Chapter 2, Problem 39RE

a

To determine

To determine the interval in which the polynomial function will definitely have real root with the help of graph and intermediate value theorem.

Answer to Problem 39RE

The required intervals are: $x\in (-3,-2)$ , $x\in (-1,0)$ and $x\in (0,1)$ .

Explanation of Solution

Given:

  $f(x)=x^3+2x^2-x-1$

Graph:

Using graphing utility, the graph of above function is,

  

Interpretation:

Using the graph, consider interval of length 1, then

f (-3) = -7 and f (-2) = 1

Since, $f\left(-3\right)\times f\left(-2\right)<0$

According to Intermediate value theorem, there will be a real root present in the interval $x\in (-3,-2)$ .

Again,

f (-1) = 1 and f (0) = -1

Since, $f\left(-1\right)\times f\left(0\right)<0$

According to Intermediate value theorem, there will be a real root present in the interval $x\in (-1,0)$ .

And,

f (0) = -1 and f (1) = 1

Since, $f\left(0\right)\times f\left(1\right)<0$

According to Intermediate value theorem there will be a real root present in the interval $x\in (0,1)$

Conclusion:

Hence, in the intervals, $x\in (-3,-2)$ , $x\in (-1,0)$ and $x\in (0,1)$ , three real roots are definitely found.

b

To determine

To estimate the real roots of the given function using the graph and verify it using the table properties of the graphing tool.

Answer to Problem 39RE

The real zeroes are: -2.25,-0.55 and 0.8.

Explanation of Solution

Given:

  $f(x)=x^3+2x^2-x-1$

Graph:

Using graphing utility, the graph of above function is,

  

Interpretation:

From the graph, it clear that the curve intersects the x -axis at the points (-2.25, 0), (-0.55, 0) and (0.8, 0), which means that, -2.25,-0.55 and 0.8 are the three real zeroes of the given polynomial function

Verification of sub-part (a):

Table feature:

  

Graph:

  

When the above table and graph are compared, it is clear that, the zeros of the function are found in the interval which was considered in sub-part (a).

Conclusion:

Hence, the real zeros of the function are -2.25,-0.55 and 0.8.