Chapter 2.2, Problem 96E

a.

To determine

To find:The interval of unit length in which zero of the given polynomial function lie definitely, using Intermediate Value Theorem and graphing utility.

Answer to Problem 96E

The required intervals are $\left[-3,-2\right]$ , $\left[-1,0\right]$ , and $\left[0,1\right]$ .

Explanation of Solution

Given:

The polynomial function is $f\left(x\right)=-2x^3-6x^2+3$ .

Formula/ concept used:

The Intermediate Value Theorem states that $f\left(b\right)$ Let “ a ” and “ b ”be real numbers such that $a<b$ . If a polynomial function such that $f\left(a\right)\ne f\left(b\right)$ , then in interval $\left[a,b\right]$ takes on every value between $f\left(a\right)$ and $f\left(b\right)$ . If $f\left(a\right)$ and $f\left(b\right)$ are of opposite sign then a zeros surely lie in interval $\left[a,b\right]$ .

Graph:

The graph of function $f\left(x\right)=-2x^3-6x^2+3$ is shown in Figure-1 here.

We have $f(a=-3)=3$ and $f(b=-2)=-5$ , thus $f\left(-3\right)>0$ and $f(-2)<0$ , hence, there exists a zero of $f\left(x\right)=-2x^3-6x^2+3$ in the intervalof unit length $\left[-3,-2\right]$ .We see that $f\left(-2.810\right)=0$ , i.e., $x=-2.810$ is zero of polynomial function $f\left(x\right)=-2x^3-6x^2+3$ .

  

We have $f(a=-1)=-1$ and $f(b=0)=3$ , thus $f\left(-1\right)<0$ and $f\left(0\right)>0$ , hence, there exists a zero of $f\left(x\right)=-2x^3-6x^2+3$ in the interval of unit length $\left[-1,0\right]$ . We see that $f\left(-0.832\right)=0$ , i.e., $x=-0.832$ is zero of polynomial function $f\left(x\right)=-2x^3-6x^2+3$ .

We have $f(a=0)=3$ and $f(b=1)=-5$ , thus $f\left(0\right)>0$ and $f\left(1\right)<0$ , hence, there exists a zero of $f\left(x\right)=-2x^3-6x^2+3$ in the intervalof unit length $\left[0,1\right]$ .We see that $f\left(0.642\right)=0$ , i.e., $x=0.642$ is zero of polynomial function $f\left(x\right)=-2x^3-6x^2+3$ .

Conclusion:

There exists a zero of $f\left(x\right)=-2x^3-6x^2+3$ in each interval $\left[-3,-2\right]$ , $\left[-1,0\right]$ , and $\left[0,1\right]$ .

b.

To determine

To find: The real zeros of polynomial function given in part (a) using graphing utility.

Answer to Problem 96E

The zeros of function $f\left(x\right)=-2x^3-6x^2+3$ are $x=-2.810$ , $x=-0.832$ , and $x=0.642$ .

Explanation of Solution

Given:

The polynomial function and the graph of function $f\left(x\right)=-2x^3-6x^2+3$ in part (a).

Concept used:

The value of x where the graph of function intersect or touch the x -axis is the zero of function.

Calculations:

From graph Figure-1, in part (a) we see that graph of function $f\left(x\right)$ intersect x -axis at points $\left(-2.810,0\right)$ , $\left(-0.832,0\right)$ , and $\left(0.642,0\right)$ .

Hence, $x=-2.810$ , $x=-0.832$ , and $x=0.642$ are the zeros of function $f\left(x\right)=-2x^3-6x^2+3$ .

Conclusion:

The points $x=-2.810$ , $x=-0.832$ , and $x=0.642$ are zeros of function $f\left(x\right)=-2x^3-6x^2+3$ .

c.

To determine

To verify: The answers of part (a) using table feature of graphing utility.

Explanation of Solution

Given:

The zeros of function $f\left(x\right)=-2x^3-6x^2+3$ lie in intervals $\left[-3,-2\right]$ , $\left[-1,0\right]$ , and $\left[0,1\right]$ .

Verification:

From the graph in Figure-1, we obtain a table of some solutions of the given polynomial function $f\left(x\right)=-2x^3-6x^2+3$ as:

  

There is change of the function $f\left(x\right)$ in intervals $\left[-3,-2\right]$ , $\left[-1,0\right]$ , and $\left[0,1\right]$ each of length 1, these results are same as of part (a).

Thus, the results of part (a) stands verified.