Find the Inverse of Exponential Functions

This tutorial explains how to find the inverse of exponential functions and how to determine their domain and range. Each example is solved step by step to help you understand the process clearly.

Example 1

Find the inverse function, its domain, and its range for:

\[ f(x) = e^{x-3} \]
View Step-by-Step Solution
  • The given function is exponential. Its domain is \( (-\infty, +\infty) \) and its range is \( (0, +\infty) \).
  • Write the function as an equation: \[ y = e^{x-3} \]
  • Take the natural logarithm of both sides: \[ x - 3 = \ln y \quad \Rightarrow \quad x = \ln y + 3 \]
  • Interchange \(x\) and \(y\) to obtain the inverse function: \[ f^{-1}(x) = \ln x + 3 \]

Conclusion: The domain of \(f^{-1}\) is \( (0, +\infty) \) and its range is \( (-\infty, +\infty) \).

Example 2

Find the inverse function, its domain, and its range for:

\[ f(x) = 2e^{2x+3} + 4 \]
View Step-by-Step Solution
  • The domain of \(f\) is \( (-\infty, +\infty) \). Since \(e^{2x+3} > 0\), the range of \(f\) is \( (4, +\infty) \).
  • Write the function as an equation and solve for \(x\): \[ y = 2e^{2x+3} + 4 \] \[ e^{2x+3} = \frac{y-4}{2} \] \[ 2x + 3 = \ln\left(\frac{y-4}{2}\right) \] \[ x = \frac{1}{2}\left[\ln\left(\frac{y-4}{2}\right) - 3\right] \]
  • Interchange \(x\) and \(y\) to obtain the inverse: \[ f^{-1}(x) = \frac{1}{2}\left[\ln\left(\frac{x-4}{2}\right) - 3\right] \]

Conclusion: The domain of \(f^{-1}\) is \( (4, +\infty) \) and its range is \( (-\infty, +\infty) \).

Example 3

Find the inverse function, its domain, and its range for:

\[ f(x) = 2e^{x^2 - 1} + 2, \quad x \ge 0 \]
View Step-by-Step Solution
  • The function is not one-to-one on \(\mathbb{R}\), but restricting the domain to \(x \ge 0\) makes it one-to-one.
  • Domain: \([0, +\infty)\).
  • Since \(x^2 \ge 0\), we have \[ x^2 - 1 \ge -1 \] Taking exponentials and simplifying gives the range: \[ f(x) \ge 2e^{-1} + 2 \] Hence the range is \([2e^{-1} + 2, +\infty)\).
  • Solve for \(x\): \[ y = 2e^{x^2 - 1} + 2 \] \[ x^2 = \ln\left(\frac{y-2}{2}\right) + 1 \] Since \(x \ge 0\), \[ x = \sqrt{\ln\left(\frac{y-2}{2}\right) + 1} \]
  • Interchange \(x\) and \(y\) to obtain the inverse: \[ f^{-1}(x) = \sqrt{\ln\left(\frac{x-2}{2}\right) + 1} \]

Conclusion: The domain of \(f^{-1}\) is \([2e^{-1} + 2, +\infty)\) and its range is \([0, +\infty)\).

Practice Exercises

Find the inverse function, its domain, and its range for each of the following functions:

  1. \( f(x) = -e^{x+4} \)
  2. \( g(x) = 2 - e^{(4x-2)/3} \)
  3. \( h(x) = -e^{2x^2 - 5} + 3, \; x \le 0 \)

Answers and Solutions

  1. \( f^{-1}(x) = \ln(-x) - 4 \)
    Domain: \( (-\infty, 0) \), Range: \( (-\infty, +\infty) \)
  2. \( g^{-1}(x) = \frac{3}{4}\ln(2-x) + \frac{1}{2} \)
    Domain: \( (-\infty, 2) \), Range: \( (-\infty, +\infty) \)
  3. \( h^{-1}(x) = -\sqrt{\tfrac{1}{2}\ln(3-x) + \tfrac{5}{2}} \)
    Domain: \( (-\infty, -e^{-5} + 3] \), Range: \( (-\infty, 0] \)

More References and Links