Question

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find the compositions.
$f(x) = 3x + 2, g(x) = x^2 - 4$
(a) $(f \circ g)(x) = \square$
(b) $(g \circ f)(x) = \square$
(c) $(f \circ g)(-1) = \square$
(d) $(g \circ f)(2) = \square$

Explanation:

Part (a)

Step 1: Recall the definition of function composition

To find \((f \circ g)(x)\), we need to substitute \(g(x)\) into \(f(x)\). That is, \((f \circ g)(x)=f(g(x))\).

Step 2: Substitute \(g(x)=x^{2}-4\) into \(f(x)\)

Given \(f(x) = 3x+2\), we replace \(x\) in \(f(x)\) with \(g(x)=x^{2}-4\). So we have:
\(f(g(x))=3(g(x)) + 2\)
Substitute \(g(x)=x^{2}-4\) into the above formula:
\(f(g(x))=3(x^{2}-4)+2\)

Step 3: Simplify the expression

First, distribute the 3: \(3(x^{2}-4)=3x^{2}-12\)
Then add 2: \(3x^{2}-12 + 2=3x^{2}-10\)

Step 1: Recall the definition of function composition

To find \((g \circ f)(x)\), we need to substitute \(f(x)\) into \(g(x)\). That is, \((g \circ f)(x)=g(f(x))\).

Step 2: Substitute \(f(x)=3x + 2\) into \(g(x)\)

Given \(g(x)=x^{2}-4\), we replace \(x\) in \(g(x)\) with \(f(x)=3x + 2\). So we have:
\(g(f(x))=(f(x))^{2}-4\)
Substitute \(f(x)=3x + 2\) into the above formula:
\(g(f(x))=(3x + 2)^{2}-4\)

Step 3: Simplify the expression

First, expand \((3x + 2)^{2}\) using the formula \((a + b)^{2}=a^{2}+2ab + b^{2}\), where \(a = 3x\) and \(b = 2\). So \((3x+2)^{2}=(3x)^{2}+2\times(3x)\times2+2^{2}=9x^{2}+12x + 4\)
Then subtract 4: \(9x^{2}+12x + 4-4=9x^{2}+12x\)

Step 1: Use the result from part (a)

From part (a), we know that \((f \circ g)(x)=3x^{2}-10\). To find \((f \circ g)(-1)\), we substitute \(x=-1\) into the expression for \((f \circ g)(x)\).

Step 2: Substitute \(x = - 1\) into \(3x^{2}-10\)

\((f \circ g)(-1)=3\times(-1)^{2}-10\)

Step 3: Simplify the expression

First, calculate \((-1)^{2}=1\), then \(3\times1 = 3\)
Then \(3-10=-7\)

Answer:

\(3x^{2}-10\)

Part (b)