Question

homework 16: section 3.4
score: 67.5/160 answered: 7/16
question 7
given that $f(x)=x^2 - 3x$ and $g(x)=x + 9$, calculate
(a) $(f\circ g)(x)=$
(b) $(g\circ f)(x)=$
(c) $(f\circ f)(x)=$
(d) $(g\circ g)(x)=$

Explanation:

Part (a)

Step1: Recall composition definition

The composition \((f \circ g)(x)\) means \(f(g(x))\). So we substitute \(g(x)\) into \(f(x)\).
Given \(f(x)=x^2 - 3x\) and \(g(x)=x + 9\), we replace \(x\) in \(f(x)\) with \(g(x)=x + 9\).

Step2: Substitute and simplify

\[

$$\begin{align*} f(g(x))&=(x + 9)^2-3(x + 9)\\ &=x^2+18x + 81-3x-27\\ &=x^2+(18x-3x)+(81 - 27)\\ &=x^2+15x + 54 \end{align*}$$

\]

Step1: Recall composition definition

The composition \((g \circ f)(x)\) means \(g(f(x))\). So we substitute \(f(x)\) into \(g(x)\).
Given \(g(x)=x + 9\) and \(f(x)=x^2-3x\), we replace \(x\) in \(g(x)\) with \(f(x)=x^2-3x\).

Step2: Substitute and simplify

\[

$$\begin{align*} g(f(x))&=(x^2-3x)+9 \end{align*}$$

\]
(We just substitute \(x^2 - 3x\) for \(x\) in \(g(x)=x + 9\), so we get \(x^2-3x + 9\))

Step1: Recall composition definition

The composition \((f \circ f)(x)\) means \(f(f(x))\). So we substitute \(f(x)\) into \(f(x)\) again.
Given \(f(x)=x^2-3x\), we replace \(x\) in \(f(x)\) with \(f(x)=x^2-3x\).

Step2: Substitute and simplify

\[

$$\begin{align*} f(f(x))&=(x^2-3x)^2-3(x^2-3x)\\ &=x^4-6x^3 + 9x^2-3x^2 + 9x\\ &=x^4-6x^3+(9x^2-3x^2)+9x\\ &=x^4-6x^3+6x^2 + 9x \end{align*}$$

\]

Answer:

\(x^2 + 15x + 54\)

Part (b)