need help with precalculus (check my answer)

Question

f(x) =  ; g(x) = 8x - 13

Find f(g(x)).

f(x) =  ; g(x) = 8x - 11

Find f(g(x)).

f(x) = 5x + 3; g(x) = 6x - 5

Find f/g

my answer : (f/g)(x) =  domain {x|x ≠ }

Answer 1

f(x) = sqr(x+9) g(x) = 8x-13

f(g(x) = f(8x-13) = sqr(8x-13+9)= sqr(8x-4)

domain is x > or = 1/2

f(x) = sqr(x+7), g(x) = 8x-11

f(g(x) = f(8x-1) = sqr(8x-11+7) = sqr(8x-4)

domain is x> or = 1/2

range is any non-negative number (the square root sign means positive square root)

graphically the function f is a rightward opening parabola with vertex (-7, 0)

f(x)=5x+3, g(x) = 6x-5

f/g = f(x)/g(x) = (5x+3)/(6x-5)

domain is any real number except where x makes a denominator = 0. x cannot = 5/6

6(5/6)-5 = 5-5=0

graphically the function has a discontinuity at x=5/6

and an asymptote is x=5/6. the function is undefined at x=5/6. f(x) approaches infinity and negative infinity at x=5/6

Answer 2

a) If f(x) = √(x + 9) and g(x) = 8x - 13 then f(g(x)) = √(8x - 4) with a domain restricted to x > 1/2

b) If f(x) = √(x + 7) and g(x) = 8x - 11 then f(g(x)) = √(8x - 4) with a domain restricted to x > 1/2

c) If f(x) = 5x + 3 and g(x) = 6x - 5 then f/g = (5x + 3)/(6x - 5) with a domain restricted to x ≠ 5/6. Note that the domain restriction comes from the fact that the denominator cannot be zero. That means 6x - 5 ≠ 0 or 6x ≠ 5 or x ≠ 5/6