Chapter 4.8, Problem 26E

To determine

To sketch: the graph of the function, and to indentify the $x$ intercepts and the points where the local maximums and local minimums occur, to find the intervals for which the function is increasing or decreasing.

Answer to Problem 26E

  $x$ intercepts are approximately $x=-2.16,x=1$ and $x=1.75$ .

Local maximum is at approximately $\left(-1.63,10.47\right)$ .

Local minimum is at approximately $\left(1.46,-1.68\right)$ .

The graph it is very clear that the function is increasing when $x<-1.63$ and $x>1.46$ and decreasing when $-1.63<x<1.46$ .

Explanation of Solution

Given information:

Given function,

  $f\left(x\right)=x^5-4x^3+x^2+2$

Calculation:

Consider the function $f\left(x\right)=x^5-4x^3+x^2+2$ :

Input function using key $X\wedge 5-4X\wedge 3+X{x}^2+2$ .

  

Adjust the WINDOW and hit GRAPH key.

  

The graph has three $x$ intercepts and 2 turning points.

Press 2^ndTRACE to go the CALCULATE menu and highlight the ZERO and press ENTER.

  

When asked for left bound and right bound, move cursor to the left or right of observed zero location.

When asked for Guess, simply Hit ENTER key.

  

Therefore, $x$ intercepts are approximately $x=-2.16,x=1$ and $x=1.75$ .

Press 2^nd TRACE to go to the CALCULATE menu and highlight the Maximum and press ENTER.

  

When asked for left bound and right bound, move cursor to the left or right of observed maximum location.

When asked for guess, simply hit ENTER key.

  

Therefore, local maximum is at approximately $\left(-1.63,10.47\right)$ .

Press 2^nd TRACE to go to the CALCULATE menu and highlight the Minimum and press ENTER.

  

When asked for left bound and right bound, move cursor to the left or right of observed maximum location.

When asked for guess, simply hit ENTER key.

  

Therefore, local minimum is at approximately $\left(1.46,-1.68\right)$ .

From the graph it is very clear that the function is increasing when $x<-1.63$ and $x>1.46$ and decreasing when $-1.63<x<1.46$ .