Chapter 4.1, Problem 95AYU

To determine

To analyze: the given polynomial function

Answer to Problem 95AYU

  

  

Explanation of Solution

Given:

  $f\left(x\right)=4x-x^3$

Calculation:

Let us consider the function $f\left(x\right)=4x-x^3$

Polynomial function $f\left(x\right)$ can be factored as

  $\begin{array}{l}f\left(x\right)=x\left(4-x^2\right)\hfill \\ =x\left(2-x\right)\left(2+x\right)\hfill \end{array}$

Step1: Determine the end behavior of the graph of the function.

Expand the polynomial to write it in the form

  $\begin{array}{l}f\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{...}+a_{1}x+a_0\\ f\left(x\right)=x\left(2-x\right)\left(2+x\right)\\ =x\left(4+2x-2x-x^2\right)\\ =4x-x^3\end{array}$

The polynomial function f is of degree 3. The graph of f behaves like $y=-x^3$ for large values of $\left|x\right|$ .

Step2: find the x -and y -intercepts of the graph of the function

The y -intercept is $f\left(0\right)=0$ . To find the x -intercepts, solve $f\left(x\right)=0$ .

  $\begin{array}{l}f\left(x\right)=0\\ x\left(2-x\right)\left(2+x\right)=0\\ x=-2,0,2\end{array}$

The x -intercepts are -2, 0, and 2.

Step3: Determine the zeros of the function and their multiplicity. Use this information to determine whether the graph crosses or touches the x -axis at each x -intercept.

The zeros offare -2, 0 and 2. The zero -2 is a zero of multiplicity l , so the graph of f crosses the x -axis at $x=-2$ . The zero 0 is a zero of multiplicity l , so the graph of f crosses the x -axis at $x=0$ . The zero 2 is a zero of multiplicity l , so the graph of f crosses the x -axis at $x=2$ .

Step 4: Determine the maximum number of turning points on the graph of the function. While the polynomial function is of degree 3 (step1), the graph of the function will have at most $3-1=2$ turning points.

Step5: Determine the behavior of the graph of f near each x -intercept.

The two x -intercepts are -2,0 and 2.

  $\begin{array}{l}\text{Near }-2:f\left(x\right)=x\left(2-x\right)\left(2+x\right)\\ \approx \left(-2\right)\left(2-\left(-2\right)\right)\left(2+x\right)\\ =\left(-2\right)\left(2+2\right)\left(2+x\right)\\ =-8\left(2+x\right)\text{ A line with slope -8}\\ =-8x-16\\ \text{Near 0: }f\left(x\right)=x\left(2-x\right)\left(2+x\right)\\ \approx x\left(2-0\right)\left(2+0\right)\\ =4x\text{ A line with slope 4}\\ \text{Near 2: }f\left(x\right)=x\left(2-x\right)\left(2+x\right)\\ \approx 2\left(2-x\right)\left(2+2\right)\\ =8\left(2-x\right)\text{ A line with slope -8}\\ =-8x+16\end{array}$

Step 6: Put all the information from Steps 1 through 5 together to obtain the graph of f Figure 1(a) illustrates the information obtained from Steps 1 through 5. Evaluatef at -1.16, 1.16 to help establish the scale on the y -axis. The graph of f is given in Figure1(b).

  

  

Conclusion:

Therefore, the polynomial function $f\left(x\right)=4x-x^3$ is analysed with graph.