Question

1. which is the inverse of the function $f(x) = 4x^2$?

$y = 4x^2$
$y = \pm 2\sqrt{x}$
$y = \frac{\pm \sqrt{x}}{4}$
$y = \frac{\pm \sqrt{x}}{2}$

Explanation:

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

We start with the function \( f(x) = 4x^2 \), so we write it as \( y = 4x^2 \).

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

To find the inverse, we swap the roles of \( x \) and \( y \). This gives us \( x = 4y^2 \).

Step3: Solve for \( y \)

First, divide both sides of the equation \( x = 4y^2 \) by 4: \( \frac{x}{4} = y^2 \). Then, take the square root of both sides. Remember that when we take the square root, we need to consider both the positive and negative roots, so we have \( y = \pm\sqrt{\frac{x}{4}} \). Simplifying the square root of \( \frac{x}{4} \), we know that \( \sqrt{\frac{x}{4}}=\frac{\sqrt{x}}{\sqrt{4}}=\frac{\sqrt{x}}{2} \). So, \( y = \frac{\pm\sqrt{x}}{2} \).

Answer:

D. \( y = \frac{\pm\sqrt{x}}{2} \) (assuming the last option is D, with the expression \( y = \frac{\pm\sqrt{x}}{2} \))