Integrated Math 3 / Unit 9: Sequences, Series and Functions Review
9.4 Composition of Functions
Notes
The composition $(f \circ g)(x) = f(g(x))$ means apply $g$ first, then $f$.
- Order matters. With $f(x) = 2x + 1$ and $g(x) = x^2$: $(f \circ g)(x) = 2x^2 + 1$ but $(g \circ f)(x) = (2x + 1)^2$.
- To evaluate, work from the inside out: $f(g(3)) = f(9) = 19$.
- The domain of $f(g(x))$ requires $x$ to be in the domain of $g$, and $g(x)$ to be in the domain of $f$.
Practice
Let $f(x) = 2x + 1$ and $g(x) = x^2$. What is $f(g(3))$?
$g(3) = 9$, then $f(9) = 2(9) + 1 = 19$.
Let $f(x) = 2x + 1$ and $g(x) = x^2$. What is $(g \circ f)(x)$?
Replace $x$ in $g$ with $2x + 1$: $(2x + 1)^2$.
Let $\displaystyle f(x) = \sqrt{x}$ and $g(x) = x - 4$. What is the domain of $(f \circ g)(x)$?
$\displaystyle (f \circ g)(x) = \sqrt{x - 4}$, so $x - 4 \geq 0$.
Ready for more? Take the Unit 9 test.