What is the inverse of the given relation? $$ y=7 x^{2}-3 $$ $$ y= \pm \sqrt{\frac{x+3}{7}} $$ $$ x=\sqrt{\frac{y+3}{7}} $$ $$ y^{2}=\frac{x-3}{7} $$ $$ y= \pm \sqrt{\frac{x-3}{7}} $$

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1. Start with the original relation:  
   $$
   y = 7x^2 - 3
   $$

2. To find the inverse, swap $$ x $$ and $$ y $$:  
   $$
   x = 7y^2 - 3
   $$

3. Solve for $$ y^2 $$ by adding 3 to both sides:  
   $$
   x + 3 = 7y^2
   $$  
   Then divide by 7:  
   $$
   y^2 = \frac{x+3}{7}
   $$

4. Take the square root of both sides to solve for $$ y $$:  
   $$
   y = \pm \sqrt{\frac{x+3}{7}}
   $$

Thus, the inverse of the given relation is:  
$$
y = \pm \sqrt{\frac{x+3}{7}}
$$
The inverse of the relation $$ y = 7x^2 - 3 $$ is $$ y = \pm \sqrt{\frac{x+3}{7}} $$.

What is the inverse of the given relation?

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