To find the inverse of the function $f(x) = (2x + 1)^2$ for $x \geq -\frac{1}{2}$, let's proceed step by step:

Step 1: Set $y = f(x)$

We start by setting $y = (2x + 1)^2$.

$y = (2x + 1)^2$

Step 2: Solve for $x$ in terms of $y$

To find the inverse, we need to express $x$ in terms of $y$.

1. Take the square root of both sides: $\sqrt{y} = 2x + 1$ Since $x \geq -\frac{1}{2}$, we will only take the positive square root.

2. Solve for $x$: $2x = \sqrt{y} - 1$ $x = \frac{\sqrt{y} - 1}{2}$

Step 3: Replace $y$ with $x$

The inverse function $f^{-1}(x)$ is given by: $f^{-1}(x) = \frac{\sqrt{x} - 1}{2}$

Domain of the Inverse Function

Since the original function $f(x) = (2x + 1)^2$ is defined for $x \geq -\frac{1}{2}$, the range of $f(x)$ starts from $(2(-\frac{1}{2}) + 1)^2 = 0$. Therefore, the inverse function is defined for $x \geq 0$.

Thus, the inverse function is: $f^{-1}(x) = \frac{\sqrt{x} - 1}{2}, \quad x \geq 0$

Would you like more details or further clarification?

Related Questions:

1. What are the general steps to find the inverse of a quadratic function?
2. How does the domain restriction $x \geq -\frac{1}{2}$ affect the inverse function?
3. Can you explain why we take the positive square root when solving for $x$?
4. What is the geometric interpretation of a function and its inverse on a graph?
5. How do we verify if a function and its inverse are correct by composition?

Tip:

To confirm whether two functions are inverses, check if $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$.