How do I find the inverse functions from abs(x^2 - 3)? WolframAlpha gives 4 expressions.
Yes, it does! And the reason is exactly what Prove It said- the function is not "one to one". If you were to graph it you would see that there are some horizontal lines that cross the graph 4 times- so 4 different values of x that give the same y value.
Specifically, $\displaystyle y= x^2- 3$ is a parabola that crosses the x-axis at $\displaystyle (-\sqrt{3}, 0)$, goes down to (0, -3), back up to the x-axis at $\displaystyle (\sqrt{3}, 0)$, and then up. Taking the absolute value then "folds" that portion below the x-axis up above it.
If $\displaystyle x\le -\sqrt{3}$, then $\displaystyle x^2\ge 3$ so $\displaystyle x^2- 3\ge 0$. Taking the absolute value of a positive number doesn't change it so, for $\displaystyle x\le -\sqrt{3}$, the function is $\displaystyle y= x^2- 3$. To find the inverse function, swap x and y and solve for y: $\displaystyle x= y^2- 3$ so $\displaystyle y^2= x+ 3$ and $\displaystyle y= \pm\sqrt{x+ 3}$. But in this case, y (which was x but got swapped) is less than or equal to -3. y is negative so $\displaystyle y= -\sqrt{x+ 3}$.
If $\displaystyle -\sqrt{3}< x\le 0$, then $\displaystyle x^2< 3$ so $\displaystyle x^2- 3< 0$. Taking the absolute value multiplies that by -1 so, for $\displaystyle -\sqrt{3}< x< 0$, $\displaystyle y= -(x^2- 3)= 3- x^2$. Again, we swap x and y and solve for y. $\displaystyle x= 3- y^2]$ so $\displaystyle y^2= 3- x$ and $\displaystyle y= \pm\sqrt{3- x}$. Because, here, x< 0 becomes y< 0, we must take the negative sign. $\displaystyle y= -\sqrt{3- x}$.
If $\displaystyle 0< x\le \sqrt{3}$, we have the same situation as the previous case except that y is positive so we take the positive sign on the square root: $\displaystyle y= \sqrt{3- x}$.
If $\displaystyle \sqrt{3}< x$, we have the same situation as the first case except that y is positive so we take the positive sign on the square root: $\displaystyle y= \sqrt{x- 3}$.