inverse functions

Quote:

Originally Posted by Stuck Man

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, $y= x^2- 3$ is a parabola that crosses the x-axis at $(-\sqrt{3}, 0)$ , goes down to (0, -3), back up to the x-axis at $(\sqrt{3}, 0)$ , and then up. Taking the absolute value then "folds" that portion below the x-axis up above it.

If $x\le -\sqrt{3}$ , then $x^2\ge 3$ so $x^2- 3\ge 0$ . Taking the absolute value of a positive number doesn't change it so, for $x\le -\sqrt{3}$ , the function is $y= x^2- 3$ . To find the inverse function, swap x and y and solve for y: $x= y^2- 3$ so $y^2= x+ 3$ and $y= \pm\sqrt{x+ 3}$ . But in this case, y (which was x but got swapped) is less than or equal to -3. y is negative so $y= -\sqrt{x+ 3}$ .

If $-\sqrt{3}< x\le 0$ , then $x^2< 3$ so $x^2- 3< 0$ . Taking the absolute value multiplies that by -1 so, for $-\sqrt{3}< x< 0$ , $y= -(x^2- 3)= 3- x^2$ . Again, we swap x and y and solve for y. $x= 3- y^2]$ so $y^2= 3- x$ and $y= \pm\sqrt{3- x}$ . Because, here, x< 0 becomes y< 0, we must take the negative sign. $y= -\sqrt{3- x}$ .

If $0< x\le \sqrt{3}$ , we have the same situation as the previous case except that y is positive so we take the positive sign on the square root: $y= \sqrt{3- x}$ .

If $\sqrt{3}< x$ , we have the same situation as the first case except that y is positive so we take the positive sign on the square root: $y= \sqrt{x- 3}$ .