Pre Calculus Homework

Question

Consider the following functions.

f(x) = 6x − 5, g(x) = x2

Find (f ∘ g)(x).

Find the domain of (f ∘ g)(x). (Enter your answer using interval notation.)

Find (g ∘ f)(x).

Find the domain of (g ∘ f)(x). (Enter your answer using interval notation.)

Find (f ∘ f)(x).

Find the domain of (f ∘ f)(x). (Enter your answer using interval notation.)

Find (g ∘ g)(x).

Find the domain of (g ∘ g)(x). (Enter your answer using interval notation.)

Answer 1

Domain* - basically wherever x values your function can go, so 1/x does not exist at x = 0, so the domain would be (-∞, 0) U (0, ∞)

f(x) = 6x − 5, g(x) = x²

a)

(f ∘ g)(x) = f(g(x)) = f(x²) = 6x² - 5

Domain: (-∞, ∞)

b)

(g ∘ f)(x) = g(f(x)) = g(6x-5) = (6x-5)² = 36x² -60x + 25

Domain: (-∞, ∞)

c)

(f ∘ f)(x) = f(f(x)) = f(6x-5) = 6(6x - 5) - 5 = 36x - 30 - 5 = 36x - 35

Domain: (-∞, ∞)

d)

(g ∘ g)(x) = g(g(x)) = g(x²) = (x²)² = x⁴

Domain: (-∞, ∞)

Answer 2

(f _º g)(x) means f(g(x)) meaning you plug in the function g(x) into f(x) in place of the "x".

So, start with the function f(x) which is 6x - 5 and instead of writing "x", write x², which is g(x) to get:

(f _º g)(x) = f(g(x)) = 6(x²) - 5

There are no restrictions for the domain so x can be any real number meaning D: (-∞, ∞)

The same holds true for (g _º f)(x), which is g(f(x)). Start with the function g(x) which is x², and then re-write it but instead of using "x", use f(x) which is 6x - 5 meaning (g _º f)(x) = g(f(x)) = (6x - 5)² which again has a domain with no restrictions for "x" so that D: (-∞, ∞)

(f _º f)(x) = f(f(x)) = 6(6x - 5) - 5 again with a domain of (-∞, ∞)

Do the same for (g _º g)(x) which is the same as g(g(x))