Find Inverse Of Exponential Functions

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 = e^x-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

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

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: e^{x 2 - 1} ? e ^-1 (the exponential function being an increasing function)
  multiply by +2 to both sides of the above inequality to obtain: 2 e^{x 2 - 1} ? 2 e ^-1
  add +2 to both sides of the above inequality to obtain: 2 e^{x 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

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