1) `F(x) = 3x^2 + 4x`

`y = 3x^2+4x` Write in y = format.

`x = 3y^2 + 4y` Exchange x and y

`x/3 = y^2 + 4/3y` Divide by 3

`x/3 + (2/3)^2 = y^2 + 4/3y + (2/3)^2` We complete the square`(1/2b)^2 `

`x/3 + 4/9 = (y + 2/3)^2` Now we can take the square root

`y+2/3 = +-sqrt(x/3+4/9)` Note the +- this is not a function

`y = -2/3+-sqrt(x/3+4/9)` Subtract 2/3 from both sides

`y = (-2+-sqrt(3x+4))/3` Simplify

` F^-1(x) = (-2+-sqrt(3x+4))/3`

Write in inverse fromat.

2) `G(x) = 4x-4`

`y = 4x-4` Write in y format

`x = 4y - 4` Exchange x and y

`x + 4 = 4y` Solve for y

`y = 1/4x + 1`

`G^-1(x) = 1/4x + 1`

It is assumed that you want the inverse functions of f(x) = 3, f(x) = 3x^2+4x and g(x)=4x-4.

For the functions f(x) = 3 and f(x) = 3x^2 + 4x the inverse function does not exist as for any value of x the value of f(x) should be unique.

For f(x) = 3, any value of x gives the result 3. The inverse of f(x) would have 3 taking on one of an infinite number of alternatives.

For f(x) = 3x^2 + 4x, 2 two values of x give the same value of f(x). The inverse of f(x) would have two options for every number x. This is not possible.

Only the inverse of g(x) = 4x - 4 can be determined.

g(x) = 4x - 4

=> x = `(g(x) + 4)/4`

substitute x with `g^-1(x)` and `g(x)` with x

=> `g^-1(x) = (x + 4)/4`

**Only the inverse of g(x) = 4x - 4 exists and **`g^-1(x) = (x + 4)/4.`