Chapter 4, Problem 16RE

To determine

To analyze:the given polynomial function.

Answer to Problem 16RE

  

Explanation of Solution

Given: $f\left(x\right)=-4x^3+4x$

Calculation:

Here the given function is:

  $f\left(x\right)=-4x^3+4x$

Step $1$ : End Behavior: the graph $f$ resembles that of the power function $y=x^3$ for larger values of $\left|x\right|$ .

Step $2$ : The $y$ intercept is $f\left(0\right)=-4{\left(0\right)}^3+4\left(0\right)=0$ . The $x$ intercepts satisfy the equation

  $\begin{array}{l}f\left(0\right)=-4x^3+4x=-4x\left(x^2-1\right)=0\\ So\\ -4x=0 or x^2-1=0\\ x=0 or x=1,-1\end{array}$

  $x-intercepts$ are $-1,0 and 1$ .

Step $3$ : The intercepts $-1,0 and 1$ are zeros of multiplicity 1, so graph of will cross-axis $-1,0 and 1$ .

Step $4$ : The graph of $f$ will contain at most two turning points

Step $5$ : The two $x-intercepts$ are $-4$ and $2$ .

Near $-1$ :

  $\begin{array}{l}f\left(x\right)=-4x^3+4x=-4x\left(x^2-1\right)\\ \approx -4\left(-1\right)\left(x^2-1\right)=4\left(x^2-1\right)\end{array}$ A parabola opening up

Near $0$ :

  $\begin{array}{l}f\left(x\right)=-4x^3+4x=-4x\left(x^2-1\right)\\ \approx -4x\left(0-1\right)=4x\end{array}$ A line with slope $4$

Near $1$ :

  $\begin{array}{l}f\left(x\right)=-4x^3+4x=-4x\left(x^2-1\right)\\ \approx -4\left(1\right)\left(x^2-1\right)=4\left(x^2-1\right)\end{array}$ A parabola opening down

Step $6$ : After combining steps $1$ through $5$ , graph as shown below:

  

Conclusion:

Therefore,the given polynomial function is analyzed.