Solutions to Problem Set 31

Math 101-06

11-29-2017

[Inequalities]

1. Solve the following inequality. Write your answer using either inequality notation or interval notation.

$$x^2 - 5 x - 14 > 0$$

Let $f(x) = x^2 - 5 x - 14 = (x -
   7)(x + 2)$ .

$f(x) = 0$ for $x = 7$ and $x = -2$ .

$f(x)$ is defined for all x.

Plug test numbers in $x < -2$ , $-2 < x < 7$ , and $x > 7$ into f:

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & Test point x & & -3 & & 0 & & 8 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & $f(x)$ & & 10 & & -14 & & 10 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

$$\hbox{\epsfxsize=2.5in \epsffile{inequalities-brackets1.eps}}$$

I want the values of x for which $x^2 - 5 x - 14$ is positive. The solution is $x <
   -2$ or $x > 7$ . In interval notation, this is $(-\infty, -2) \cup (7, \infty)$ .


2. Solve the following inequality. Write your answer using either inequality notation or interval notation.

$$4 x - x^2 \le 0$$

Let $f(x) = 4 x - x^2 = x(4 -
   x)$ .

$f(x) = 0$ for $x = 0$ and $x = 4$ .

$f(x)$ is defined for all x.

Plug test numbers in $x < 0$ , $0 < x < 4$ , and $x > 4$ into f:

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & Test point x & & -1 & & 1 & & 5 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & $f(x)$ & & -5 & & 3 & & -5 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

$$\hbox{\epsfxsize=2.5in \epsffile{inequalities-brackets2.eps}}$$

I want the values of x for which $4 x - x^2$ is negative or equal to 0. The solution is $x
   \le 0$ or $x \ge 4$ . In interval notation, this is $(-\infty, 0] \cup [4, \infty)$ .


3. Solve the following inequality. Write your answer using either inequality notation or interval notation.

$$x^3 + 5 x < 6 x^2$$

I need to have 0 on one side, so rewrite the inequality as

$$x^3 - 6 x^2 + 5 x < 0.$$

Let $f(x) = x^3 - 6 x^2 + 5 x =
   x(x - 1)(x - 5)$ .

$f(x) = 0$ for $x = 0$ , $x = 1$ , and $x = 5$ .

$f(x)$ is defined for all x.

Plug test numbers in $x < 0$ , $0 < x < 1$ , $1 < x < 5$ , and $x > 5$ into f:

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & Test point x & & -1 & & 0.5 & & 2 & & 6 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $f(x)$ & & -12 & & 1.125 & & -6 & & 30 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

$$\hbox{\epsfxsize=2.5in \epsffile{inequalities-brackets3.eps}}$$

I want the values of x for which $x^3 - 6 x^2 + 5 x$ is negative. The solution is $x <
   0$ or $1 < x < 5$ . In interval notation, this is $(-\infty, 0) \cup (1, 5)$ .


4. Solve the following inequality. Write your answer using either inequality notation or interval notation.

$$\dfrac{3 - x}{x + 8} < 0$$

Let $f(x) = \dfrac{3 - x}{x +
   8}$ .

$f(x) = 0$ for $x = 3$ .

$f(x)$ is undefined for $x = -8$ .

Plug test numbers in $x < -8$ , $-8 < x < 3$ , and $x > 3$ into f:

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & Test point x & & -9 & & 0 & & 4 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & $f(x)$ & & -12 & & 0.375 & & $-0.08333 \ldots$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

$$\hbox{\epsfxsize=2.5in \epsffile{inequalities-brackets4.eps}}$$

I want the values of x for which $\dfrac{3 - x}{x + 8}$ is negative. The solution is $x <
   -8$ or $x > 3$ . In interval notation, this is $(-\infty, -8) \cup (3, \infty)$ .


5. Solve the following inequality. Write your answer using either inequality notation or interval notation.

$$\dfrac{x(x - 4)}{x + 2} > 0$$

Let $f(x) = \dfrac{x(x - 4)}{x +
   2}$ .

$f(x) = 0$ for $x = 0$ and $x = 4$ .

$f(x)$ is undefined for $x = -2$ .

Plug test numbers in $x < -2$ , $-2 < x < 0$ , $0 < x < 4$ , and $x > 4$ into f:

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & Test point x & & -3 & & -1 & & 1 & & 5 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $f(x)$ & & -21 & & 5 & & -1 & & $0.71428 \ldots$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

$$\hbox{\epsfxsize=2.5in \epsffile{inequalities-brackets5.eps}}$$

I want the values of x for which $\dfrac{x(x - 4)}{x + 2}$ is positive. The solution is $-2 <
   x < 0$ or $x > 4$ . In interval notation, this is $(-2, 0) \cup (4, \infty)$ .


6. Solve the following inequality. Write your answer using either inequality notation or interval notation.

$$\dfrac{x^2 - 6 x + 9}{x^2 + 5 x} > 0$$

Let $f(x) = \dfrac{x^2 - 6 x +
   9}{x^2 + 5 x} = \dfrac{(x - 3)^2}{x(x + 5)}$ .

$f(x) = 0$ for $x = 3$ .

$f(x)$ is undefined for $x = 0$ and $x = -5$ .

Plug test numbers in $x < -5$ , $-5 < x < 0$ , $0 < x < 3$ , and $x > 3$ into f:

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & Test point x & & -6 & & -1 & & 1 & & 4 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $f(x)$ & & 13.5 & & -4 & & $0.66666 \ldots$ & & $0.02777 \ldots$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

$$\hbox{\epsfxsize=2.5in \epsffile{inequalities-brackets6.eps}}$$

I want the values of x for which $\dfrac{x^2 - 6 x + 9}{x^2 + 5 x}$ is positive. The solution is $x < -5$ , $0 < x < 3$ , or $x > 3$ . In interval notation, this is $(-\infty, -5) \cup (0, 3) \cup (3,
   \infty)$ .


Strength doesn't come from physical capacity. It comes from indomitable will. - Mahatma Gandhi


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