Inverse and Composite Functions [H]
Inverse and Composite Functions
Two operations on functions extend the basic function idea: the inverse function reverses what \( f \) does, and a composite function chains two functions together. Both require care with the order in which operations are applied.
Inverse Functions
The inverse of \( f \), written \( f^{-1} \), is the function that undoes \( f \). If \( f(3) = 11 \) then \( f^{-1}(11) = 3 \). The key property is:
\[ f(f^{-1}(x)) = x quad \text{and} \quad f^{-1}(f(x)) = x \]Method for finding \( f^{-1}(x) \):
- Write \( y = f(x) \).
- Rearrange to make \( x \) the subject (i.e. express \( x \) in terms of \( y \)).
- Replace \( y \) with \( x \) — the result is \( f^{-1}(x) \).
Example: \( f(x) = 3x - 5 \). Let \( y = 3x - 5 \). Then \( x = \dfrac{y+5}{3} \). So \( f^{-1}(x) = \dfrac{x+5}{3} \).
Composite Functions
The composite function \( fg(x) \) means "apply \( g \) first, then apply \( f \) to the result":
\[ fg(x) = f(g(x)) \]The order matters: \( fg(x) eq gf(x) \) in general. Think of it as reading right to left — the rightmost function is applied first.
\[ \text{If } f(x) = x^2 + 1 \text{ and } g(x) = 2x: \] \[ fg(x) = f(g(x)) = f(2x) = (2x)^2 + 1 = 4x^2 + 1 \] \[ gf(x) = g(f(x)) = g(x^2+1) = 2(x^2+1) = 2x^2 + 2 \]These are different unless specific values of \( x \) happen to make them equal.
Worked Examples
Find \( f^{-1}(x) \) for \( f(x) = \dfrac{2x+1}{x-3} \) and state any domain restrictions.
Let \( y = \frac{2x+1}{x-3} \). Multiply both sides by \( (x-3) \):
\[ y(x-3) = 2x+1 \implies yx - 3y = 2x+1 \implies x(y-2) = 3y+1 \] \[ x = \frac{3y+1}{y-2} \implies f^{-1}(x) = \frac{3x+1}{x-2} \]\( f \) is undefined at \( x = 3 \) (domain of \( f \): \( x \neq 3 \)). \( f^{-1} \) is undefined at \( x = 2 \) (domain of \( f^{-1} \): \( x \neq 2 \)).
Given \( f(x) = x^2 - 1 \) and \( g(x) = 3x + 2 \), find \( fg(x) \), \( gf(x) \) and evaluate \( fg(2) \).
\[ fg(x) = f(g(x)) = f(3x+2) = (3x+2)^2 - 1 = 9x^2 + 12x + 3 \] \[ gf(x) = g(f(x)) = g(x^2-1) = 3(x^2-1)+2 = 3x^2-1 \] \[ fg(2) = 9(4)+12(2)+3 = 36+24+3 = 63 \]Given \( f(x) = 2x + 3 \) and \( g(x) = x^2 \), solve \( fg(x) = 21 \). Also solve \( f^{-1}(x) = g(1) \).
\[ fg(x) = f(g(x)) = f(x^2) = 2x^2 + 3 = 21 \implies x^2 = 9 \implies x = \pm 3 \] \[ f^{-1}(x): \text{ let } y = 2x+3 \implies x = \frac{y-3}{2} \implies f^{-1}(x) = \frac{x-3}{2} \] \[ g(1) = 1. \quad f^{-1}(x) = 1 \implies \frac{x-3}{2} = 1 \implies x = 5 \]Key Takeaways
- To find \( f^{-1}(x) \): write \( y = f(x) \), solve for \( x \) in terms of \( y \), then replace \( y \) with \( x \).
- Key property: \( f(f^{-1}(x)) = x \). Use this to verify an inverse.
- \( fg(x) = f(g(x)) \): apply \( g \) first (rightmost), then \( f \). Order is crucial — \( fg eq gf \) in general.
- The domain of \( f^{-1} \) equals the range of \( f \), and vice versa.
- To solve \( fg(x) = k \): find a formula for \( fg(x) \) first, then solve the resulting equation.