Question

find the inverse of each function.
$y = x^2 - 5$
$y = \sqrt{x} + 3$
$y = \frac{2x}{5}$

Explanation:

For \( y = x^2 - 5 \)

Step1: Swap \( x \) and \( y \)

\( x = y^2 - 5 \)

Step2: Solve for \( y \)

Add 5 to both sides: \( x + 5 = y^2 \)
Take square roots: \( y=\pm\sqrt{x + 5} \), with the domain \( x\geq - 5 \) (since the original function \( y=x^{2}-5\) has a range \( y\geq - 5\), so the inverse function has domain \( x\geq - 5\))

Step1: Swap \( x \) and \( y \)

\( x=\sqrt{y}+3 \)

Step2: Solve for \( y \)

Subtract 3 from both sides: \( x - 3=\sqrt{y} \)
Square both sides: \( y=(x - 3)^{2} \), with the domain \( x\geq3 \) (since the original function \( y = \sqrt{x}+3\) has a domain \( x\geq0\) and range \( y\geq3\), so the inverse function has domain \( x\geq3\))

Step1: Swap \( x \) and \( y \)

\( x=\frac{2y}{5} \)

Step2: Solve for \( y \)

Multiply both sides by 5: \( 5x = 2y \)
Divide both sides by 2: \( y=\frac{5x}{2} \)

Answer:

\( y=\pm\sqrt{x + 5},x\geq - 5 \)

For \( y=\sqrt{x}+3 \)