Answer
(a.)$(f\circ g)(x) =14x^2-62$
(b.)$(g\circ f)(x) =98x^2+28x-7$
(c.)$(f\circ g)(2) =-6$.
Work Step by Step
The given functions are
$f(x)=7x+1$ and $g(x)=2x^2-9$.
(a) $(f\circ g)(x) = f(g(x))$
Replace $x$ with $g(x)$ in the function $f(x)$.
$f(g(x))=7(g(x))+1$
Plug value of $g(x)$ in the right hand side.
$f(g(x))=7(2x^2-9)+1$
$f(g(x))=14x^2-63+1$
$f(g(x))=14x^2-62$.
Hence, $(f\circ g)(x) =14x^2-62$.
(b) $(g\circ f)(x) = g(f(x))$
Replace $x$ with $f(x)$ in the function $g(x)$.
$g(f(x))=2(f(x))^2-9$
Plug value of $f(x)$ in the right hand side.
$g(f(x))=2(7x+1)^2-9$
$g(f(x))=2(49x^2+1+14x)-9$
$g(f(x))=98x^2+2+28x-9$
$g(f(x))=98x^2+28x-7$.
Hence, $(g\circ f)(x) =98x^2+28x-7$.
(c) Replace $x$ with $2$ in the part (a) solution.
$(f\circ g)(2) =14(2)^2-62$
$(f\circ g)(2) =14(4)-62$
$(f\circ g)(2) =56-62$
$(f\circ g)(2) =-6$.
