Question

Let f(x) = Võ - 6 and g(x) = x2 + 5.(fog)(x) =.(gof)(x) =

Answer 1

Recall the following:

`\begin{gathered} (f\circ g)(x)=f(g(x)), \\ (g\circ f)(x)=g(f(x)). \end{gathered}`

Therefore:

`\begin{gathered} (f\circ g)(x)=f(x^2+5)=\sqrt[]{(x^2+5)}-6, \\ (g\circ f)(x)=g(\sqrt[]{x}-6)=(\sqrt[]{x}-6)^2+5. \end{gathered}`

Answer:

Simplifying the above equations we get:

`\begin{gathered} (f\circ g)(x)=\sqrt[]{x^2+5}-6, \\ (g\circ f)(x)=(\sqrt[]{x}-6)^2+5=x-12\sqrt[]{x}+36+5=x-12\sqrt[]{x}+41, \\ (g\circ f)(x)=x-12\sqrt[]{x}+41. \end{gathered}`