Pre Calc HW - Functions Question!?

Question

Left f (x) = x^2 + 1. Find a function "g" so that:
(a) (fg)(x) = x^4 - 1
(b) (f + g)(x) = 3x^2
(c) (f/g)(x) = 1
(d) f(g(x)) = 9x^4 +1
(e) g(f(x)) = 9x^4 + 1

Answer 1

a.) (f * g)(x) may be rewritten as f(x) * g(x)
f(x) * g(x) = x^4 - 1
(x^2 + 1) * g(x) = x^4 - 1
g(x) = x^4 - 1 / x^2 + 1
g(x) = (x^2 - 1) (x^2 + 1) / x^2 + 1
g(x) = x^2 - 1 (solution!)

b.) (f + g)(x) may be rewritten as f(x) + g(x)
f(x) + g(x) = 3x^2
x^2 + 1 + g(x) = 3x^2
g(x) = 2x^2 - 1 (solution!)

c.) (f/g)(x) may be rewritten as f(x) / g(x)
f(x) / g(x) = 1
x^2 + 1 / g(x) = 1
x^2 + 1 = g(x) (solution!)

d.) f(g(x)) means you substitute g(x) for x in f(x).
f(g(x)) = 9x^4 + 1 = g(x)^2 + 1
g(x)^2 = 9x^4
g(x) = 3x^2 (solution!)

e.) g(f(x)) means you substitute f(x) for x in g(x).
g(f(x)) = 9x^4 + 1
g(x^2 + 1) = 9x^4 + 1

To undo g(x), you need to find it's inverse. Do that by switching x and y in the function:
y = x^2 + 1
x = y^2 + 1
x - 1 = y^2
(x - 1)^1/2 = y (inverse!)

Now, insert f-1(x) into g(f(x)) to undo f(x) and leave g(x):
9((x - 1)^1/2)^4 + 1
9(x-1)^2 + 1
9(x^2 - 2x + 1) + 1
9x^2 - 18x + 9 + 1
g(x) = 9x^2 - 18x + 10 (solution!)

check:
g(f(x) = 9(x^2 + 1)^2 - 18(x^2 +1) + 10
= 9(x^4 + 2x^2 + 1) - 18(x^2 +1) + 10
= 9x^4 + 18x^2 - 18x^2 + 9 - 18 + 10
= 9x^4 + 1 (check!)