 # Find Inverse Of Cube Root Functions

Find the inverse of cube root functions as well as their domain and range; examples with detailed solutions. In what follows, the symbol 3√ is used to indicate the principal cube root.

## Example 1

Find the inverse function, its domain and range, of the function given by

f(x) = 3√(2 x - 1)

Solution to example 1

• The domain D and range R of the given function are given by:
D: (- ∞ , + ∞) and R: (- ∞ , + ∞)
• In order to find the inverse, we first write the function as an equation as follows
y = 3√(2 x - 1)
• Then solve it starting by cubing both sides
y 3 = ( 3√(2 x - 1) ) 3
• Simplify and solve for x
y 3 = 2 x - 1
x = (1 / 2)(y 3 + 1)
• Change x into y and y into x to obtain the inverse function.
f -1(x) = y = (1 / 2)(x3 + 1)
The domain and range of the inverse function are respectively the range and domain of the given function f. Hence
domain and range of f -1 are given by: domain: (- ∞ , + ∞) range: (- ∞ , + ∞)

## Example 2

Find the inverse, its domain and range, of the function given by
f(x) = 3√(x / 3 - 1) - 4

Solution to example 2

• The domain and range of the given function are given by
D: (- ∞ , + ∞) and R: (- ∞ , + ∞)
• Write the given function as an equation.

y = 3√(x / 3 - 1) - 4
which can be writeen as: 3√(x / 3 - 1) = 4 + y
• Cube both sides of the above equation and simplify.
( 3√(x / 3 - 1) ) 3 = (4 + y) 3
(x / 3 - 1) = (4 + y) 3
• Solve for x.
x / 3 = (4 + y) 3 + 1
• Which gives
x = 3 ( (4 + y) 3 + 1 ) = 3 (4 + y) 3 + 3
• Interchange x and y to obtain the inverse function
f -1(x) = y = 3 (4 + x) 3 + 3
The domain and range of the inverse function are respectively the range and domain of the given function f. Hence
domain and range of f -1 are given by: domain: (- ∞ , + ∞) range: (- ∞ , + ∞)

## Example 3

Find the inverse, its domain and range, of the function given by
f(x) = 3√(4 x2 + 8) + 2 ; x ≥ 0

Solution to example 3

• Although the formula of the given function indicates an even function (not a one to one), the explicit given domian (x ≥ 0) make the given function a one to one.
Domain of f: [ 0 ; ∞) , given.
Range:
For x in the domain [ 0 ; ∞) , the range of 4 x2 + 8 is given by [8,+∞)
which gives a range of 3√(4 x2 + 8) in the interval [3√8 , +∞) or [2 , +∞)
and finally, taking into account the shift +2, the range for the given function is given by [4 , +∞)
• To find the inverse, we first write the given function f as an equation

y = 3√(4 x2 + 8) + 2
• Which may be written as
y - 2 = 3√(4 x2 + 8)
• Cube both side and simplify
(y - 2)3 = ( 3√(4 x2 + 8) )3
(y - 2)3 = 4 x2 + 8
• Solve for x
x2 = (1 / 4) ( (y - 2)3 - 8 )
x = ~+mn~ (1 / 4) √ ( (y - 2)3 - 8 )
• The domain of f is given by [ 0 ; ∞) and therefore x is selected to be given by
x = (1 / 4) √ ( (y - 2)3 - 8 )
• Interchange x and y to obtain the inverse function
f -1(x) = y = (1 / 4) √ ( (x - 2)3 - 8 )
The domain and range of f -1 are respectively given by the range and domain of f found above
Domain of f -1 is given by: [4 , + ∞) and its range is given by: [0 , + ∞)

## Exercises

Find the inverse, its domain and range, of the functions given below
1. f(x) = -
3√(- x + 3)
2. g(x) =
3√(x2 + 2x + 4) ; x ≥ - 1 (Hint: start by finding the vertex of x2 + 2x + 4 to understand the given domain x ≥ - 1)