consider the functions f(x)=x^3-3 and g(x)=3 sqrt x+3:
a. find f(g(x))
b. find g(f(x))
c. determine whether the functions f and g are inverses of each other.
I really need help with these, I don't get it at all.
I'll do f(g). Then g(f) should be no trouble.
You know f(x) = x^3-3, so
f(g) = g^3-3
But what's g? g(x) = 3√(x+3). So,
f(g) = g^3-3 = (3√(x+3))^3 - 3
= 27(x+3)√(x+3)-3
= 3(9(x+3)^(3/2) - 1)
now, g(f) = 3√(f+3)
...
To be honest, I really don't know how to do any of it:(
a. To find f(g(x)), we need to substitute g(x) into the function f(x). Remember that g(x) = 3√(x+3). So, replace x in f(x) with g(x):
f(g(x)) = (g(x))^3 - 3 = (3√(x+3))^3 - 3
Now, simplify the expression:
f(g(x)) = (3√(x+3))^3 - 3 = 27(x+3)^(3/2) - 3
b. To find g(f(x)), we need to substitute f(x) into the function g(x). Remember that f(x) = x^3 - 3. So, replace x in g(x) with f(x):
g(f(x)) = 3√(f(x)+3) = 3√((x^3 - 3) + 3) = 3√(x^3) = 3x√x = 3x^(3/2)
c. To determine whether f and g are inverses of each other, we need to check if the composition of f(g(x)) and g(f(x)) gives the original x. In other words, does f(g(x)) = x and g(f(x)) = x?
Let's check:
f(g(x)) = 27(x+3)^(3/2) - 3 != x
g(f(x)) = 3x^(3/2) != x
Since both f(g(x)) and g(f(x)) do not equal x, we can conclude that f and g are not inverses of each other.