If $f (x) = 2 x + 8$ $f(x)=2x+8$ and $g (x) = 3 x - 2$ $g(x)=3x-2$ , how do you find $f (g (1))$ $f(g(1))$ ?

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Answer 1 · 2017-03-12T12:25:02

The answer is $= 10$ $=10$

This is a composition of functions

$f (x) = 2 x + 8$ $f(x)=2x+8$

$g (x) = 3 x - 2$ $g(x)=3x-2$

$f (g (x)) = f (3 x - 2) = 2 (3 x - 2) + 8$ $f(g(x))=f(3x-2)=2(3x-2)+8$

$= 6 x - 4 + 8 = 6 x + 4$ $=6x-4+8=6x+4$

Therefore,

$f (g (1)) = (6 \cdot 1) + 4 = 10$ $f(g(1))=(6*1)+4=10$

If f(x)=2x+8f(x)=2x+8f(x)=2x+8 and g(x)=3x-2g(x)=3x−2g(x)=3x-2, how do you find f(g(1))f(g(1))f(g(1))?