Q. 29 - Precalculus Enhanced with Graphing Utilities

Question

In problems 25-32, use the given functions f and g.

$$\begin{gathered} \left(a\right) f\left(x\right)=0 \\ \left(b\right) g\left(x\right)=0 \\ \left(c\right) f\left(x\right)=g\left(x\right) \\ \left(d\right)f\left(x\right)>0 \end{gathered}$$

$$\begin{gathered} \left(e\right) g\left(x\right)\le 0 \\ \left(f\right) f\left(x\right)>g\left(x\right) \\ \left(g\right) f\left(x\right)\ge 1 \end{gathered}$$

$$\begin{gathered} f\left(x\right)=x^2-4 \\ g\left(x\right)=-x^2+4 \end{gathered}$$

Step-by-Step Solution

Answer

The solution set is,

$$\begin{gathered} a) \{-2,2\left\} \\ b\right) \{-2,2\} \\ c) \{-2,2\} \end{gathered}$$

$$\begin{gathered} d\left)\right\{x|x<-2 or x>2\} \\ e) \{x|x\le -2 or x\ge 2\} \\ f) \{x|x<-2 or x>2\} \\ g) \{x|x\le -\sqrt{5} or x\ge \sqrt{5}\} \end{gathered}$$

Part (a) Step 1. Given Information

The given function is $f\left(x\right)=x^2-4$.

Part (a) Step 2. PLot the function and observe

Plot the graph of the function.

From the graph, it is observed that $f\left(x\right)=0$ when $x=\{-2,2\}$.

Part (b) Step 1. Given information

The given function is $g\left(x\right)=-x^2+4$.

Part (b) Step 2. Plot the function and observe

Plot the function in the graph obtained for the first function.

From the graph, it is observed that $g\left(x\right)=0$, when $x=\{-2,2\}$.

Part (c) Step 1. Given Information

The functions are $f\left(x\right)=x^2-4,g\left(x\right)=-x^2+4$

Part (c) Step 2. Read the obtained graphs

• For $f\left(x\right)=g\left(x\right)$, the curves must intersect each other.
• From the obtained graphs, it is observed that the graphs intersect at $(-2,0),(2,0)$.
• So, $f\left(x\right)=g\left(x\right)$ at $x=\{-2,2\}$.

Part (d) Step 1. Given Information

The function is $f\left(x\right)=x^2-4$

Part (d) Step 2. Find the region above the horizontal axis.

• $f\left(x\right)>0$ when the curve of the function is above the horizontal axis.
• According to the graph obtained in step 2 of part (b), the curve is above the horizontal axis when $x<-2 or x>2$.
• So, the solution set is $\left\{x\right|x<-2 or x>2\}$.

Part (e) Step 1. Given Information

The given equation is $g\left(x\right)=-x^2+4$

Part (e) Step 2. Find the region on or below the horizontal axis.

• $g\left(x\right)\le 0$when the curve of the function is on or below the horizontal axis. 
• From the graph, the line is on or below the axis for $x\le -2 or x\ge 2$.
• So, the solution set is $\left\{x\right|x\le -2 or x\ge 2\}$.

Part (f) Step 1. Given Information

The given equations are $f\left(x\right)=x^2-4,g\left(x\right)=-x^2+4$.

Part (f) Step 2. Observe from the graph

• The inequality $f\left(x\right)>g\left(x\right)$ represents the when the curve $f\left(x\right)$ (represented by black color) lies above the curve $g\left(x\right)$(represented by red color) on the graph.
• From the graph, it can be observed that $f\left(x\right)>g\left(x\right)$ for $x<-2 or x>2$.
• So, the solution set of the inequality is $\left\{x\right|x<-2 or x>2\}$.

Part (g) Step 1. Given information

The given equation is $-x^2+4$.

Part (g) Step 2. Read the graph

• The inequality holds when the curve $f\left(x\right)$ lies on or above the value 1 on the vertical axis.
• Find the value of $x$ when $f\left(x\right)=1$.

$\begin{array}{rcl}{x}^2-4& =& 1\\ x^2& =& 5\\ x& =& \sqrt{5},-\sqrt{5}\end{array}$

• From the graph, the curve is above 1 when $x\le -\sqrt{5} or x\ge \sqrt{5}$.
• So, the solution set is $\left\{x\right|x\le -\sqrt{5} or x\ge \sqrt{5}\}$.