Practice 1 Solutions

Graph the inverse on the provided coordinate plane.

Graph the inverse on the coordinate plane. Explain whether or not the graph and its inverse represent functions.

The given graph is a function because it passes the VLT. The inverse of the graph is not a function because it fails the VLT.

The given graph and the inverse are functions because they pass the VLT.

The given graph and the inverse are functions because they pass the VLT.

Find the inverse of the function algebraically. Then graph the function and its inverse on the coordinate plane. Explain whether or not the graph and its inverse represent functions.

$k (x) = 3 (x + 1) - 4$
$\begin{array}{rcl} x & = & 3 (y + 1) - 4 \\ x & = & 3 y + 3 - 4 \\ x + 1 & = & 3 y \\ y & = & \frac{1}{3} (x + 1) \end{array}$
$k^{- 1} (x) = \frac{1}{3} (x + 1)$  

The given equation and the inverse are functions because they pass the VLT.

$h (x) = x^{2}$  
$x = y^{2}$
$y = \pm \sqrt{x}$

The given equation is a function, but the inverse is not a function because it does not pass the VLT.

Note

9) The inverse of $x^{2}$ is $\pm \sqrt{x}$ . To plot on the coordinate plane, first graph $y = \sqrt{x}$ and then graph $y = - \sqrt{x}$ . The inverse will not be a function. Because the inverse is not a function, do not use $f^{- 1} (x)$ .

f (x) = 2
 x = 2
The given equation is a function, but the inverse is not a function because it does not pass the VLT.

$g (x) = - \frac{4}{3} x - 4$  
$\begin{array}{rcl} x & = & - \frac{4}{3} y - 4 \\ x + 4 & = & - \frac{4}{3} y \\ - \frac{3}{4} (x + 4) & = & y \end{array}$  
$\begin{array}{rcl} g^{- 1} (x) & = & - \frac{3}{4} x - 3 \end{array}$  

The given equation and the inverse are functions because they both pass the VLT.

For the given graph: 

- Name the domain and range for the graph and its inverse. 
- Explain whether or not the graph represents a function. 
- If the graph is a function, determine if it is one-to-one. 
- If the graph is a function, determine if the inverse is also a function.

Note

When checking the graphs using the VLT and HTL, you do not need to sketch the lines on the graph, but you can if you choose. You do not need to graph the inverse of the function, but you may choose to if you would like to see what it looks like.

Given 
Domain: $\{\begin{array}{rcl} x | x & \in & R \end{array}\}$  
Range: $\{\begin{array}{rcl} y | y & \in & R \end{array}\}$  

Inverse 
Domain: $\{\begin{array}{rcl} x | x & \in & R \end{array}\}$  
Range: $\{\begin{array}{rcl} y | y & \in & R \end{array}\}$  

f (x) is a function, but is not one-to-one because it fails the HLT. The inverse is not a function because it fails the VLT.

Given 
Domain: $\{\begin{array}{rcl} x | x & \in & R \end{array}\}$  
Range: $\{\begin{array}{rcl} y | y & \in & R \end{array}\}$  

Inverse 
Domain: $\{\begin{array}{rcl} x | x & \in & R \end{array}\}$  
Range: $\{\begin{array}{rcl} y | y & \in & R \end{array}\}$  

f (x) is a function, and is one-to-one. The inverse is also a function.

Given 
Domain: $\{\begin{array}{rcl} x | x & \in & R \end{array}\}$  
Range: $\{\begin{array}{rcl} y | y & \geq & 3 \end{array}\}$  

Inverse 
Domain: $\{\begin{array}{rcl} x | x & \geq & 3 \end{array}\}$  
Range: $\{\begin{array}{rcl} y | y & \in & R \end{array}\}$  

f (x) is a function, but is not one-to-one. The inverse is not a function.

Given 
Domain: $\{\begin{array}{rcl} x | x & \geq & 0 \end{array}\}$  
Range: $\{\begin{array}{rcl} y | y & \geq & 2 \end{array}\}$  

Inverse 
Domain: $\{\begin{array}{rcl} x | x & \geq & 2 \end{array}\}$  
Range: $\{\begin{array}{rcl} y | y & \geq & 0 \end{array}\}$  

f (x) is a function, and is one-to-one. The inverse is also a function.

To continue, return to the Online Lesson.