Explore: Inverses of Functions Algebraically Solutions

Inverses of Functions Algebraically Solutions

The inverse of a function can be found algebraically by interchanging, or switching, the domain and range values.
When given a function, find its inverse by switching the variable x with the variable y .
Then solve for the inverse by isolating y in the new equation, which gives you the inverse of the original function.
To check that you have found the inverse, determine if $f (a) = b and g (b) = a$ .

Note

You can work with any value of x if you check the function and inverse. The values in the examples have been chosen so that you can discuss the same values. You can also complete the checks with ordered pairs $(a, b)$ and $(b, a)$ using mental math.

Function notation f(x) is used to represent functions and their inverses (when the inverse is also a function).
The notation for the inverse of a function is $f^{- 1} (x)$ , and is read “the inverse of the function” or “f inverse.”
The “–1” in the $f^{- 1} (x)$ notation is superscript ; not an exponent. It cannot be written as a fraction. $f^{- 1} (x) \neq \frac{1}{f (x)}$
It is important to note that, as with relations, not every inverse of a function will also be a function .

Example 6

Find the inverse of the function. Then check your work using f(6).

$f (x) = \frac{3}{x - 4}$

Plan
Write as a “y =” equation
Switch x and y
Solve for y
Check

Implement

$y = \frac{3}{x - 4} x = \frac{3}{y - 4}$

$x (y - 4) = 3 y - 4 = \frac{3}{x} y = \frac{3}{x} + 4 or f^{- 1} (x) = \frac{3}{x} + 4$

Explain

Write as y =
Switch x and y
Solve for y

Check

$f (6) = \frac{3}{6 - 4} f (6) = \frac{3}{2}$

$f^{- 1} (\frac{3}{2}) = \frac{3}{\frac{3}{2}} + 4 = 2 + 4 f^{- 1} (\frac{3}{2}) = 6$

Find f(6)

Find $f^{- 1} (\frac{3}{2})$

The inverse is correct because $f (6) = \frac{3}{2}$ and $f^{- 1} (\frac{3}{2}) = 6 .$

Note

Recall that the two relations R and Q are inverses of one another, if and only if every ordered pair in R is $(a, b)$ and every ordered pair in Q is $(b, a)$ where $\{a, b \in ℂ\} .$

Example 7

Find the inverse of the function. Check your work.

$h (x) = \frac{5}{8} x - 7$

$\begin{array}{rcl} y & = & \frac{5}{8} x - 7 \\ x & = & \frac{5}{8} y - 7 \\ x + 7 & = & \frac{5}{8} y \\ (\frac{8}{5}) (x + 7) & = & \frac{5}{8} y (\frac{8}{5}) \\ y & = & \frac{8}{5} x + \frac{56}{5} \end{array}$

$Check h (8) h (8) = \frac{5}{8} (8) - 7 = - 2 y = \frac{8}{5} (- 2) + \frac{56}{5} = \frac{40}{5} = 8$

Example 8

Find the inverse of the function. Check your work.

$g (x) = \frac{1}{4} x^{2} + 5, \{x | x \geq 0\}$

$\begin{array}{rcl} y & = & \frac{1}{4} x^{2} + 5 \\ x & = & \frac{1}{4} y^{2} + 5 \\ x - 5 & = & \frac{1}{4} y^{2} \\ (4) (x - 5) & = & \frac{1}{4} y^{2} (4) \\ \sqrt{4 (x - 5)} & = & \sqrt{y^{2}} \\ y & = & 2 \sqrt{x - 5} \end{array}$

$Check g (2) g (2) = \frac{1}{4} {(2)}^{2} + 5 = \frac{4}{4} + 5 = 6 y = 2 \sqrt{(6) - 5} = 2 \sqrt{1} = 2$

Note

Only $x \geq 0$ needs to be considered from $g (x)$ , therefore the ± symbol is not needed when the square root is taken.