Q. 26 - Precalculus Enhanced with Graphing Utilities

Question

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

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

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

$$\begin{gathered} f\left(x\right) = -x^2+3 \\ g\left(x\right)=-3x+3 \end{gathered}$$

Step-by-Step Solution

Answer

The required solution sets are 

(a) $x=-\sqrt{3},\sqrt{3}$

(b) $x=1$

(c) $x=0,3$

(d) $\left\{x:-\sqrt{3}<x<\sqrt{3}\right\}$

(e) $x\ge 1$

(f) $\left\{x:0<x<3\right\}$

(g) $\left\{x:-\sqrt{2}\le x\le \sqrt{2}\right\}$

Part (a) Step 1. Given information

• The given functions are $$\begin{gathered} f\left(x\right) = -x^2+3 \\ g\left(x\right)=-3x+3 \end{gathered}$$.
• The equation is $$\begin{gathered} f\left(x\right) = -x^2+3 \\ g\left(x\right)=-3x+3 \end{gathered}$$.

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

• Plot the graph of the function.

• From the graph, it can be observed that $f\left(x\right)=0$ when $x=-\sqrt{3},\sqrt{3}$.

Part (b) Step 1. Given information

• The given functions are $$\begin{gathered} f\left(x\right) = -x^2+3 \\ g\left(x\right)=-3x+3 \end{gathered}$$
• The equation is $g\left(x\right)=0$.

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

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

• From the graph, it can be observed that $g\left(x\right)=0$when $x=1$.

Part (c) Step 1. Given information

• The given functions are $$\begin{gathered} f\left(x\right) = -x^2+3 \\ g\left(x\right)=-3x+3 \end{gathered}$$
• 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 $(0,3) and (3,-6)$.
• So, $f\left(x\right) = g\left(x\right)$at $x=0,3$.

Part (d) Step 1. Given information

• The given functions are $$\begin{gathered} f\left(x\right) = -x^2+3 \\ g\left(x\right)=-3x+3 \end{gathered}$$
• 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 is above the horizontal axis.
• According to the graph obtained in step 2 of part (b), the curve is above the horizontal axis when $-\sqrt{3}<x<\sqrt{3}$.
• So, the solution set is $\left\{x:-\sqrt{3}<x<\sqrt{3}\right\}$.

Part (e) Step 1. Given information

• The given functions are $$\begin{gathered} f\left(x\right) = -x^2+3 \\ g\left(x\right)=-3x+3 \end{gathered}$$
• The inequality is $g\left(x\right)\le 0$.

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

• $g\left(x\right)\le 0$ 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 $x\ge 1$.

Part (f) Step 1. Given information

• The given functions are $$\begin{gathered} f\left(x\right) = -x^2+3 \\ g\left(x\right)=-3x+3 \end{gathered}$$
• 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 $0<x<3$.
• So, the solution set of the inequality is $\left\{x:0<x<3\right\}$.

Part (g) Step 1. Given information

• The given functions are $$\begin{gathered} f\left(x\right) = -x^2+3 \\ g\left(x\right)=-3x+3 \end{gathered}$$
• The inequality is $f\left(x\right)\ge 1$.

Part (g) Step 2. Read the graph

• The inequality holds when the curve lies above the value 1 on the vertical axis.
• From the graph, the curve is above 1 when $-\sqrt{2}\le x\le \sqrt{2}$.
• So, the solution set of the inequality is $\left\{x:-\sqrt{2}\le x\le \sqrt{2}\right\}$.