Q. 31 - Precalculus Enhanced with Graphing Utilities

Question

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

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

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

Step-by-Step Solution

Answer

The solution set is,

$$\begin{gathered} a\left) \\ b\right) \\ c\left) \\ d \\ e\right) \\ f\left) \\ g\right)\left\{x\right|x\le \frac{-1-\sqrt{13}}{2} or x\ge \frac{-1+\sqrt{13}}{2}\} \end{gathered}$$

Part a) Given Information

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

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 $x=\{-1,2\}$.

Part (b) Step 1. Given information

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

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 $x=\{-1,1\}$.

Part (c) Step 1. Given Information

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

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 $(0,-2)$.
• So, $f\left(x\right)=g\left(x\right)$ at $x=0$.

Part (d) Step 1. Given Information

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

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 $x<-1 or x>2$.
• So, the solution set is $\left\{x\right|x<-1 or x>2\}$.

Part (e) Step 1. Given Information

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

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 $-2\le x\le 1$.
• So, the solution set is $\left\{x\right|-2\le x\le 1\}$.

Part (f) Step 1. Given Information

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

Part (f) Step 2. Observe from the graph

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

Part (g) Step 1. Given information

The given equation is $f\left(x\right)=x^2-x-2$

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 $f\left(x\right)=1$.

$\begin{array}{rcl}{x}^2-x-2& =& 1\\ x^2-x-3& =& 0\\ x& =& \frac{-\left(1\right)\pm \sqrt{{(-1)}^2-4\left(1\right)(-3)}}{2\left(1\right)}\\ & =& \frac{-1\pm \sqrt{13}}{2}\end{array}$

• From the graph, the curve is above 1 when $x\le \frac{-1-\sqrt{13}}{2} or x\ge \frac{-1+\sqrt{13}}{2}$.
• So, the solution set is $\left\{x\right|x\le \frac{-1-\sqrt{13}}{2} or x\ge \frac{-1+\sqrt{13}}{2}\}$.