find the compositions. $f(x) = 3x + 2, g(x) = x^2 - 4$ (a) $(f \\circ g…
\(3x^{2}-10\) ### Part (b)
\(3x^{2}-10\) ### Part (b)
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) $(g \\circ f)(2) = \\square$
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) $(g \\circ f)(2) = \\square$
To find \((f \circ g)(x)\), we need to substitute \(g(x)\) into \(f(x)\). That is, \((f \circ g)(x)=f(g(x))\).
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:
\(f(g(x))=3(x^{2}-4)+2\)
First, distribute the 3: \(3(x^{2}-4)=3x^{2}-12\)
Then add 2: \(3x^{2}-12 + 2=3x^{2}-10\)
To find \((g \circ f)(x)\), we need to substitute \(f(x)\) into \(g(x)\). That is, \((g \circ f)(x)=g(f(x))\).
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:
\(g(f(x))=(3x + 2)^{2}-4\)
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: \(9x^{2}+12x + 4-4=9x^{2}+12x\)
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 \((f \circ g)(x)\).
\((f \circ g)(-1)=3\times(-1)^{2}-10\)
First, calculate \((-1)^{2}=1\), then \(3\times1 = 3\)
Then \(3-10=-7\)
\(3x^{2}-10\)
To find \((f \circ g)(x)\), we need to substitute \(g(x)\) into \(f(x)\). That is, \((f \circ g)(x)=f(g(x))\).
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:
\(f(g(x))=3(x^{2}-4)+2\)
First, distribute the 3: \(3(x^{2}-4)=3x^{2}-12\)
Then add 2: \(3x^{2}-12 + 2=3x^{2}-10\)
\(3x^{2}-10\)
To find \((g \circ f)(x)\), we need to substitute \(f(x)\) into \(g(x)\). That is, \((g \circ f)(x)=g(f(x))\).
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:
\(g(f(x))=(3x + 2)^{2}-4\)
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: \(9x^{2}+12x + 4-4=9x^{2}+12x\)
\(9x^{2}+12x\)
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 \((f \circ g)(x)\).
\((f \circ g)(-1)=3\times(-1)^{2}-10\)
First, calculate \((-1)^{2}=1\), then \(3\times1 = 3\)
Then \(3-10=-7\)
\(-7\)
From part (b), we know that \((g \circ f)(x)=9x^{2}+12x\). To find \((g \circ f)(2)\), we substitute \(x = 2\) into the expression for \((g \circ f)(x)\).
\((g \circ f)(2)=9\times(2)^{2}+12\times2\)
First, calculate \((2)^{2}=4\), then \(9\times4 = 36\) and \(12\times2=24\)
Then \(36 + 24=60\)
\(60\)
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) $(g \\circ f)(2) = \\square$
Top-left cell: 180 Top-right cell: 6 Bottom-left cell: 600 Bottom-right cell: 20 Final product: 806
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