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.
Examples with Detailed Solutions
Example 1
Find the inverse of the quadratic function in vertex form:
\[
f(x) = 2(x - 2)^2 + 3, \quad x \le 2
\]
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 solution:
\[
x = 2 - \sqrt{\frac{y - 3}{2}}
\]
- Swap \(x\) and \(y\) to get the inverse function:
\[
y = 2 - \sqrt{\frac{x - 3}{2}}, \quad f^{-1}(x) = 2 - \sqrt{\frac{x - 3}{2}}
\]
Example 2
Find the inverse of the quadratic function:
\[
f(x) = -2x^2 + 4x + 2, \quad x \ge 1
\]
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:
\[
x = 1 + \sqrt{\frac{4 - y}{2}}
\]
- Swap \(x\) and \(y\) to get the inverse function:
\[
y = 1 + \sqrt{\frac{4 - x}{2}}, \quad f^{-1}(x) = 1 + \sqrt{\frac{4 - x}{2}}
\]
Exercises
Find the inverse of the following quadratic functions:
- \[
f(x) = (x - 3)^2 + 3, \quad x \ge 3
\]
- \[
g(x) = -x^2 + 4x - 4, \quad x \le 2
\]
Answers
- \[
f^{-1}(x) = 3 + \sqrt{x - 3}
\]
- \[
g^{-1}(x) = 2 - \sqrt{-x}
\]
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