Section 1.7 - Inequalities

An inequality says that one expression is greater than, greater than, or equal to, less than, or less than or equal to another.

Property 2 (Properties of Inequality)

Let \(a\), \(b\), and \(c\) represent real numbers.

1. If \(a<b\), then \(a+c<b+c\).

2. If \(a<b\) and \(c>0\), then \(ac<bc\).

3. If \(a<b\) and \(c<0\), then \(ac>bc\).

(Replace \(<\) with {\(>\), \(\le\), or \(\ge\)} and results are similar. Except for when looking at conditions on \(c\).)

Linear Inequalities

Definition 2

A linear inequality in one variable is an inequality that can be written in the form

\[ax+b>0,\]

where \(a\) and \(b\) are real numbers and \(a\ne0\).

(Replace \(<\) with {\(>\), \(\le\), or \(\ge\)} and results are similar.)

Set-Builder Notation	Verbal Description	Interval Notation
\(\{x\,|\,x>a\}\)	The set of real numbers greater than \(a\).	\((a,\infty)\)
\(\{x\,|\,x\ge a\}\)	The set of real numbers greater than or equal to \(a\).	\([a,\infty)\)
\(\{x\,|\,x<a\}\)	The set of real numbers less than \(a\).	\((-\infty,a)\)
\(\{x\,|\,x\le a\}\)	The set of real numbers less than or equal to \(a\).	\((-\infty,a]\)
\(\{x\,|\,a<x<b\}\)	The set of real numbers between \(a\) and \(b\).	\((a,b)\)
\(\{x\,|\,a\le x\le b\}\)	The set of real numbers between \(a\) and \(b\), inclusive.	\([a,b]\)
\(\{x\,|\,a<x\le b\}\)	The set of real numbers greater than \(a\) and less than or equal to \(b\).	\((a,b]\)
\(\{x\,|\,a\le x<b\}\)	The set of real numbers greater than or equal to \(a\) and less than \(b\).	\([a,b)\)
\(\{x\,|\,x<a\text{ or }x>b\}\)	The set of real numbers less than \(a\) or greater than \(b\).	\((-\infty,a)\cup(b,\infty)\)
\(\{x\,|\,x\in\mathbb{R}\}\)	The set of all real numbers	\((-\infty,\infty)\)

Example 14

In Business, the Profit function is the difference between the Revenue function and the Cost function, i.e.,

\[P(x)=R(x)-C(x)\]

where \(x\) is the number of units produced and sold.

Ideally business would like Profit to be positive:

\[P(x)>0.\]

Use \(P(x)=R(x)-C(x)\) and rewrite \(P(x)>0\) in terms of \(R(x)\) and \(C(x)\).

Solution:

\[\begin{align*} P(x) & >0\iff R(x)-C(x)>0\\ & \iff R(x)>C(x) \end{align*}\]

Let \(R(x)=45x\) and \(C(x)=30x+5250\). Find when the profit is positive.

Solution:

\[\begin{align*} R(x) & >C(x)\\ 45x & >30x+5250\\ 15x & >5250\\ x & >\frac{5250}{15}\\ x & >350 \end{align*}\]

Interpret the results.

Solution:

We expect to maintain positive profits when more than 350 units are produced and sold. (Why?)

Quadratic Inequality

Definition 3

A quadratic inequality is an inequality that can be written in the form

\[ax^{2}+bx+c*0,\]

where \(*\) is \(\{<,>,\le,\ge\}\), \(a,b,c\in\mathbb{R}\), and \(a\ne0\).

Example 15

Solve the quadratic inequality

\[x^{2}-x-12<0\]

Solution:

Note \(x^{2}-x-12<0\) occurs at the red curve:

Notice that if we first know when \(x^{2}-x-12=0\) we can start to figure out when \(x^{2}-x-12<0\) .

When is \(x^{2}-x-12=0\)?

\[\begin{split}\begin{aligned} x^{2}-x-12 & =0\\ (x-4)(x+3) & =0\\ x & =4\\ x & =-3\end{aligned}\end{split}\]

When is \(x^{2}-x-12\) positive?

• If \(x\) is \(-4\) then what? (why ask about when \(x=-4\)?)

• If \(x\) is \(5\) then what? (why ask about when \(x=5\)?)

what is \(x^{2}-x-12\) negative?

• If \(x=0\) then what? (why ask when \(x=0\)?)

This method can be generalized in the following way.

Algorithm (Solving quadratic inequality)

1. What was the first thing we did to solve the original inequality? (\(ax^{2}+bx+c=0\))

2. After finding when \(ax^{2}+bx+c=0\) we identified how many subintervals? (three (or two) subintervals)

3. In those three subintervals we did what in those subintervals? (we tested if \(ax^{2}+bx+c\) is positive or negative)

Example 16

Solve

\[2x^{2}+5x-12\ge0\]

Solution:

First, we will solve

\[\begin{split}\begin{aligned} 2x^{2}+5x-12 & =0\\ 2x^{2}+8x-3x-12 & =0\\ 2x(x+4)-3(x+4) & =0\\ (x+4)(2x-3) & =0\\ x & =\{-4,\frac{3}{2}\}\end{aligned}\end{split}\]

Second, identify three subintervals: \((-\infty,-4]\cup[-4,\frac{3}{2}]\cup[\frac{3}{2},\infty)\).

