To find the inverse of $f(x) = (x+1)^2$ restricted to the domain $[-1, \infty)$, let's proceed step by step.

Step 1: Restate the function

The function is: $f(x) = (x+1)^2$ and its domain is restricted to $x \geq -1$.

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

Let $y = f(x) = (x+1)^2$. Solve for $x$ in terms of $y$: $y = (x+1)^2$ $\sqrt{y} = |x+1| \quad \text{(take the square root on both sides)}$ Since the domain is $x \geq -1$, $x+1 \geq 0$, so: $\sqrt{y} = x+1$ $x = \sqrt{y} - 1$

Step 3: Write the inverse

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

Step 4: Restrict the domain of the inverse

Since $f(x)$ is defined on $[-1, \infty)$, its range is: $y = (x+1)^2, \quad x \geq -1 \implies y \geq 0$ Thus, the domain of the inverse $f^{-1}(x)$ is $[0, \infty)$.

Final Answer:

$f^{-1}(x) = \sqrt{x} - 1, \quad x \geq 0$

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Would you like additional examples or clarifications?

Here are some related questions to deepen understanding:

1. Why is the restriction $x \geq -1$ necessary for $f(x)$ to have an inverse?
2. What is the graphical interpretation of $f(x)$ and its inverse?
3. How do we confirm $f^{-1}(x)$ is the correct inverse function?
4. Can $f(x)$ have a one-to-one inverse without restricting its domain?
5. What happens to the inverse if the domain of $f(x)$ is further restricted?

Tip: Always check that the domain of the original function matches the range of its inverse, and vice versa.