Question

find the inverse function.
$f(x) = x^2 + 21$
$f^{-1}(x) = ?\sqrt{x + \square}$

Explanation:

Step1: Replace \( f(x) \) with \( y \)

We start by writing the function as \( y = x^2 + 21 \).

Step2: Swap \( x \) and \( y \)

To find the inverse, we swap the roles of \( x \) and \( y \), so we get \( x = y^2 + 21 \).

Step3: Solve for \( y \)

First, subtract 21 from both sides: \( x - 21 = y^2 \). Then, take the square root of both sides. Since the original function \( f(x)=x^2 + 21 \) is a parabola opening upwards, its inverse (for the principal square root, considering the domain where the function is one - to - one, usually for \( x\geq0 \) or \( x\leq0 \)) will involve a square root. So \( y=\pm\sqrt{x - 21} \), but looking at the given form \( f^{-1}(x)=\sqrt[?]{x + \square} \), we consider the positive square root (assuming the domain of \( f(x) \) is \( x\geq0 \) so that it is one - to - one). The square root has an index of 2 (since it's a square root, \( \sqrt[n]{a} \) with \( n = 2 \) is the square root). And from \( x=y^2 + 21\Rightarrow y^2=x - 21 \), but in the given form it's \( x+\square \), we must have made a mistake in the sign when rearranging. Wait, no, let's re - arrange correctly. If \( y=x^2 + 21 \), then \( x^2=y - 21 \), so \( x=\pm\sqrt{y - 21} \). Then, replacing \( x \) with \( f^{-1}(y) \) and \( y \) with \( x \), we have \( f^{-1}(x)=\pm\sqrt{x - 21} \). But the given form is \( f^{-1}(x)=\sqrt[?]{x+\square} \). Wait, maybe there was a typo in the problem's form, but if we consider the form \( f^{-1}(x)=\sqrt[2]{x-21} \) (since square root has index 2) or adjusting to the given form: if we rewrite \( x=y^2 + 21\Rightarrow y^2=x - 21\Rightarrow y=\pm\sqrt{x - 21} \). But the given form has \( x+\square \), so maybe the original function was \( f(x)=x^2-21 \), but no, the given function is \( f(x)=x^2 + 21 \). Wait, let's check the form again. The form is \( f^{-1}(x)=\sqrt[?]{x+\square} \). Let's assume that maybe the problem has a sign error, but following the steps:

From \( y=x^2 + 21 \), swap \( x \) and \( y \): \( x = y^2+21 \), then \( y^2=x - 21 \), so \( y=\pm\sqrt{x - 21} \). The index of the square root is 2 (because it's a square root, \( \sqrt[n]{a} \) with \( n = 2 \) is the square root, and we can write \( \sqrt[2]{a} \) although we usually write \( \sqrt{a} \)). And the term inside the square root is \( x-21 \), but the given form is \( x+\square \), so if we consider that maybe the problem intended to have \( x - 21 \) as \( x+(- 21) \), then the index is 2 and the square is \( - 21 \).

Answer:

The index of the square root (the value of \(? \)) is 2, and the value of the square (the \( \square \)) is \( - 21 \). So \( f^{-1}(x)=\sqrt[2]{x-21} \) (or \( \pm\sqrt{x - 21} \), but based on the given form with a single square root, likely the positive square root with index 2 and \( \square=-21 \)). So the index is 2 and the constant term is - 21. So in the form \( f^{-1}(x)=\sqrt[2]{x+(-21)} \), so \(? = 2 \) and the square is \( - 21 \).