If $f (x) = {(x - 2)}^{2}$ $f(x) = (x - 2)^2$ ; $g (x) = x + 3$ $g (x) = x +3$ , then what is (f + 2g)(4)?

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Answer 1 · 2017-05-26T15:12:16

18

First, find the double of $g (x)$ $g(x)$ :

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

Now add this to $f (x)$ $f(x)$ :

$f + 2 g = {(x - 2)}^{2} + 2 (x + 3)$ $f+2g=(x-2)^2+2(x+3)$

Now plug in $4$ $4$ :

$(f + 2 g) (4) = [{(4 - 2)}^{2} + 2 (4 + 3)]$ $( f + 2g ) ( 4 ) = [(4-2)^2+2(4+3)]$

This results in:

$\Rightarrow [4 + 14] = 18$ $rArr [4+14] = 18$

If f(x)=(x−2)2f(x) = (x - 2)^2; g(x)=x+3g (x) = x +3 , then what is (f + 2g)(4)?