Chapter 2.2, Problem 60E

To determine

To approximate: The zeros (up to three decimal places) and relative extrema of polynomial function, using graphing utility.

Answer to Problem 60E

The real zeros of polynomial function $f\left(x\right)=-\frac{3}{8}{x}^4-x^3+2x^2+5$ are $x=-4.142$ , and $x=1.934$ .Therelative minima of the function exist at point $\left(0,5\right)$ , and $f_{\min }=5$ . The relative maxima of the function exit at points $\left(0.915,5.646\right)$ and $\left(-2.915,19.688\right)$ , these maxima are $f_{\max .}\left(x=0.915\right)=5.646$ , and $f_{\max .}\left(x=-2.915\right)=19.688$ .

Explanation of Solution

Given:

The polynomial function is $f\left(x\right)=-\frac{3}{8}{x}^4-x^3+2x^2+5$ .

Concept used:

If $\left(z,0\right)$ is an x-intercept of the polynomial function $f\left(x\right)$ then $f\left(z\right)=0$ , and $x=z$ is a zero of $f\left(x\right)$ . The vertex $\left(x,f_{extrema}\right)$ of the graph of $f\left(x\right)$ represents an extrema of the function.

Calculations:

The graph of a polynomial function $f\left(x\right)=-\frac{3}{8}{x}^4-x^3+2x^2+5$ is shown in Figure-1.

From the graph, we observe that the graph of function $f\left(x\right)=-\frac{3}{8}{x}^4-x^3+2x^2+5$ intersects x-axis at points $\left(-4.142,0\right)$ and $\left(1.934,0\right)$ .Thus, we have $f\left(-4.142\right)=0$ and $f\left(1.934\right)=0$ .

Therefore, $x=-4.142$ , and $x=1.934$ are zeros of the function $f\left(x\right)=-\frac{3}{8}{x}^4-x^3+2x^2+5$ .

The vertices of the graph of $f\left(x\right)=-\frac{3}{8}{x}^4-x^3+2x^2+5$ are $\left(0,5\right)$ , $\left(0.915,5.646\right)$ and $\left(-2.915,19.688\right)$ .

  

Since $\left(0,5\right)$ is the vertex of the graph that is opening upwards hence there existrelative minima of the function at points $\left(0,5\right)$ , and $f_{\min }=5$ .

Since $\left(0.915,5.646\right)$ and $\left(-2.915,19.688\right)$ are the vertices of the graph that is opening downwards hence there exist the maxima of the function at points $\left(0.915,5.646\right)$ and $\left(-2.915,19.688\right)$ , these maxima are $f_{\max .}\left(x=0.915\right)=5.646$ , and $f_{\max .}\left(x=-2.915\right)=19.688$ .

Conclusion:

The real zeros of polynomial function $f\left(x\right)=-\frac{3}{8}{x}^4-x^3+2x^2+5$ are $x=-4.142$ , and $x=1.934$ .Therelative minima of the function exist at point $\left(0,5\right)$ , and $f_{\min }=5$ . The relative maxima of the function exit at points $\left(0.915,5.646\right)$ and $\left(-2.915,19.688\right)$ , these maxima are $f_{\max .}\left(x=0.915\right)=5.646$ , and $f_{\max .}\left(x=-2.915\right)=19.688$ .