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

1. $y=x^2+3$ 

   Hint (a):

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

   Step 1(a):

   $x=y^2+3$

   Step 2(a):

   $x-3=y^2$

   Step 3(a):

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

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

   Hint (b):

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

   Step 1(b):

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

   Step 2(b):

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

   Step 3(b):

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

   Answer (b):

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

3. $y = \sqrt { 5 x - 6 }$ 

   Hint (c):

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

   Answer (c):

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