Test 10 (Lessons 19-20): Functions and Their Inverses Solutions

Find the inverse of the relation. Show this as a table and a mapping.
R = {(–3, 4), (0, 3), (4, 1), (5, –2)}

$R^{- 1}$

x	y
–2	5
1	4
3	0
4	–3

Is the relation a function? Explain. Is the inverse of the relation a function? Explain.

Sample: The relation is a function because all of the domain values are unique. The inverse of the relation is also a function for the same reason. When the domain does not repeat, a function exists.

Find the inverse of the function.

$f (x) = - \frac{2}{3} x + 5$

$\begin{array}{rcl} y = - \frac{2}{3} x + 5 x = - \frac{2}{3} y + 5 (Solve for y) x - 5 = - \frac{2}{3} y \\ - \frac{3}{2} (x - 5 = - \frac{2}{3} y) \\ y = - \frac{3}{2} x + \frac{15}{2} \end{array}$

$f^{- 1} (x) = - \frac{3}{2} x + \frac{15}{2}$

$j (x) = \sqrt[3]{7 - x} - 3$

$\begin{array}{rcl} y & = & \sqrt[3]{7 - x} - 3 \\ x & = & \sqrt[3]{7 - y} - 3 \\ x + 3 & = & \sqrt[3]{7 - y} \\ {(x + 3)}^{3} & = & {(\sqrt[3]{7 - y})}^{3} \\ {(x + 3)}^{3} & = & 7 - y \\ {(x + 3)}^{3} - 7 & = & - y \\ y & = & - {(x + 3)}^{3} + 7 \end{array}$

$\begin{array}{rcl} j^{- 1} (x) & = & - {(x + 3)}^{3} + 7 \end{array}$
$or j^{- 1} (x) = - x^{3} - 9 x^{2} - 27 x - 20$

Determine the inverse of the function algebraically. Then graph the function and its inverse on the coordinate plane.

$g (x) = x^{2} + 3$ where $x \geq 0$

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

$\begin{array}{rcl} g^{- 1} (x) & = & \sqrt{x - 3} \end{array}$

Determine the inverse of the function algebraically. Then graph the function and its inverse on the coordinate plane.

$f (x) = 2 x - 1$

$\begin{array}{rcl} y & = & 2 x - 1 \\ x & = & 2 y - 1 \\ x + 1 & = & 2 y \\ y & = & \frac{x + 1}{2} \end{array}$

$\begin{array}{rcl} f^{- 1} (x) & = & \frac{x + 1}{2} or f^{- 1} (x) = \frac{1}{2} x + \frac{1}{2} \end{array}$

Name the domain and range for the given function and its inverse.

$\begin{array}{l} Given \\ Domain : \{x | x \in ℝ, x \geq - 4\} \\ Range : \{y | y \in ℝ, x \geq - 2\} \end{array}$

$\begin{array}{l} Inverse \\ Domain : \{x | x \in ℝ, x \geq - 2\} \\ Range : \{y | y \in ℝ, x \geq - 4\} \end{array}$

Note

You may draw the inverse, but it is not required.

Explain whether or not the function is one-to-one and whether or not the inverse is also a function.

Sample: The function is one-to-one. It passes the HLT. The inverse is also a function because the given passes the HLT, which makes the inverse pass a VLT.

Graph the inverse.

Explain how you know that the inverse in problem 11 is also a function.

Sample: The graph is a function because it passes the vertical line test (VLT). The graph of f (x) also passes the horizontal line test (HLT) which makes $f^{- 1} (x)$ also a function.