# Math Help - exercises (functions )

1. ## exercises (functions )

1. If f(x) = x^2 + 3x - 7 and g(x) = 3x - 4, then g(f(-3)) =

2. If F(x) = √X^2=1 G(x) = 3x - 2 then a formula for (FoG)(x) is :

3. If F(x) = x/x-1 G(x) = 2x + 1 then a formula for (FoG)(x) is

4. If f(x) = g(x) = √x-1 [f(x)] = 1/X^2+1 then g[f(x)] =

2. 1.

f(x) = x^2 + 3x - 7
g(x) = 3x - 4

g(f(x)) = 3(x^2 + 3x - 7) - 4 = 3x^2 + 9x - 25
g(f(-3)) = 3(-3)^2 + 9(-3) - 25 = -25

3. 4. If f(x) = √x-1 and g(x) =1/X^2+1 then g[f(x)] =

4. How about trying yourself? You are told what g(x) is. To find g(f(x)), replace the "x" in that formula by the formula for f(x).

You learn mathematics by doing it, not by watching someone else do it!

5. Originally Posted by dragovx
1. If f(x) = x^2 + 3x - 7 and g(x) = 3x - 4, then g(f(-3)) =

2. If F(x) = √X^2=1 G(x) = 3x - 2 then a formula for (FoG)(x) is :

3. If F(x) = x/x-1 G(x) = 2x + 1 then a formula for (FoG)(x) is

4. If f(x) = g(x) = √x-1 [f(x)] = 1/X^2+1 then g[f(x)] =

I'm not fimmiliar with the $(FoG)(x)$ notation, but I will assume it means $f(g(x))$, if it means otherwise please tell me and I will repost the correct answers. I'll do number 2 and 3 since the others are already done:

$[2] f(g(x) = \sqrt{(3x-2)^2} = \sqrt{9x^2 -12x + 4}$

$[3] f( g(x) ) = \frac{2x+1}{2x+1-1} = \frac{2x+1}{2x}$