Third,

\[\begin{split}\begin{aligned} 2*(-5)^{2}+5*(-5)-12 & =13\\ 2*(0)^{2}+5*(0)-12 & =-12\\ 2*(2)^{2}+5*(2)-12 & =6\end{aligned}\end{split}\]

From the test values we see the solution set is:

\[\{x\,|\,x\le-4\lor\frac{3}{2}\le x\}=(-\infty,-4]\cup[\frac{3}{2},\infty)\]

Example 17

If an object is launched from ground level with an initial velocity of 144 feet per second, its height \(s\) in feet \(t\) seconds after launching is

\[s=-16t^{2}+144t.\]

When will the object be greater than 128 feet above ground level?

Solution:

Solve the inequality

\[128>-16t^{2}+144t\]

First, we will solve

\[\begin{split}\begin{aligned} 128 & =-16t^{2}+144t\\ 16t^{2}-144t+128 & =0\\ 16\left(t-8\right)\left(t-1\right) & =0\end{aligned}\end{split}\]

The solution set is \(t=\{1,8\}\).

Second, identify the three subintervals

\[[0,1],[1,8],[8,9].\]

Third, test values at each subinterval.

\[\begin{split}\begin{aligned} -16(0)^{2}+144*(0) & =0<128\\ -16(2)^{2}+144*(2) & =224>128\checkmark\\ -16(9)^{2}+144*(9) & =0<128\end{aligned}\end{split}\]

This means the solution set for the inequality

\[\{x\,|\,1\le x\le8\}=[1,8]\]

Other Inequalities

Example 18

Solve

\[\dfrac{6x+1}{2x-3}<6\]

Solution:

First, we solve

\[\dfrac{6x+1}{2x-3}=6\]

and note that \(\dfrac{6x+1}{2x-3}\) is defined for all \(x\) such that \(x\ne\frac{3}{2}\). Since \(x\ne\frac{3}{2}\) we can validly multiply by \(2x-3\) to both sides of the original equation. (It is important to make this observation before multiplying by \(2x-3\) to both sides.)

\[\begin{split}\begin{aligned} \dfrac{6x+1}{2x-3} & =6\\ \cancel{(2x-3)}\left(\dfrac{6x+1}{\cancel{2x-3}}\right) & =6(2x-3)\\ 6x+1 & =12x-18\\ \cancel{6x-6x}+1+18 & =12x-6x\cancel{-18+18}\\ 19 & =6x\\ x & =\{\frac{19}{6}\}.\end{aligned}\end{split}\]

Important to identify:

\[\dfrac{6x+1}{2x-3}=6\iff x=\frac{19}{6}\]

\[\dfrac{6x+1}{2x-3}\text{ is undefined when $x=\frac{3}{2}.$}\]

Second, identify the different subintervals

\[(-\infty,\frac{3}{2}),(\frac{3}{2},\frac{19}{6}],[\frac{19}{6},\infty)\]

Third, test each of the subintervals

\[\begin{split}\begin{aligned} \dfrac{6*(0)+1}{2*(0)-3} & =-\frac{1}{3}<6\checkmark\\ \dfrac{6*(2)+1}{2*(2)-3} & =13>6\\ \dfrac{6*(6.5)+1}{2*(6.5)-3} & =4<6\checkmark\end{aligned}\end{split}\]

This means the solution set for the inequality

\[\{x\,|\,x<\frac{3}{2}\lor x>\frac{19}{6}\}=(-\infty,\frac{3}{2})\cup[\frac{19}{6},\infty)\]

Example 19

Solve

\[-3\le\dfrac{3x-4}{-5}<4\]

Solution:

In this case there is two inequalities to solve

\[-3\le\dfrac{3x-4}{-5}\text{ and }\dfrac{3x-4}{-5}<4\]

First, solve the following two cases:

Solve \(-3=\dfrac{3x-4}{-5}\)

\[\begin{split}\begin{aligned} -3 & =\dfrac{3x-4}{-5}\\ (-5)\cdot\left(-3\right) & =\left(\dfrac{3x-4}{\cancel{-5}}\right)\cancel{(-5)}\\ 15 & =3x-4\\ 15+4 & =3x\cancel{-4+4}\\ 19 & =3x\\ x & =\frac{19}{3}\end{aligned}\end{split}\]

