Chapter 11.1, Problem 40E

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 40E

The function has no $x-\mathrm{int}ercepts$ .

The local maximum of this function is at $\left(-1.7,-4\right)$ and $\left(1.7,-4\right)$ .

The local minimum of this function is at $\left(0,-13\right)$ .

The function is increasing when $x<-1.7$ and $0<x<1.7$ and decreasing when $-1.7<x<0$ and $x>1.7$ .

Explanation of Solution

Given:

  $f\left(x\right)=-x^4+6x^2-13$

Calculation:

To graph the function $f\left(x\right)=-x^4+6x^2-13$ , make a table of values and plot the corresponding points. Connect the points with a smooth curve and check the end behavior.

$x$	$-2$	$-1$	$0$	$1$	$2$
$f\left(x\right)$	$-5$	$-8$	$-13$	$-8$	$-5$

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

  

From the graph,it is clear that the function has no $x-\mathrm{int}ercepts$ .

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 $\left(-1.7,-4\right)$ and $\left(1.7,-4\right)$ .

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 $\left(0,-13\right)$ .

The function is increasing when $x<-1.7$ and $0<x<1.7$ and decreasing when $-1.7<x<0$ and $x>1.7$ .

Conclusion:

Hence, the function has no $x-\mathrm{int}ercepts$ .

The local maximum of this function is at $\left(-1.7,-4\right)$ and $\left(1.7,-4\right)$ .

The local minimum of this function is at $\left(0,-13\right)$ .

The function is increasing when $x<-1.7$ and $0<x<1.7$ and decreasing when $-1.7<x<0$ and $x>1.7$ .