Question

if $f(x) = x^2 + 7$, $g(x) = x - 8$, and $h(x) = \sqrt{x}$, then
$(f \circ g)(x) = \square$
$(g \circ f)(x) = \square$
$(h \circ g)(x) = \square$

Explanation:

Step1: Find \((f \circ g)(x)\)

To find the composition \((f \circ g)(x)\), we substitute \(g(x)\) into \(f(x)\). So we replace every \(x\) in \(f(x)\) with \(g(x)=x - 8\).
\(f(g(x))=(x - 8)^{2}+7\)
Expand \((x - 8)^{2}\) using the formula \((a - b)^{2}=a^{2}-2ab + b^{2}\), where \(a = x\) and \(b = 8\).
\((x - 8)^{2}=x^{2}-16x + 64\)
Then \(f(g(x))=x^{2}-16x + 64+7=x^{2}-16x + 71\)

Step2: Find \((g \circ f)(x)\)

To find \((g \circ f)(x)\), we substitute \(f(x)\) into \(g(x)\). So we replace every \(x\) in \(g(x)\) with \(f(x)=x^{2}+7\).
\(g(f(x))=(x^{2}+7)-8\)
Simplify the expression: \(g(f(x))=x^{2}+7 - 8=x^{2}-1\)

Step3: Find \((h \circ g)(x)\)

To find \((h \circ g)(x)\), we substitute \(g(x)\) into \(h(x)\). So we replace every \(x\) in \(h(x)\) with \(g(x)=x - 8\).
\(h(g(x))=\sqrt{x - 8}\)

Answer:

\((f \circ g)(x)=x^{2}-16x + 71\)
\((g \circ f)(x)=x^{2}-1\)
\((h \circ g)(x)=\sqrt{x - 8}\)