Question

compositions of functions

• the composition of functions is the process of combining two or more functions into a single function.

representations

• in $f(g(x))$, $g(x)$ is the input of $f(x)$
• in $(f \circ g)(x)$, $g(x)$ is the input of $f(x)$
• in $g(f(x))$, $f(x)$ is the input of $g(x)$
• in $(g \circ f)(x)$, $f(x)$ is the input of $g(x)$

examples - directions: perform the indicated operation

1. $f(x) = -2x + 2$, $g(x) = x - 3$ find $f(g(x))$
2. $f(x) = -3x + 1$, $g(x) = 4x + 3$ find $(f \circ g)(x)$
3. $f(x) = -2x + 2$, $g(x) = x - 3$ find $g(f(x))$
4. $f(x) = -3x + 1$, $g(x) = 4x + 3$ find $(g \circ f)(x)$

Explanation:

1. For $f(g(x))$ (Problem 1):

Step1: Substitute $g(x)$ into $f(x)$

$g(x)=x-3$, so replace $x$ in $f(x)$ with $x-3$:
$f(g(x)) = -2(x-3) + 2$

Step2: Expand and simplify

$$\begin{align*} f(g(x)) &= -2x + 6 + 2 \\ &= -2x + 8 \end{align*}$$

2. For $(f \circ g)(x)$ (Problem 2):

Step1: Substitute $g(x)$ into $f(x)$

$g(x)=4x+3$, so replace $x$ in $f(x)$ with $4x+3$:
$(f \circ g)(x) = -3(4x+3) + 1$

Step2: Expand and simplify

$$\begin{align*} (f \circ g)(x) &= -12x - 9 + 1 \\ &= -12x - 8 \end{align*}$$

3. For $g(f(x))$ (Problem 3):

Step1: Substitute $f(x)$ into $g(x)$

$f(x)=-2x+2$, so replace $x$ in $g(x)$ with $-2x+2$:
$g(f(x)) = (-2x+2) - 3$

Step2: Simplify the expression

$$\begin{align*} g(f(x)) &= -2x + 2 - 3 \\ &= -2x - 1 \end{align*}$$

4. For $(g \circ f)(x)$ (Problem 4):

Step1: Substitute $f(x)$ into $g(x)$

$f(x)=-3x+1$, so replace $x$ in $g(x)$ with $-3x+1$:
$(g \circ f)(x) = 4(-3x+1) + 3$

Step2: Expand and simplify

$$\begin{align*} (g \circ f)(x) &= -12x + 4 + 3 \\ &= -12x + 7 \end{align*}$$

Answer:

1. $f(g(x)) = -2x + 8$
2. $(f \circ g)(x) = -12x - 8$
3. $g(f(x)) = -2x - 1$
4. $(g \circ f)(x) = -12x + 7$