Inverse, domain and range of f and f-1, of f(x)=x-1/x+2?

Answer 1

f's range is all the numbers except 0 (where 1/x is undefined).

f's domain is all the numbers.


f's inverse is f^-1(x) = -1 + x/2 +/- 1/2 * Sqrt[ 8 - 4x +x^2 ]

The derivation of this is shown below.

The domain of f^-1 is all the numbers (though for many numbers only one reading of the +/- yields a value).
The range of f^-1 is all the numbers except 0 (for the same reason that f's domain didn't include 0).

Here's the derivation for f^-1(x) (basically, you just substitute y for f(x) and then solve for x in terms of y) :

y = x - 1/x + 2
xy = x^2 + 2x - 1
0 = x^2 + (2-y)x - 1

Then, using the quadratic formula:

x = { -(2-y) +/- Sqrt[ (2-y)^2 + 4] } / 2
= -1 + y/2 +/- 1/2 * Sqrt[ (2-y)^2 + 4]
= -1 + y/2 +/- 1/2 * Sqrt[ 8 - 4y +y^2 ]

Answer 2

Need to correct "Justin's" comments . Interchange the words "domain" and "range " for the function f(x). Then for the inverse functions use the plus sign to correspond to range of the inverse function x > 0 and negative sign for x < 0. And for "Helmut" the range of the inverse cannot contain y = - 2, since it is not in the domain of f(x).

Answer 3

f(x) - (x-1/x) = 2

Answer 4

f(x) = (x - 1) / (x + 2)
domain: all x ≠ - 2
Range: all y
x = (y - 1) / (y + 2)
xy + 2x = y - 1
y(1 - x) = 2x + 1
y = (2x + 1) / (1 - x)
f^-1(x) = (2x + 1) / (1 - x)
domain: all x ≠ 1
Range: all y