Q. 28 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 25–32, use the given functions f and g.

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

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

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

Step-by-Step Solution

Answer

The required solutions are:

$$\begin{gathered} \left(a\right) x=\{-2,2\} \\ \left(b\right) x=-2 \\ \left(c\right) x=\{-2,3\} \\ \left(d\right) \left\{x|-2<x<2\right\} \end{gathered}$$

$$\begin{gathered} \left(e\right) x\ge -2 \\ \left(f\right) \left\{x|-2<x<3\right\} \\ \left(g\right) \left\{x|-\sqrt{3}\le x\le \sqrt{3}\right\} \end{gathered}$$

Part (a) Step 1. Given information

• The given functions are $$\begin{gathered} f\left(x\right)=-x^2+4 \\ g\left(x\right)=-x-2 \end{gathered}$$.
• The equation is $f\left(x\right)=0$.

Part (a) Step 2: Plot the graph and observe

• Plot the graph of the function f.

• From the graph, it can be observed that $f\left(x\right)=0$when $x=-2,2$.

Part (b) Step 1. Given information

• The given functions are $$\begin{gathered} f\left(x\right)=-x^2+4 \\ g\left(x\right)=-x-2 \end{gathered}$$.
• The equation is $g\left(x\right)=0$.

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

• Plot the line g in the graph obtained for the first function.

• From the graph, it can be observed that  $g\left(x\right)=0$ when $x=-2$.

Part (c) Step 1. Given information

• The given functions are$$\begin{gathered} f\left(x\right)=-x^2+4 \\ g\left(x\right)=-x-2 \end{gathered}$$$$\begin{gathered} f\left(x\right)=-x^2+4 \\ g\left(x\right)=-x-2 \end{gathered}$$.
• The equation is $f\left(x\right)=g\left(x\right)$

Part (c) Step 2. Read the Graph

• For $f\left(x\right)=g\left(x\right)$, the curves of both the functions must intersect.
• From the graph, it can be observed that the functions intersect at $(-2,0),(3,-5)$.
• So, the given equation holds when $x=\left\{-2,3\right\}$.

Part (d) Step 1. Given information

• The given functions are $$\begin{gathered} f\left(x\right)=-x^2+4 \\ g\left(x\right)=-x-2 \end{gathered}$$.
• The inequality is $f\left(x\right)>0$.

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

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

Part (e) Step 1. Given information

• The given functions are $$\begin{gathered} f\left(x\right)=-x^2+4 \\ g\left(x\right)=-x-2 \end{gathered}$$.
• The inequality is $g\left(x\right)\le 0$.

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

• The inequality holds when the line of the function is on or below the horizontal axis.
• From the graph, the line is on or below the axis for $x\ge -2$.

Part (f) Step 1. Given information

• The given functions are $$\begin{gathered} f\left(x\right)=-x^2+4 \\ g\left(x\right)=-x-2 \end{gathered}$$.
• The inequality is $f\left(x\right)>g\left(x\right)$.

Part (f) Step 2. Read the graph

• The inequality holds when the curve lies above the line on the graph.
• From the graph, it can be observed that the curve is above the line when $-2<x<3$.
• So, the solution set of the inequality is$\left\{x|-2<x<3\right\}$.

Part (g) Step 1. Given information

• The given functions are $$\begin{gathered} f\left(x\right)=-x^2+4 \\ g\left(x\right)=-x-2 \end{gathered}$$.
• The inequality is $f\left(x\right)\ge 1$.

Part (g) Step 2. Read the graph

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

$\begin{array}{rcl}-x^2+4& =& 1\\ -x^2& =& -3\\ x& =& -\sqrt{3},\sqrt{3}\end{array}$

• From the graph, the curve is above 1 when $\left\{x|-\sqrt{3}\le x\le \sqrt{3}\right\}$.