Attempted solution:
y=x/(xˆ2-4)
let y = inverse of y
x=y/(yˆ2-4)
xyˆ2-4x = y
xyˆ2-y-4x = 0
a = x, b = -1, c = -4x
y = (1 +/- sqrt(1-(4x)(-4x)))/(2x)
= (1+/- sqrt(1+16xˆ2))/2x, which is my solution
actual solution is:
y= (1-sqrt(16+xˆ2))/2x when x≠0
y= 0 when x = 0
I don't understand why the + sign has been discarded from the solution when x≠0 and I don't understand why y = 0 when x = 0. You could set y to 0 and solve for x in the original equation, but is that what is supposed to be done?
Thanks.
You simply must consider things up front so that they are not a surprise along the way.
Start from the top:
y = x/(x^2 - 4) perhaps it is wiser to think about f(x) = x/(x^2 - 4). This may help as we think on Domain and Range.
1) Discard x = 2 and x = -2. The Domain of f(x) is the Real Numbers with 2 and -2 excluded.
2) y = 0 is an asymptote for the graph. One might be tempted di discard y = 0 from the Domain, but remember that asymptotes don't really care about more centralized behavior. We have f(0) = 0.
3) Having determined that the function f(x) is in three pieces, one is caused to wonder if there is an easy INVERSE FUNCTION. A quick check of the graph indicates there is not. We'll have to chop it up a bit. Vertical asumptotes are a great place to start making divisions of the Range. There may be entirely different solutions for inverses of the three pieces. It's possible that two pieces will work together reasonably well. Thinking on it a bit, I think the middle portion should have one solution and the two outer portions another.
4) Very nicely, you have found the inverse relation of the two ourter portions. Good work. The Domain of the outer portions is x < -2 or x > 2. Notice how [-2,2] is not included. The range of the outer portions is all Real Numbers except y = 0. we can expect, then, that a proper inverse would have a Domain of all Real Numbers with zero (0) removed and a Range of y>2 or y < -2. See if these last characteristics fit your solution with the "+". Notice how, since x is already not zero, it is nto necessary to exclude it artificially and it is inappropriate to define it.
5) You "Actual Solution" seems to be missing some stuff. It has the inverse of only the middle section of f(x). If this is what is wanted, then the "actual solution" is right on.
Really, you have to think about these things. You can't just swap and solve willy-nilly.
From your first post:
(assuming what you wrote has a typo, by the way).
Clearly (1, -1/3) is a point on the graph of y = f(x). Therefore (-1/3, 1) is a point on the graph of . So the positive root in the above is rejected. Therefore
Clearly (0, 0) is a point on the graph of y = f(x). Therefore (0, 0) is a point on the graph of . But has the indeterminant form 0/0. Therefore you must define what happens at x = 0 separately.
As explained in...
http://www.mathhelpforum.com/math-he...ns-185019.html
... if exists the limit...
(1)
... then has inverse function and is...
(2)
In our case is so that the limit (1) is . Now is...
(3)
... so that the result is 'all right' ...
Kind regards