Expert: Sherman D. - 11/6/2004

Question
f(x)=2x-1
g(x)= cube root of x+1
h(x)= x^2+2x

1) (f/g)(x)
2) (f open circle g)(x)
3) g^-1(x)

Answer
f(x) = 2x - 1
g(x) = cbrt(x + 1)
h(x) = x^2 + 2x

1.)
(f/g)(x) = (2x - 1)/(cbrt(x + 1))
same as saying
(2x - 1)/((x + 1)^(1/3))
multiply top and bottom by (x + 1)^(2/3)
((2x - 1)(x + 1)^(2/3))/(x + 1)
(f/g) = ((2x - 1)(x + 1)^(2/3))/(x - 1)

2.)
(f o g)(x) = f(g(x))
f(cbrt(x + 1)) = 2(cbrt(x + 1)) - 1
f(g(x)) = 2cbrt(x + 1) - 1

3.)
g^-1(x)
y = cbrt(x + 1)
x = cbrt(y + 1)
cube both sides
x^3 = y + 1
subtract both sides by 1
y = x^3 - 1
same as saying
g^-1(x) = x^3 - 1

cbrt stands for cubed root

if you are wanting something other than that, just let me know. Also whats up with having h(x) and then not having a problem involving h(x).