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

Attachment 29990

and came up with

Attachment 29991

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

Printable View

- Jan 7th 2014, 10:46 AMalexpasty2013Is 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

Attachment 29990

and came up with

Attachment 29991

Don't feel the x should be under the root somehow. Any ideas? - Jan 7th 2014, 11:46 AMHallsofIvyRe: Is my inverse right?
No, that is not correct. For one thing, in the original function, , when x= 4, y= 0. But with your function, , 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? - Jan 8th 2014, 02:45 AMalexpasty2013Re: Is my inverse right?
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. - Jan 8th 2014, 05:38 AMDevenoRe: Is my inverse right?
This is how I do it:

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

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

. Wrong one.

. Bingo!

So it ought to be that:

.

Just as a safeguard, let's see if :

- Jan 8th 2014, 05:46 AMHallsofIvyRe: Is my inverse right?
Of course, deveno's is the same as . But note it is 12- x, not x- 12.

- Jan 9th 2014, 08:40 AMHallsofIvyRe: Is my inverse 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

Quote:

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.