Q. 30 - Precalculus Enhanced with Graphing Utilities

Question

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

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

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

$$\begin{gathered} f\left(x\right)=x^2-2x+1 \\ g\left(x\right)=-x^2+1 \end{gathered}$$

Step-by-Step Solution

Answer

The solution set is

$$\begin{gathered} a) \{1\left\} \\ b\right)\{-1,1\} \\ c) \{1\left\} \\ d\right)\left\{x\right|x<1 or x>1\left\} \\ e\right)\left\{x\right|x\le -1 or x\ge 1\left\} \\ f\right) \left\{x\right|x<0 or x>1\left\} \\ g\right)\left\{x\right|x\le 0 or x\ge 2\} \end{gathered}$$

Part (a) Step 1. Given information

The given function is $f\left(x\right)=x^2-2x+1$.

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

Plot the graph of the function.

From the graph, is it observed that $f\left(x\right)=0$when $x=\left\{1\right\}$

Part (b) Step 1. Given information

The given function is $g\left(x\right)=-x^2+1$.

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 $f\left(x\right)=x^2-2x+1,g\left(x\right)=-x^2+1$.

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 $(1,0)$.
• So, $f\left(x\right)=g\left(x\right)$ at $x=1$.

Part (d) Step 1. Given Information

The function is $f\left(x\right)=x^2-2x+1$.

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

Part (e) Step 1. Given Information

The given equation is $g\left(x\right)=-x^2+1$.

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

Part (f) Step 1. Given Information

The given equations are $f\left(x\right)=x^2-2x+1,g\left(x\right)=-x^2+1$.

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 green color) on the graph.
• From the graph, it can be observed that $f\left(x\right)>g\left(x\right)$ for $x\le 0 or x\ge 1$.
• So, the solution set of the inequality is $x\le 0 or x\ge 1$.

Part (g) Step 1. Given information

The given equation is $f\left(x\right)=x^2-2x+1$

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

$\begin{array}{rcl}{x}^2-2x+1& =& 1\\ x^2-2x& =& 0\\ x(x-2)& =& 0\\ x& =& 0,2\end{array}$

• From the graph, the curve is above 1 when $x\le 0 or x\ge 2$.
• So, the solution set is $\left\{x\right|x\le 0 or x\ge 2\}$.