For f(x)=x^3+7 and g(x)=∛(x-7), find:

Question

f(x)+g(x)

f(x)-g(x)

f(x)⋅g(x)

f(g(x))

g(f(x))

Are f(x) and g(x) inverses? Why or why not?

Please explain how you do these with steps

Answer 1

f(x) = x^3 + 7

g(x) = (x-7)^1/3

they're inverses of each other. f(g(x)) = g(f(x)) = x

a couple ways you could show this

one is take y = x^3 + 7 and switch x and y, then solve for the new y which is g(x)

x = y^3 +7

y^3 = x-7

y = cube root of x-7 = (x-7)^1/3 = g(x) = f^-1(x)

for an inverse function, x and y are switched. If (x,y) is a point on f(x) then (y,x) is a point on g(x)

another way is as follows

f(g(x)) = f([(x-7)^1/3]^3 +)= x-7+7 = x

if you start with x and go back to x, they're inverse functions

It's like a multiplicative inverse. 2x/2 = x. f(x) =2x, g(x)= x/2. Multiply by 2, then divide by 2 you go back to x

or an additive inverse. x +2 -2 = 0

or the inverse of a cube root is a cube. cube root of x-7 cubed = x-7

or in this example of f(x) and g(x), the inverse is a combination of additive inverses and cubic inverses

the following though, don't involve properties of inverses, just simple addition, subtraction or multiplication

f(x) + g(x) = x^3 + 7 + (x-7)^1/3

f(x) - g(x) = x^3 + 7 - (x-7)^1/3

f(x) times g(x) = (x^3+7)(x-7)^1/3 Probably that's enough

or if you really want to expand that further

f(x) times g(x) = x^3[(x-7)^1/3] + 7[(x-7)^1/3] = (x^10- 7x^9)^1/3 + (343x - 2401)^1/3

Answer 2

f(x)=x³+7 and g(x)=∛(x-7)

f(x)+g(x) = x³+7 + ∛(x-7) (add the two expressions)

f(x)-g(x) = x³+7 - ∛(x-7) (subtract the two expressions)

f(x)⋅g(x) = (x³+7)∛(x-7) = x³∛(x-7) + 7∛(x-7) (multiply the two expressions)

f(g(x)) = ∛(x-7)³ + 7 = (x-7) + 7 = x (plug in ∛(x-7) everywhere you see an x in f(x))

g(f(x)) = ∛((x³+7)-7) = ∛(x³) = x (plug in x³+7 everywhere you see an x in g(x))

Since f(g(x)) = g(f(x)) = x, the functions are inverses.