Inverse of Quadratic Functions

Learn how to find the inverse of quadratic functions with restricted domains. This tutorial includes step-by-step examples with detailed solutions.

Example 1

Find the inverse of the quadratic function in vertex form:

\[ f(x) = 2(x - 2)^2 + 3, \quad x \le 2 \]
View Step-by-Step Solution
  • The function is a quadratic with a restricted domain, so it is one-to-one. Write the function as an equation: \[ y = 2(x - 2)^2 + 3 \]
  • Solve for \(x\): \[ (x - 2)^2 = \frac{y - 3}{2} \] \[ x - 2 = \pm \sqrt{\frac{y - 3}{2}} \] \[ x = 2 \pm \sqrt{\frac{y - 3}{2}} \]
  • Since \(x \le 2\), we take the negative branch: \[ x = 2 - \sqrt{\frac{y - 3}{2}} \]
  • Swap \(x\) and \(y\) to get the inverse function: \[ f^{-1}(x) = 2 - \sqrt{\frac{x - 3}{2}} \]

Example 2

Find the inverse of the quadratic function:

\[ f(x) = -2x^2 + 4x + 2, \quad x \ge 1 \]
View Step-by-Step Solution
  • Write \(f\) in vertex form by completing the square: \[ f(x) = -2(x^2 - 2x) + 2 = -2((x - 1)^2 - 1) + 2 = -2(x - 1)^2 + 4 \]
  • The graph confirms it is one-to-one. Solve for \(x\): \[ y = -2(x - 1)^2 + 4 \] \[ (x - 1)^2 = \frac{4 - y}{2} \] \[ x - 1 = \pm \sqrt{\frac{4 - y}{2}} \] \[ x = 1 \pm \sqrt{\frac{4 - y}{2}} \]
  • Since \(x \ge 1\), choose the positive branch: \[ x = 1 + \sqrt{\frac{4 - y}{2}} \]
  • Swap \(x\) and \(y\) to get the inverse function: \[ f^{-1}(x) = 1 + \sqrt{\frac{4 - x}{2}} \]

Exercises

Find the inverse of the following quadratic functions:

  1. \[ f(x) = (x - 3)^2 + 3, \quad x \ge 3 \]
  2. \[ g(x) = -x^2 + 4x - 4, \quad x \le 2 \]
View Answers & Solutions
  1. \( f^{-1}(x) = 3 + \sqrt{x - 3} \)
  2. \( g^{-1}(x) = 2 - \sqrt{-x} \)

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