Find Inverse Of Exponential Functions

Examples, with detailed solutions, on how to find the inverse of exponential functions and also their domain and range.

Example 1

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

f(x) = ex-3

Solution to example 1
  • Note that the given function is a an exponential function with domain (-∞ , + ∞) and range (0, +∞). We first write the function as an equation as follows
    y = ex-3
  • Take the ln of both sides to obtain
    x-3 = ln y or x = ln y + 3
  • Change x into y and y into x to obtain the inverse function.
    f -1(x) = y = ln x + 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: (0,+ ∞) range: (-∞ , + ∞)



Example 2

Find the inverse, its domain and range, of the function given by
f(x) = 2 e(2 x + 3) + 4

Solution to example 2

  • Let us first find the domain and range of the given function.
    Domain of f: (-∞ , + ∞)
    Range: for x in the domain, the range of e(2 x + 3) is given by (0,+∞)
    The range of 2 e(2 x + 3) is also given by (0,+∞)
    The range of f(x) = e(2 x + 3) + 4 is (4,+∞) because the +4 shifts up the graph of the function
  • Find the inverse of f, write f as an equation and solve for x.

    y = 2 e(2 x + 3) + 4
    2 e(2 x + 3) = y - 4
    e(2 x + 3) = (y - 4)/2
    Take the ln of both sides to obtain
    2x + 3 = ln ((y - 4)/2)
    and finally x = (1/2) (ln ((y - 4)/2) - 3)
  • Change x into y and y into x to obtain the inverse function.
    f-1(x) = y = (1/2) (ln ((x - 4)/2) - 3)
    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: (-∞ , + ∞)

Example 3

Find the inverse, its domain and range, of the function given by
f(x) = 2 e(x 2 - 1) + 2 , for x ≥ 0

Solution to example 3

  • It is easy to show that function f given by the formula above is an even function and therefore not a one to one if the domain is R. However the domain in our case is given by x ≥ 0 which makes the given function a one to one function and therefore has inverse.
    Domain of f: [0 , + ∞) , given
    Range: for x in the domain [0, + ∞) , the range of x 2 is given by [0,+∞) which can be written as
    x 2 ≥ 0
    subtract -1 to both sides to obtain: x 2 - 1≥ - 1
    take the exponential of both sides to obtain: ex 2 - 1 ≥ e -1 (the exponential function being an increasing function)
    multiply by +2 to both sides of the above inequality to obtain: 2 ex 2 - 1 ≥ 2 e -1
    add +2 to both sides of the above inequality to obtain: 2 ex 2 - 1 + 2≥ 2 e -1 + 2
    the left hand side of the above inequality is the given function, hence the range of the given function is given by : [2 e -1 + 2, + ∞)
  • Find the inverse of f, write f as an equation and solve for x.

    y = 2 e(x 2 - 1) + 2
    2 e(x 2 - 1) = y - 2
    e(x 2 - 1) = (y - 2)/2
    Take the ln of both sides to obtain
    x 2 - 1 = ln ((1/2)(y - 2))
    and finally x = + or - sqrt[ln ((1/2)(y - 2)) + 1]
    Since x ≥ 0 (given domain), we have x = sqrt[ln ((1/2)(y - 2)) + 1]
  • Change x into y and y into x to obtain the inverse function.
    f-1(x) = y = sqrt[ln ((1/2)(x - 2)) + 1]
    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: [2 e -1 + 2, + ∞) and its range is given by: [0, + ∞)



Exercises

Find the inverse, its domain and range, of the functions given below
1. f(x) = -ex + 4
2. g(x) = 2 - e(4x - 2) / 3
3. h(x) = - e(2 x 2 - 5) + 3, for x ≤ 0


Answers to above exercises
1. f -1(x) = ln( -x) - 4 ; domain: (-∞ , 0) Range: (-∞ , +∞)
2. g -1(x) = (3/4) ln (2 - y) +1/2 ; domain: (-∞ , 2) Range: (-∞ , +∞)
3. h -1(x) = - sqrt[(1/2) ln (3 - y) + 5/2]; domain: (-∞ , - e(-5) + 3) Range: (-∞ , +∞)

More links and references related to the inverse functions.

More References and Links to Inverse Functions


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