Question

algebra 2a semester online practice
complete this assessment to review what you’ve learned. it will not count toward your grade.
find the inverse of the dilated quadratic function $f(x) = 8x^2$ with domain restricted to $x \geq 0$. (1 poi
$\circ$ $f^{-1}(x) = \sqrt{x + 8}$
$\circ$ $f^{-1}(x) = \sqrt{\frac{x}{8}}$
$\circ$ $f^{-1}(x) = \sqrt{8x}$
$\circ$ $f^{-1}(x) = \sqrt{x} + 8$

Explanation:

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

We start with the function \( f(x) = 8x^2 \), so we write \( y = 8x^2 \).

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

To find the inverse, we swap the roles of \( x \) and \( y \), getting \( x = 8y^2 \).

Step3: Solve for \( y \)

First, divide both sides by 8: \( \frac{x}{8}=y^2 \). Then, since the original domain is \( x\geq0 \), the inverse function will have a range \( y\geq0 \), so we take the square root of both sides: \( y = \sqrt{\frac{x}{8}} \).

Step4: Replace \( y \) with \( f^{-1}(x) \)

We get \( f^{-1}(x)=\sqrt{\frac{x}{8}} \).

Answer:

\( f^{-1}(x)=\sqrt{\frac{x}{8}} \) (the second option)