Chapter 2.8, Problem 21E

Solution

a.

To Solve:

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

  $x\in ℝ$

Given:

  $f\left(x\right)=\left(x^2+4\right)\left(2x^2+3\right)$

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 $x^2+4$ of the polynomial function $f\left(x\right)=\left(x^2+4\right)\left(2x^2+3\right)$ .

  $\begin{array}{cc}& x^2+4=0\\ \Rightarrow & x^2=-4\end{array}$

Note that $x^2+4$ 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 $2x^2+3$ of the polynomial function $f\left(x\right)=\left(x^2+4\right)\left(2x^2+3\right)$ .

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

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

The polynomial function $f\left(x\right)=\left(x^2+4\right)\left(2x^2+3\right)$ has no zero in $ℝ$ . Thus, it can be said that $f\left(x\right)$ cannot change its sign.

Determine the sign of $f\left(x\right)=\left(x^2+4\right)\left(2x^2+3\right)$ in $ℝ$ . Take a test point $x=0\in ℝ$ and evaluate $f\left(x\right)$ :

  $\begin{array}{cc}& f\left(x\right)=\left(x^2+4\right)\left(2x^2+3\right)\\ \Rightarrow & f\left(0\right)=\left(0+4\right)\left(0+3\right)\\ \Rightarrow & f\left(0\right)=\left(4\right)\left(3\right)\\ \Rightarrow & f\left(0\right)=12\\ \Rightarrow & f\left(0\right)>0\end{array}$

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

Conclusion:

The set $x\in ℝ$is the solution for the inequality $f\left(x\right)>0$ .

b.

To Solve:

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

  $x\in ℝ$

Given:

  $f\left(x\right)=\left(x^2+4\right)\left(2x^2+3\right)$

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 $f\left(x\right)=\left(x^2+4\right)\left(2x^2+3\right)$ has no zero in $ℝ$ .

  $f\left(x\right)$ is positive in $ℝ$ . The solution to $f\left(x\right)>0$ is $x\in ℝ$ .

Calculations:

  $f\left(x\right)$ is positive 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, $ℝ$ .

Conclusion:

The solution of $f\left(x\right)\ge 0$ is $x\in ℝ$ .

c.

To Solve:

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

No solution.

Given:

  $f\left(x\right)=\left(x^2+4\right)\left(2x^2+3\right)$

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 $x\in ℝ$ .

Calculations:

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

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

Conclusion:

There is no solution to $f\left(x\right)<0$ .

d.

To Solve:

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

No Solution.

Given:

  $f\left(x\right)=\left(x^2+4\right)\left(2x^2+3\right)$

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 $x\in ℝ$ .

Calculations:

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

It follows that the solution to $f\left(x\right)\le 0$ is the empty set .

Conclusion:

There is no solution to $f\left(x\right)\le 0$ .