# Mr. G: please check me on this...

[ Follow Ups ] [ Post Followup ] [ Main Message Board ] [ FAQ ]

Posted by Joel on October 15, 2002 at 19:31:16:

In Reply to: inverse operations posted by Madison on October 15, 2002 at 17:36:53:

: g(x)=2 + sqrt[x + 1]
: find the inverse of this, if it exists as a function.
: also, state the restrictions. i dont know how to tell what the restrictions of the inverse are. reply ASAP

because if I'm overlooking something I'm sure he'll spot it.

In the meantime, to find the inverse, let g(x)=y , so
y = 2 + sqrt(x+1)
Now reverse the variables and make it
x = 2 + sqrt(y+1)
Now solve for y
x - 2 = sqrt(y+1)
square both sides:
x^2 - 4x + 4 = y + 1
y = x^2 - 4x + 3 This is the inverse function. Call it h(x).
To check it, you know that if h(x) is the inverse of g(x), then g(h(x)) = x so try it:
g(h(x)) = 2 + sqrt[(x^2-4x+3)+1]
g(h(x)) = 2 + sqrt[x^2-4x+4]
g(h(x)) = 2 + sqrt[(x-2)^2]
g(h(x)) = 2 + x - 2
g(h(x)) = x
So it works.

As to restrictions, y = x^2 - 4x + 3 by itself does not need any restrictions to be a valid function. But your original function had restrictions. Your g(x) was valid only for x >= -1. Also, only the positive square root of (x+1) was allowed, which meant that y had to be >= 2. So I'm guessing that we have to keep the same restrictions, but apply them to the opposite variables, so we would restrict h(x) to x >= 2. y >= -1 is assured by the function itself (in h(x), no value of x will result in a y < -1 ) so presumably that doesn't have to be stated explicitly.
Correct, Mr. G?

• yes! T.Gracken 09:53:22 10/16/02 (1)
• Thanks. Joel 11:29:32 10/16/02 (0)

Name:
E-Mail:

Subject: