Solved Problems: Finding Inverses of Quadratic Functions

📚 Understanding Inverse Functions

In mathematics, finding the inverse of a function essentially reverses the roles of input and output. Given a function $f(x)$, its inverse, denoted as $f^{-1}(x)$, satisfies the property that $f^{-1}(f(x)) = x$ and $f(f^{-1}(x)) = x$. This means if you apply a function and then its inverse, you get back the original input.

📜 A Brief History

The concept of inverse functions has evolved alongside the development of algebra and calculus. While the explicit notation and formal definition came later, the idea of reversing mathematical operations dates back to ancient problem-solving techniques. As mathematicians explored more complex functions, the need to understand and define inverse functions became crucial for solving equations and analyzing mathematical relationships.

✨ Key Principles for Finding Inverses of Quadratic Functions

Quadratic functions, given by the form $f(x) = ax^2 + bx + c$, present a unique challenge when finding inverses because they are not one-to-one over their entire domain. This means that to find a true inverse function, we often need to restrict the domain. Here's how to do it:

• 🔍 Check for One-to-One: Ensure the function is one-to-one or restrict the domain to make it so. This often involves restricting to $x \geq -b/2a$ or $x \leq -b/2a$.
• 📝 Replace $f(x)$ with $y$: This makes the algebraic manipulation easier.
• 🔄 Swap $x$ and $y$: This is the key step in finding the inverse.
• ➗ Solve for $y$: Isolate $y$ to express it in terms of $x$. This will give you the inverse function.
• ✅ Replace $y$ with $f^{-1}(x)$: Use the proper notation for the inverse function.

🧮 Example 1: Finding the Inverse of $f(x) = x^2$ for $x \geq 0$

Since $f(x) = x^2$ is not one-to-one over its entire domain, let's restrict it to $x \geq 0$.

1. Replace $f(x)$ with $y$: $y = x^2$
2. Swap $x$ and $y$: $x = y^2$
3. Solve for $y$: $y = \sqrt{x}$ (Since we restricted $x \geq 0$, we only take the positive square root).
4. Replace $y$ with $f^{-1}(x)$: $f^{-1}(x) = \sqrt{x}$

📈 Example 2: Finding the Inverse of $f(x) = (x - 2)^2$ for $x \geq 2$

Restrict the domain to $x \geq 2$ to make the function one-to-one.

1. Replace $f(x)$ with $y$: $y = (x - 2)^2$
2. Swap $x$ and $y$: $x = (y - 2)^2$
3. Solve for $y$: $\sqrt{x} = y - 2$, so $y = \sqrt{x} + 2$
4. Replace $y$ with $f^{-1}(x)$: $f^{-1}(x) = \sqrt{x} + 2$

🧪 Example 3: Finding the Inverse of $f(x) = 2x^2 + 4x - 6$ for $x \geq -1$

First, complete the square to rewrite the function in vertex form: $f(x) = 2(x^2 + 2x) - 6 = 2(x^2 + 2x + 1) - 6 - 2 = 2(x + 1)^2 - 8$ Since we are given $x \geq -1$, the function is one-to-one on this domain.

1. Replace $f(x)$ with $y$: $y = 2(x + 1)^2 - 8$
2. Swap $x$ and $y$: $x = 2(y + 1)^2 - 8$
3. Solve for $y$: $x + 8 = 2(y + 1)^2$ \\ $\frac{x + 8}{2} = (y + 1)^2$ \\ $\sqrt{\frac{x + 8}{2}} = y + 1$ \\ $y = \sqrt{\frac{x + 8}{2}} - 1$
4. Replace $y$ with $f^{-1}(x)$: $f^{-1}(x) = \sqrt{\frac{x + 8}{2}} - 1$

✍️ Conclusion

Finding the inverse of a quadratic function involves restricting its domain to ensure it is one-to-one, swapping the variables, and solving for the new dependent variable. With practice and a careful application of algebraic principles, you can confidently find the inverses of various quadratic functions.