If you could me? i have been having a really hard time so far. i wiuld reallly appreciate if you could explain how to do the problems.
you!

I will set up question 1 and do question 2, which is easier to solve.

Question 1

To determine if a function is one-to-one, we must prove that f(a) = f(b). In other words, if you get the same y value by plugging a into the function as you do by plugging b, then the function is one-to-one. I will partially do the first one and let you do the rest on your own.

f(x) = (x - 1)/(3x + 3)

f(a) = (a - 1)/(3(a) + 3)

f(a) = (a - 1)/(3a + 3)

f(a) = (a - 1)/[3(a + 1)]

Now do f(b).

f(b) = (b - 1)/[3(b + 1)]

Set f(a) = f(b). If after doing the math you end up with a = b, then you will know that this function is one-to-one.

Take it from here and do the same for all functions listed.

Question 2

Let cbrt = cube root for short.

cbrt{x - 2} + 8

Let y = f(x)

y = cbrt{x- 2} + 8

Switch x and y.

x = cbrt{y - 2} + 8

x - 8 = cbrt{y - 2}

Cube both sides.

(x - 8)^3 = [cbrt{y - 2}]^3

(x - 8)^3 = y - 2

Solve for y.

(x - 8)^3 + 2 = y

Replace y with f^(-1) x.

f^(-1) x = (x - 8)^3 + 2

Choice 1