Given that f(x) = 3x + 1 and g(x) = the quantity of 4x plus 2 over 3, solve for g(f(0)).

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Question

Given that f(x) = 3x + 1 and g(x) = the quantity of 4x plus 2 over 3, solve for g(f(0)).

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anonymous · Answer

$$\frac{ 4x + 2 }{ 3 }$$

solomonzelman · Answer

4x+2
----
3                is the g(x)

to find   f(g(x)),  plug in  what the f(x) equals for x,  INTO g(x).

Can you do that ?

solomonzelman · Answer

I mean that would be g(f(x)),   not f(g(x))

solomonzelman · Answer

$\LARGE\color{black}{    g(x)=\frac{4x+2}{3}  }$


$\LARGE\color{black}{    g(f(x))=\frac{4(3x+1)+2}{3}= \frac{12x+4+2}{3}= \frac{12x+6}{3} }$

$\LARGE\color{black}{  = \frac{3(4x+2)}{3}=4x+2 }$

solomonzelman · Answer

$\LARGE\color{black}{    g(f(0))=4(0)+2=?}$

anonymous · Answer

sorry was afk. $$g(f(0)) = 4(0) + 2 = 2? $$

anonymous · Answer

Ok so A. these type of functions have always been difficult to me, thanks for the help