1. From your work, I will assume that
f(x) = ³√(x-5)
then f(g(x)) = f(g(x))=³√((x^3+1)-5)
= f(g(x))=³√(x^3 - 4)
test it with some value of x, say x=2
g(2) = 8+1 = 9
f(9) = ³√(9-5) = ³√4
using my answer of f(g(x)) = ³√(x^3 - 4) = ³√(2^3-4) = ³√4
my answer is correct, you just needed the brackets
g(f(x)) = (³√(x-5)^3 + 1 = x-5 + 1 = x - 4
2. Again you will need brackets
f(g(x)) = √(2x-3)
g(f(x)) = yours is correct
3. yes, simplify the exponents by using the exponent rule (x^a)^b = x^(ab)
f(g(x)) = (x^6)^(23) = x^4
g(f(x)) = (x^(2/3))^6 = x^4
4. Did you mean
f(x) = 1/(x^2 - 1) or the way you typed it ?
I need some help with composite functions
where i have to find the f(g(x)) and g(f(x)) on most of them I just couldn't figure out how to multiply....
1) f(x) = ³√x-5 , g(x) = x^3 + 1
So I started on this one and i got confused with the cubed
f(g(x))=³√(x^3+1)-5
then for g(f(x)) = (³√x-5)^3+1
I couldn't figure out how to solve with the cubes please help :) thank you
2) f(x)=√x , g(x) = 2x-3
f(g(x))= √2x-3
and
g(f(x))= 2(√x)-3
Am I done here for finding the f(g(x)) and g(f(x)) ? Or do I have to go a step further?
3) f(x)= x^2/3 , g(x) = x^6
f(g(x))= (x^6)^2/3
and
g(f(x))= (x^2/3)^6
how would I multiply this?
4) f(x)= 3/x^2-1 , g(x) = x+1
f(g(x))= 3/(x+1)^2-1
and
g(f(x))= 3/x^2-1 + 1
1 answer