I don't understand how to do these questions. it's a lot of questions that I keep getting wrong.

Question

1. f(x) = 3x+5 and g(x) = 3x^2 +3x

After simplifying: (f o g )=

1. f(x) = 7x+7 and g(x) = 3 - x^2

Calculate: a. f (g(0))=

b. g (f (0))=

1. f(x) = x^4 +2, g(x) = x-9, h(x)= square root of x then

f o g o h(x)=

1. f(x)= 1/x-5 and g(x)=2/x +5

Then (f o g)(x)=

( g o f)(x)=

1. f(x)= 1/x-3 and g(x)= 5x +7

Then (f o g)(2)=

(f o g)(x)=

1. f(x)= 7 - square root of x^2 +1 and g(x)=x-7

Then (f o g)(1)=

(f o g)(x)=

1. f(x)= 5x+5. Determine (f o f o f) (x)

( f o f o f) (x)=

Answer 1

f =3x+5, g=3x^2+3x

f(g(x))= 3(3x^2+3x)+5

= 9x^2+9x+5

f=7x+7, g=3-x^2

f(g(0))=f(3)=21+7=28

g(f(0)= g(7)=3-49=-46

f=x^4+2 g=x-9, h=sqr(x)

f(g(h)))= f(g(sqr(x^4+2)=f((sqr(x^4+2)-9))

=(sqr(x^4+2)-9)^4+2

f=1/(x-5), g=2/(x+5)

f(g(x))= f(2/(x+5))=1/((2/(x+5))-5)= 1/((2-5x-25)/(x+5))

=(x+5)/(-5x-23)

g(f(x))=g(1/(x-5))=2/((1/(x-5) +5)= 2/((1+5x-25)/(x-5))

= (x-5)/(5x-24)

f=1/(x-3), g=5x+7

f(g(2))= f(17)= 1/(17-3)=1/14

f(g(x)=f(5x+7)= 1/(5x+4)

f=5x+5

f(f(f(x)))= f(f(5x+5))=f(5(5x+5)+5)=f(25x+30)= 5(25x+30)+5

= 125x+155

f=7-sqr(x^2+1), g=x-7

f(g(1))=f(-6)= 7-sqr37

f(g(x))=f(x-7))=7-sqr((x-7)^2+1)

UNLESS you really meant

f=7-sqr(x^2)+1, then

f(g(0))= 7-7+1=1

f(g(x))=f(x-7)= 8- sqr(x^2-14x+49)