### Home > CCA2 > Chapter Ch5 > Lesson 5.1.3 > Problem5-51

5-51.

Find the inverse of each of the following functions by first switching $x$ and $y$ and then solving for $y$.

1. $y=x^2+3$

Interchange $x\ \text{and}\ y$. Then solve your inverse (new) equation for $y$.

$x=y^2+3$

$x-3=y^2$

$\pm\sqrt{x-3}=y$

2. $y = ( \frac { 1 } { 4 } x + 6 ) ^ { 3 }$

Interchange $x\ \text{and}\ y$. Solve for $y$ in the new equation.

$x = \left( \frac{1}{4}y+6\right)^3$

$\sqrt[3]{x}=\frac{1}{4}y+6$

$\sqrt[3]{x}-6=\frac{1}{4}y$

$y = 4(\sqrt[3]{x}-6)$

3. $y = \sqrt { 5 x - 6 }$

To undo square rooting, square both sides of the equation.

$y=\frac{x^2+6}{5}$