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

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Answer 1 · 2018-03-02T23:42:48

The result is $- 8$ $-8$ .

First, compute $g (- 2)$ $g(-2)$ :

$\Rightarrow g (x) = 3 x - 2$ $color(white)=>g(x)=3x-2$

$\Rightarrow g (- 2) = 3 (- 2) - 2$ $=>g(-2)=3(-2)-2$

$\Rightarrow g (- 2) = - 6 - 2$ $color(white)(=>g(-2))=-6-2$

$\Rightarrow g (- 2) = - 8$ $color(white)(=>g(-2))=-8$

Now, compute $f (- 8)$ $f(-8)$ :

$\Rightarrow f (x) = 2 x + 8$ $color(white)=>f(x)=2x+8$

$\Rightarrow f (- 8) = 2 (- 8) + 8$ $=>f(-8)=2(-8)+8$

$\Rightarrow f (- 8) = - 16 + 8$ $color(white)(=>f(-8))=-16+8$

$\Rightarrow f (- 8) = - 8$ $color(white)(=>f(-8))=-8$

The result is $- 8$ $-8$ .

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(-2))f(g(−2))f(g(-2))?