(a) Let \(f(x) : (-\infty,0) \cup (0,\infty) \to \mathbb{R}\) be defined by \( f(x) = x - \frac{1}{x}.\)
Show that f(x) has no inverse function.
(b) Let \(g(x) : (0,\infty) \to \mathbb{R}\) be defined by \(g(x) = x - \frac{1}{x}.\)
Show that g(x) has an inverse function.
I'm a little confused on how to find the inverse function of f(x). I thought I could prove that two x values could both have the same y value, but I'm not really sure how. Solving for the inverse of f(x) also confused me a little. How should I proceed?
Thanks!
+128474 y = x - 1/x
y = [ x^2 -1 ] / x
xy = x^2 - 1
Trying to get x by itself and then "swap" x and y is sticky
Here's one approach by completing the square on x
yx = x^2 - 1 rearrange as
x^2 - yx - 1 = 0 complete the square
Take 1/2 of y = y/2 square it = y^2/4 add to both sides
x^2 - yx + (y^2/4) = 1 + (y^2/4) the left side factors as
( x - y/2)^2 = ( 4 + y^2) / 4
x - y/2 = ± √(4 + y^2) / 2
x = ± √(4 + y^2) / 2 + y/2
x = [ y ± √ (4 + y^2) ] / 2 "swap" x and y
y = [ x ± √ (4 + x^2 ) ] / 2 = the "inverse"
Note that if x = 1 in the first function, then y = 0
So the point ( 0, 1) should be on the "inverse" graph...so
1 = [ 0 ± √ [ 4 + 0 ] ] / 2
1 = ± √4/2
1 = ±1
This is only true if we take the positive root
So..... x - 1/x has no inverse on (-inf, 0) U (0, inf) because we don't get a unique point for the "inverse"
However....if we restrict the domain of g (x) to (0, inf) then the inverse becomes
y = [ x + √ [ 4 + x^2 ] ] / 2
Now.....note if x = 2 in g(x) then g(2) = 3/2
And the point ( 2,3/2) is on g
So....the point (3/2 , 2) should be on the inverse of g
So
2 = [ 3/2 + √ [ 4 + (3/2)^2 ] ]/ 2
4 = 3/2 + √ [ 4 + 1.5^2 ]
4 = 3/2 + √ [ 4 + 2.25 ]
5/2 = √(6.25
2.5 = 2.5 which is true
So we have an inverse if we restrict the domain to (0, inf)
Also.....look at the graph here for x -1/x
https://www.desmos.com/calculator/g4o70vl223
This graph does not pass the horizontal line test ( a requirement for an inverse)
However....if we restrict the domain to (0, inf) we will have a graph that will pass the horizontal line test, and so, it will have an inverse
