x\approx -2.8$ and $x\approx 3.2$ .

The local maximum of a function is $y-coordinate$ of the turning point, when the point is higher than all the nearby points. Hence, local maximum of this function is at $\left(-0.25,-2.87\right)$ .

The local minimum of a function is $y-coordinate$ of the turning point, when the point is lower than all the nearby points. Hence, local minimum of this function is at $\left(2.1,-9.06\right)$ .

The function is increasing when $-1.86<x<-0.25$ and $x>2.1$ and decreasing when $x<-1.86$ and $-0.25<x<2.1$

Conclusion:

Hence, $x-\mathrm{int}ercepts$ are $x\approx -2.8$ and $x\approx 3.2$ .

The local maximum of a function is at $\left(-0.25,-2.87\right)$ .

The local minimum of this function is at $\left(2.1,-9.06\right)$ .

The function is increasing when $-1.86<x<-0.25$ and $x>2.1$ and decreasing when $x<-1.86$ and $-0.25<x<2.1$