Question

which equation is the inverse of $y = 16x^2 + 1$?

• $y = \pm \sqrt{\frac{x}{16} - 1}$
• $y = \frac{\pm \sqrt{x - 1}}{16}$
• $y = \frac{\pm \sqrt{x}}{4} - \frac{1}{4}$
• $y = \frac{\pm \sqrt{x - 1}}{4}$

Explanation:

Step1: Swap x and y

To find the inverse of a function \( y = f(x) \), we first swap \( x \) and \( y \). So for the function \( y = 16x^2 + 1 \), we get \( x = 16y^2 + 1 \).

Step2: Solve for y

Subtract 1 from both sides: \( x - 1 = 16y^2 \)
Divide both sides by 16: \( \frac{x - 1}{16} = y^2 \)
Take the square root of both sides: \( y = \pm\sqrt{\frac{x - 1}{16}} \)
Simplify the square root: \( \sqrt{\frac{x - 1}{16}}=\frac{\sqrt{x - 1}}{4} \), so \( y=\pm\frac{\sqrt{x - 1}}{4} \)

Answer:

\( y = \pm\frac{\sqrt{x - 1}}{4} \) (the last option)