Chapter 2.2, Problem 59E

To determine

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

Answer to Problem 59E

The zeros of polynomial function $f\left(x\right)=2x^4-6x^2+1$ are $x=-1.68$ , $x=-0.421$ , $x=0.421$ , and $x=1.68$ .Therelative minima of the function exist at points $\left(-1.225,-3.5\right)$ and $\left(1.225,-3.5\right)$ , and $f_{\min }=3.5$ . The relative maxima of the function at exits at point $\left(0,1\right)$ and $f_{\max .}=1$ .

Explanation of Solution

Given:

The polynomial function is $f\left(x\right)=2x^4-6x^2+1$ .

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 $f\left(x\right)=2x^4-6x^2+1$ is shown in Figure-1.

From the graph, we observe that the graph of function $f\left(x\right)=2x^4-6x^2+1$ intersects x-axis at points $\left(-1.68,0\right)$

  $\left(-0.421,0\right)$

  $\left(0.421,0\right)$ and $\left(1.68,0\right)$ .

Thus, we have $f\left(-1.68\right)=0$ , $f\left(-0.421\right)=0$ , $f\left(0.421\right)=0$ and $f\left(1.68\right)=0$ .

Therefore, $x=-1.68$ , $x=-0.421$ , $x=0.421$ , and $x=1.68$ are zeros of the function $f\left(x\right)=2x^4-6x^2+1$ .

The vertices of the graph of $f\left(x\right)=2x^4-6x^2+1$ are $\left(-1.225,-3.5\right)$ , $\left(1.225,-3.5\right)$ and $\left(0,1\right)$ .

  

Since $\left(-1.225,-3.5\right)$ and $\left(1.225,-3.5\right)$ are the vertices of the graph that is opening upwards hence there existrelative minima of the function at points $\left(-1.225,-3.5\right)$ and $\left(1.225,-3.5\right)$ , and $f_{\min }=3.5$ .

Since $\left(0,1\right)$ is the vertex of the graph that is opening downwards hence there exist the maxima of the function at point $\left(0,1\right)$ and $f_{\max .}=1$ .

Conclusion:

The zeros of polynomial function $f\left(x\right)=2x^4-6x^2+1$ are $x=-1.68$ , $x=-0.421$ , $x=0.421$ , and $x=1.68$ .Therelative minima of the function exist at points $\left(-1.225,-3.5\right)$ and $\left(1.225,-3.5\right)$ , and $f_{\min }=3.5$ . The relative maxima of the function at exits at point $\left(0,1\right)$ and $f_{\max .}=1$ .