Chapter 2.8, Problem 23E

Solution

a.

To Solve:

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

  $x\in \left(-\infty ,\frac{4}{3}\right)\cup \left(\frac{4}{3},\infty \right)$

Given:

  $f\left(x\right)=\left(2x^2-2x+5\right){\left(3x-4\right)}^2$

Concepts Used:

Finding zeroes of a quadratic polynomial using completing the square method.

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 $2x^2-2x+5$ of the polynomial function $f\left(x\right)=\left(2x^2-2x+5\right){\left(3x-4\right)}^2$ .

  $\begin{array}{cc}& 2x^2-2x+5=0\\ \Rightarrow & x^2-x+\frac{5}{2}=0\\ \Rightarrow & x^2-2\left(\frac{1}{2}\right)\left(x\right)+{\left(\frac{1}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2+\frac{5}{2}=0\\ \Rightarrow & \left(x^2-2\left(\frac{1}{2}\right)\left(x\right)+{\left(\frac{1}{2}\right)}^2\right)-\frac{1}{4}+\frac{10}{4}=0\\ \Rightarrow & {\left(x-\frac{1}{2}\right)}^2+\frac{9}{4}=0\\ \Rightarrow & {\left(x-\frac{1}{2}\right)}^2=-\frac{9}{4}\end{array}$

Note that $2x^2-2x+5$ 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 $3x-4$ of the polynomial function $f\left(x\right)=\left(2x^2-2x+5\right){\left(3x-4\right)}^2$ .

  $\begin{array}{cc}& 3x-4=0\\ \Rightarrow & 3x=4\\ \Rightarrow & x=\frac{4}{3}\end{array}$

The only zero $x=\frac{4}{3}$ of the polynomial function $f\left(x\right)=\left(2x^2-2x+5\right){\left(3x-4\right)}^2$ divides the set of real numbers $ℝ$ into two open intervals, namely, $\left(-\infty ,\frac{4}{3}\right)$ , and $\left(\frac{4}{3},\infty \right)$ . Since none of these intervals include any zero of $f\left(x\right)$ , it can be said that $f\left(x\right)$ cannot change sign in any of these intervals.

Determine the sign of $f\left(x\right)=\left(2x^2-2x+5\right){\left(3x-4\right)}^2$ in the interval $\left(-\infty ,\frac{4}{3}\right)$ . Take a test point $x=0$ in the interval $\left(-\infty ,\frac{4}{3}\right)$ and evaluate $f\left(x\right)$ :

  $\begin{array}{cc}& f\left(x\right)=\left(2x^2-2x+5\right){\left(3x-4\right)}^2\\ \Rightarrow & f\left(0\right)=\left(0+5\right){\left(0-4\right)}^2\\ \Rightarrow & f\left(0\right)=\left(5\right)\left(16\right)\\ \Rightarrow & f\left(0\right)=80\\ \Rightarrow & f\left(0\right)>0\end{array}$

  $f\left(x\right)$ is positive in the interval $\left(-\infty ,\frac{4}{3}\right)$ .

Determine the sign of $f\left(x\right)=\left(2x^2-2x+5\right){\left(3x-4\right)}^2$ in the interval $\left(\frac{4}{3},\infty \right)$ . Take a test point $x=2$ in the interval $\left(\frac{4}{3},\infty \right)$ and evaluate $f\left(x\right)$ :

  $\begin{array}{cc}& f\left(x\right)=\left(2x^2-2x+5\right){\left(3x-4\right)}^2\\ \Rightarrow & f\left(2\right)=\left(8-4+5\right){\left(6-4\right)}^2\\ \Rightarrow & f\left(2\right)=\left(9\right){\left(2\right)}^2\\ \Rightarrow & f\left(2\right)=36\\ \Rightarrow & f\left(2\right)>0\end{array}$

  $f\left(x\right)$ is positive in the interval $\left(\frac{4}{3},\infty \right)$ .

Conclusion:

The polynomial function $f\left(x\right)=\left(2x^2-2x+5\right){\left(3x-4\right)}^2$ is positive in the intervals $\left(-\infty ,\frac{4}{3}\right)$ and $\left(\frac{4}{3},\infty \right)$ .

The polynomial function $f\left(x\right)=\left(2x^2-2x+5\right){\left(3x-4\right)}^2$ is zero only at $x=\frac{4}{3}$ .

  $\left(-\infty ,\frac{4}{3}\right)\cup \left(\frac{4}{3},\infty \right)$ 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:

  $f\left(x\right)=\left(2x^2-2x+5\right){\left(3x-4\right)}^2$

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:

  $\left(-\infty ,\frac{4}{3}\right)\cup \left(\frac{4}{3},\infty \right)$ is the solution set for the inequality $f\left(x\right)>0$ .

  $\left\{\frac{4}{3}\right\}$ is the solution set for $f\left(x\right)=0$

Calculations:

The solution to $f\left(x\right)\ge 0$ is the union of $\left(-\infty ,\frac{4}{3}\right)\cup \left(\frac{4}{3},\infty \right)$ and $\left\{\frac{4}{3}\right\}$ , that is

  $\left(-\infty ,\frac{4}{3}\right)\cup \left(\frac{4}{3},\infty \right)\cup \left\{\frac{4}{3}\right\}=ℝ$

Conclusion:

  $ℝ$ is the solution for the inequality $f\left(x\right)\ge 0$ .

c.

To Solve:

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

No Solution.

Given:

  $f\left(x\right)=\left(2x^2-2x+5\right){\left(3x-4\right)}^2$

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

  $x=\frac{4}{3}$

Given:

  $f\left(x\right)=\left(2x^2-2x+5\right){\left(3x-4\right)}^2$

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 $\left(-\infty ,\frac{4}{3}\right)\cup \left(\frac{4}{3},\infty \right)$ .

Calculations:

The complement of $\left(-\infty ,\frac{4}{3}\right)\cup \left(\frac{4}{3},\infty \right)$ is $ℝ-\left(-\infty ,\frac{4}{3}\right)\cup \left(\frac{4}{3},\infty \right)=\left\{\frac{4}{3}\right\}$ .

It follows that the solution to $f\left(x\right)\le 0$ is $x=\frac{4}{3}$

Conclusion:

  $x=\frac{4}{3}$ is the solution to $f\left(x\right)\le 0$ .