Algebra II Help/Functions

Hi i need some help with an algebra2 problem but not just the answer but also some explanation thanks.


Given f(x)=-2x-5 and g(x)=3/4x+6 evaluate

a. g(f(0))

b. f(g(-8))

c.g(f(x))

d. f(g(x))

Thanks a bunch

for example, a.
first take the inside function or f(0), so plug 0 into -2x-5 which gets you -5.
then plug in -5 into g(f(0), so g(-5) and that gets you 3/-26
b. plug -8 into g(x), gets you 3/-26, then you plug that into f(g(-8)) (in the g(-8)) and solve f(3/-26)
and basically, just keep doing that for the others.

I will do an example. When you have (g(f(-8)) Thats like saying g of f of -8. You need to first find f when x is negative 8. since f(X)= -2x-5, it would be 11. So now you need to find g of 11. That is like saying find g when x is 11, so plug 11 into the g equation. the answer is 56/4 and reduced 14.

You can check my math but the concept is there.

1 freebie that is all.

a. g(f(0)) replace f(0) with -2(0)-5 which = -5

so we get g(-5) = 3/4(-5)+6= -(15/4)+6 = -3.75+6 = 2.25

Enjoy the rest because you need to earn this.