# Thread: Is my inverse right?

1. ## Is my inverse right?

I had to find the inverse of a function, and it sort of feels right but looks wrong if you know what I mean.

I was given

and came up with

Don't feel the x should be under the root somehow. Any ideas?

2. ## Re: Is my inverse right?

No, that is not correct. For one thing, in the original function, $\displaystyle f(x)= -3(x-2)^2+ 12$, when x= 4, y= 0. But with your function, $\displaystyle f^{-1}(x)= 1+ \sqrt{\frac{x- 12}{3}}$, does not have a real number when x= 0. Also since the original function has limits on x, so should the inverse function.

As for x being "under the root", since in the original function x is squared, I can't see why you should expect the "opposite" of that, the square root of x, for the inverse function. But how did you get x- 12, rather than 12- x?

3. ## Re: Is my inverse right?

Originally Posted by HallsofIvy
No, that is not correct. For one thing, in the original function, $\displaystyle f(x)= -3(x-2)^2+ 12$, when x= 4, y= 0. But with your function, $\displaystyle f^{-1}(x)= 1+ \sqrt{\frac{x- 12}{3}}$, does not have a real number when x= 0. Also since the original function has limits on x, so should the inverse function.

As for x being "under the root", since in the original function x is squared, I can't see why you should expect the "opposite" of that, the square root of x, for the inverse function. But how did you get x- 12, rather than 12- x?
The domain/range (x axis) is restricted to being less than or equal to 1 or less than or equal so as to eliminate the possibility of two roots. Also an unrestricted function f(x)=-3(x-1)^2+12 has y=9 when x is 0 and x= -1 and x=3 when y = 0. So not sure where you are getting x=4 from?

I got x-12 by subtracting 12 from both sides. I then put the whole of both sides over -3 to get rid of the -3 which multiplies the brackets on the right, then put the both sides under a root to un-square the bracket. Tried replacing (x-1)^2 with (x-1)(x-1) and multiplying out and got into a bit of a pickle.

4. ## Re: Is my inverse right?

This is how I do it:

$\displaystyle y = -3(x - 1)^2 + 12$

$\displaystyle 12 - y = 3(x - 1)^2$

$\displaystyle 4 - \frac{y}{3} = (x - 1)^2$

Now here we have a problem....we don't know which square root to take, yet. More on that later.

$\displaystyle \pm \sqrt{4 - \frac{y}{3}} = x - 1$

$\displaystyle 1 \pm \sqrt{4 - \frac{y}{3}} = x$

now when x is 0, y is 9, so this helps us decide which square root we need:

$\displaystyle 1 + \sqrt{4 - \frac{9}{3}} = 1 + \sqrt{4 - 3} = 1 + 1 = 2$. Wrong one.

$\displaystyle 1 - \sqrt{4 - \frac{9}{3}} = 1 - \sqrt{4 - 3} = 1 - 1 = 0$. Bingo!

So it ought to be that:

$\displaystyle f^{-1}(x) = 1 - \sqrt{4 - \frac{x}{3}}$.

Just as a safeguard, let's see if $\displaystyle f \circ f^{-1}(x) = x$:

$\displaystyle f \circ f^{-1}(x) = f\left(1 - \sqrt{4 - \frac{x}{3}}\right) = -3\left(1 - \sqrt{4 - \frac{x}{3}} - 1\right)^2 + 12$

$\displaystyle = -3\left(-\sqrt{4 - \frac{x}{3}}\right)^2 + 12 = -3\left(4 - \frac{x}{3}\right) + 12 = -12 + x + 12 = x$

5. ## Re: Is my inverse right?

Of course, deveno's $\displaystyle 4- \frac{x}{3}$ is the same as $\displaystyle \frac{12- x}{3}$. But note it is 12- x, not x- 12.

6. ## Re: Is my inverse right?

Originally Posted by alexpasty2013
The domain/range (x axis) is restricted to being less than or equal to 1 or less than or equal so as to eliminate the possibility of two roots. Also an unrestricted function f(x)=-3(x-1)^2+12 has y=9 when x is 0 and x= -1 and x=3 when y = 0. So not sure where you are getting x=4 from?

I got x-12 by subtracting 12 from both sides. I then put the whole of both sides over -3 to get rid of the -3 which multiplies the brackets on the right,
There's your difficulty. Dividing x- 12 by -3 gives [tex]\frac{x- 12}{-3}= \frac{12- x}{3}. You don't have "-3" in the denominator so you must have 12- x, not x- 12, in the numerator

then put the both sides under a root to un-square the bracket. Tried replacing (x-1)^2 with (x-1)(x-1) and multiplying out and got into a bit of a pickle.