How do you find $[f o g] (x)$ $[fog](x)$ and $[g o f] (x)$ $[gof](x)$ given $f (x) = \frac{1}{2} x - 7$ $f(x)=1/2x-7$ and $g (x) = x + 6$ $g(x)=x+6$ ?

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Answer 1 · 2017-03-07T06:49:06

The answers are $[f o g] (x) = \frac{1}{2} x - 4$ $[fog] (x) =1/2x-4$ and $[g o f] (x) = \frac{1}{2} x - 1$ $[gof] (x)=1/2x-1$

This is a composition of functions

$f (x) = \frac{1}{2} x - 7$ $f(x)=1/2x-7$

$g (x) = x + 6$ $g(x)=x+6$

$[f o g] (x) = f (g (x)) = f (x + 6) = \frac{1}{2} (x + 6) - 7$ $[fog] (x) =f(g(x))=f(x+6)=1/2(x+6)-7$

$= \frac{1}{2} x + 3 - 7 = \frac{1}{2} x - 4$ $=1/2x+3-7=1/2x-4$

$[g o f] (x) = g (f (x)) = g (\frac{1}{2} x - 7) = \frac{1}{2} x - 7 + 6$ $[gof] (x)=g(f(x))=g(1/2x-7)= 1/2x-7+6$

$= \frac{1}{2} x - 1$ $=1/2x-1$

How do you find [fog](x)[fog](x) and [gof](x)[gof](x) given f(x)=12x−7f(x)=1/2x-7 and g(x)=x+6g(x)=x+6?