Chapter 2.8, Problem 22E

Solution

a.

To Solve:

The inequality $f\left(x\right)>0$ .

No Solution.

Given:

  $f\left(x\right)=\left(x^2+1\right)\left(-2-3x^2\right)$

Concepts Used:

The zeroes of a polynomial function are the set of zeroes of all its factors.

Polynomials are continuous everywhere in $ℝ$ . So they can change signs only at the points where they cross the $x$ -axis.

The solution to $f\left(x\right)>0$is the region where the curve $y=f\left(x\right)$ lies above the $x$ -axis. This represents the region where $f\left(x\right)$ is positive.

Calculations:

Determine the zeroes of the factor $x^2+1$ of the polynomial function $f\left(x\right)=\left(x^2+1\right)\left(-2-3x^2\right)$ .

  $\begin{array}{cc}& x^2+1=0\\ \Rightarrow & x^2=-1\end{array}$

Note that $x^2+1$ cannot be zero for any real number $x$ because there is no real number whose square is a negative number.

Determine the zeroes of the factor $-2-3x^2$ of the polynomial function $f\left(x\right)=\left(x^2+1\right)\left(-2-3x^2\right)$ .

  $\begin{array}{cc}& -2-3x^2=0\\ \Rightarrow & -3x^2=2\\ \Rightarrow & x^2=\frac{2}{-3}\end{array}$

Note that $-2-3x^2$ cannot be zero for any real number $x$ because there is no real number whose square is a negative number.

The polynomial function $f\left(x\right)=\left(x^2+1\right)\left(-2-3x^2\right)$ has no zero in $ℝ$ . Thus, it can be said that $f\left(x\right)$ cannot change its sign.

Determine the sign of $f\left(x\right)=\left(x^2+1\right)\left(-2-3x^2\right)$ in $ℝ$ . Take a test point $x=0\in ℝ$ and evaluate $f\left(x\right)$ :

  $\begin{array}{cc}& f\left(x\right)=\left(x^2+1\right)\left(-2-3x^2\right)\\ \Rightarrow & f\left(0\right)=\left(0+1\right)\left(-1+0\right)\\ \Rightarrow & f\left(0\right)=\left(1\right)\left(-1\right)\\ \Rightarrow & f\left(0\right)=-1\\ \Rightarrow & f\left(0\right)<0\end{array}$

  $f\left(x\right)$ is negative in $ℝ$ .

Conclusion:

There exists no solution for the inequality $f\left(x\right)>0$ .

b.

To Solve:

The inequality $f\left(x\right)\ge 0$ .

No Solution.

Given:

  $f\left(x\right)=\left(x^2+1\right)\left(-2-3x^2\right)$

Concepts Used:

Polynomials are continuous everywhere in $ℝ$ . So they can change signs only at the points where they cross the $x$ -axis.

The solution to $f\left(x\right)\ge 0$is the region where the curve $y=f\left(x\right)$ lies on or above the $x$ -axis. This represents the region where $f\left(x\right)$ is not negative.

For a polynomial function $f\left(x\right)$ , the solution to $f\left(x\right)>0$ is a subset of the solution to $f\left(x\right)\ge 0$ . The solution to $f\left(x\right)\ge 0$ is the union of solution to $f\left(x\right)>0$ and solution set of $f\left(x\right)=0$ .

Known from previous part:

The polynomial function $f\left(x\right)=\left(x^2+1\right)\left(-2-3x^2\right)$ has no zero in $ℝ$ .

  $f\left(x\right)$ is negative in $ℝ$ .

Calculations:

  $f\left(x\right)$ is negative in $ℝ$ so the solution set of $f\left(x\right)=0$ is null.

It follows that the solution to $f\left(x\right)\ge 0$ equals the solution to $f\left(x\right)>0$ , namely, null .

Conclusion:

There exists no solution for the inequality $f\left(x\right)\ge 0$ .

c.

To Solve:

The inequality $f\left(x\right)<0$ .

  $x\in ℝ$

Given:

  $f\left(x\right)=\left(x^2+1\right)\left(-2-3x^2\right)$

Concepts Used:

Polynomials are continuous everywhere in $ℝ$ . So they can change signs only at the points where they cross the $x$ -axis.

The solution to $f\left(x\right)\ge 0$is the region where the curve $y=f\left(x\right)$ lies on or above the $x$ -axis. This represents the region where $f\left(x\right)$ is not negative.

The solution to $f\left(x\right)>0$ is a subset of the solution to $f\left(x\right)\ge 0$ . The solution to $f\left(x\right)\ge 0$ is the union of solution to $f\left(x\right)>0$ and solution set of $f\left(x\right)=0$ .

For a polynomial function $f\left(x\right)$ , the solution to $f\left(x\right)<0$ is the complement of solution set $f\left(x\right)\ge 0$ considering $ℝ$ as universal set.

Known from previous part:

The solution to $f\left(x\right)\ge 0$ is null.

Calculations:

The complement of null is $ℝ-\varphi =ℝ$ .

It follows that the solution to $f\left(x\right)<0$ is $x\in ℝ$ .

Conclusion:

  $x\in ℝ$ is the solution to $f\left(x\right)<0$ .

d.

To Solve:

The inequality $f\left(x\right)\le 0$ .

  $x\in ℝ$

Given:

  $f\left(x\right)=\left(x^2+1\right)\left(-2-3x^2\right)$

Concepts Used:

Polynomials are continuous everywhere in $ℝ$ . So they can change signs only at the points where they cross the $x$ -axis.

The solution to $f\left(x\right)\ge 0$is the region where the curve $y=f\left(x\right)$ lies on or above the $x$ -axis. This represents the region where $f\left(x\right)$ is not negative.

The solution to $f\left(x\right)>0$ is a subset of the solution to $f\left(x\right)\ge 0$ . The solution to $f\left(x\right)\ge 0$ is the union of solution to $f\left(x\right)>0$ and solution set of $f\left(x\right)=0$ .

For a polynomial function $f\left(x\right)$ , the solution to $f\left(x\right)\le 0$ is the complement of solution set $f\left(x\right)>0$ considering $ℝ$ as the universal set.

Known from previous part:

The solution to $f\left(x\right)>0$ is null .

Calculations:

The complement of null is $ℝ-\varphi =ℝ$ .

It follows that the solution to $f\left(x\right)\le 0$ is $x\in ℝ$

Conclusion:

  $x\in ℝ$ is the solution to $f\left(x\right)\le 0$ .