Solve \(\dfrac{3x-4}{-5}=4\)

\[\begin{split}\begin{aligned} \dfrac{3x-4}{-5} & =4\\ (-5)\cdot\left(\dfrac{3x-4}{-5}\right) & =(4)\cdot(-5)\\ 3x-4 & =-20\\ 3x\cancel{-4+4} & =-20+4\\ 3x & =-16\\ x & =-\dfrac{16}{3}\end{aligned}\end{split}\]

Second, identify the different subintervals

\[(-\infty,-\frac{16}{3}),(-\frac{16}{3},\frac{19}{3}),(\frac{19}{3},\infty)\]

Third, test each subinterval for validity

\[-3\le\dfrac{3x-4}{-5}<4\]

Test \(x=-10\)

\[-3\le\dfrac{3(-10)-4}{-5}<4\]

\[-3\le\frac{34}{5}\nless4\]

This means \((-\infty,-\frac{16}{3})\) is not part of the solution set.

Test \(x=0\)

\[-3\le\dfrac{3(0)-4}{-5}<4\]

\[-3\le\frac{4}{5}<4\]

This means \([-\frac{16}{3},\frac{19}{3})\) is part of the solution set.

Test \(x=10\)

\[-3\le\dfrac{3(10)-4}{-5}<4\]

\[-3\nleq-\frac{26}{5}<4\]

This means that \([\frac{19}{3},\infty)\) is not part of the solution set.

Example 20

Solve the following rational inequality.

\[\dfrac{1-x}{x+2}<-1\]

Solution:

Note that \(\dfrac{1-x}{x+2}\) is undefined when \(x=-2\).

First, we will solve

\[\begin{split}\begin{aligned} \dfrac{1-x}{x+2} & =-1\to1-x=-(x+2)\\ & \to1-x=-x-2\\ & \to1\ne2\end{aligned}\end{split}\]

This means for all \(x\) the rational expression will not equal negative one. (Also, \(y=-1\) is a horizontal asymptote for the plot of \(y=\dfrac{1-x}{x+2}\).)

Since there isn’t an \(x\) such that \(\dfrac{1-x}{x+2}=-1\) and \(\frac{1-x}{x+2}\) is undefined at \(x=-2\). We will consider two subintervals:

\[(-\infty,-2)\cup(-2,\infty)\]

Test \(x=-10\)

\[\dfrac{1-(-10)}{(-10)+2}<-1\]

\[-\frac{11}{8}<-1\]

This means \((-\infty,-2)\) is part of the solution set.

Test \(x=0\)

\[\dfrac{1-(0)}{(0)+2}<-1\]

\[\frac{1}{2}\nless-1\]

This means \((-2,\infty)\) is not part of the solution set.

Therefore, the solution set for the inequality is

\[(-\infty,-2)\]

or the set of all \(x\) such that \(x<-2\).

Example 21

Solve the following inequality

\[\dfrac{(x^{2}-4)^{2}}{(x-5)(x-3)}<0\]

Solution:

Note,

\(\dfrac{(x^{2}-4)^{2}}{(x-5)(x-3)}\) is undefined when \(x=5\) and \(x=3\).

\(\dfrac{(x^{2}-4)^{2}}{(x-5)(x-3)}=0\) when \(x=\pm2\).

\(\dfrac{(x^{2}-4)^{2}}{(x-5)(x-3)}=\frac{(x-2)^{2}(x+2)^{2}}{(x-5)(x-3)}\)

This means we have the following subintervals

\[(-\infty,-2)\cup(-2,3)\cup(3,5)\cup(5,\infty)\]

Test \(x=-3\)

\[\frac{(x-2)^{2}(x+2)^{2}}{(x-5)(x-3)}\to\dfrac{(+)(+)}{(-)(-)}=(+).\]

This means \(\dfrac{(x^{2}-4)^{2}}{(x-5)(x-3)}\nless0\) and \((-\infty,-2)\) is not part of the solution set.

Test \(x=0\)

\[\frac{(x-2)^{2}(x+2)^{2}}{(x-5)(x-3)}\to\dfrac{(+)(+)}{(-)(-)}=(+).\]

This means \(\dfrac{(x^{2}-4)^{2}}{(x-5)(x-3)}\nless0\) and \((-2,3)\) is not part of the solution set.

Test \(x=4\)

\[\frac{(x-2)^{2}(x+2)^{2}}{(x-5)(x-3)}\to\dfrac{(+)(+)}{(-)(+)}=(-).\]

This means \(\dfrac{(x^{2}-4)^{2}}{(x-5)(x-3)}<0\) and \((3,5)\) is part of the solution set.

Test \(x=6\)

\[\frac{(x-2)^{2}(x+2)^{2}}{(x-5)(x-3)}\to\dfrac{(+)(+)}{(+)(+)}=(+).\]

This means \(\dfrac{(x^{2}-4)^{2}}{(x-5)(x-3)}\nless0\) and \((5,\infty)\) is not part of the solution set.

therefore, the solution set is

\[(3,5)\]

or the set of all \(x\) such that \(3<x<5\).