Posted by Joel on October 16, 2002 at 11:29:32:
In Reply to: yes! posted by T.Gracken on October 16, 2002 at 09:53:22:
: : : 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. you are correct (again)! the domain of the inverse is the range of the original function and the range of the inverse is the domain of the original function.
Post a Followup