Inverse Fucntions
How do I, algebraically, find the inverse function to f(x) = 2x/(x^2-1) ?
I just got the answer of this, it's x = 1/y.(+/- sqrt(-y^2+1) + 1)
But according to this equation, y=0 is undefined but f(0) = 0. Why the inverse not giving the correct answer? (The inverse matches graphically)
STEP 1 For a given f(x), set f(x) = y.
STEP 2 Isolate x in terms of y.
STEP 3 Set x = f⁻¹(y).
Note that this process does not always yield a function.
Many rational functions are not invertible.
If f(x) = 2x/(x² - 1) then to find f⁻¹(y), try to isolate the x.
y = 2x/(x² - 1)
y(x² - 1) = 2x
yx² - y = 2x
yx² - 2x - y = 0
Using the Quadratic Formula with a = y, b = -2, c = -y:
x = [2 ± √(4 + 4y²)]/2y
x = [2 ± 2√(1 + y²)]/2y
x = [1 ± √(1 + y²)]/y
f⁻¹(y) = [1 ± √(1 + y²)]/y
f⁻¹(x) = [1 ± √(1 + x²)]/x
Some follow up questions:
Is f⁻¹(x) the result of reflecting f(x) over the line y = x? Yes.
https://www.desmos.com/calculator/xdouywionu
f(x) is blue. f⁻¹(x) is red. y = x is green.
Is f⁻¹(x) a function? No. f⁻¹(x) fails the vertical line test for all nonzero x.
Does f(f⁻¹(x)) = f⁻¹(f(x)) = x?
The composition of two inverse functions always equals to x.
This is how inverse functions can be verified algebraically.
I'll omit the algebraic check here, since f⁻¹(x) is already not a function.
For more information and practice problems on inverse rational functions, check out the following pages:
http://wmueller.com/precalculus/newfunc/invrat.html
(Mueller's page is crystal clear that invertible rational functions have at most one horizontal asymptote and one singularity.)
ChiliMath - Finding the Inverse Function of a Rational Function
TutorVista - Inverse of a Rational Function
Sal made a video on Finding inverse functions: rational... but since he made a mistake in the video, here are some alternatives:
How to Find the Inverse of a Function (mathbff)
Find the Inverse of a Rational Function (Mathispower4u)
and this clear, concise series of videos on inverse functions:
Inverse Functions - The Basics! (patrickJMT)
Finding the Inverse or Showing One Does Not Exist, Ex 1 (patrickJMT)
Finding the Inverse or Showing One Does not Exist, Ex 2 (patrickJMT)
Finding the Inverse or Showing One Does not Exist, Ex 3 (patrickJMT)
Finding the Inverse or Showing One Does not Exist, Ex 4 (patrickJMT)
Don't forget to come back and check your understanding with the practice exercises at Khan Academy - Finding Inverse Functions.
Good luck!
Yes, f(x) = 2x/(x^2-1) is not invertible. But restricting it's domain to (-1,1) would allow us to conclude that the inverse of the function f is 1/y.(1 - sqrt(1+x^2)) and the graph of the same is as below.
Now, f-inverse(x) = 1/x.[1-sqrt(x^2+1)]
and f(x) = 2x/(x^2-1) for (-1 < x < 1).
f-inverse is now a completely valid function, and (f-inverse(f(x)) = x and vice versa.
The problem is f(0) = 0 but f-inverse(0) is undefined! Why/how is this happening?
Ah, now we're delving into calculus.
If f(x) is restricted to the domain (-1,1) and the range (-∞,∞), then
f⁻¹(x) is restricted to the domain (-∞,∞) and the range (-1,1).
Restriction to the range (-1,1) excludes all the points on
Equation 3
and we have only the points on
Equation 4
But this expression of f⁻¹ falsely introduces a point discontinuity at 0. The removable discontinuity is merely an algebraic artifact of the inversion process that we employed, which involved the quadratic formula and algebraic division. Consider instead:
Equation 2
This expression of f⁻¹ has no discontinuities.
Alternately, the pointwise transformation of all points on the domain (-1,1) and the range (-∞,∞) (x,y) → (y,x) on the domain (-∞,∞) and the range (-1,1) generates a continuous f⁻¹.
This is what we expect, due to the following corollary of the Intermediate Value Theorem:
If f:I→J is a strictly decreasing continuous function on I, then f⁻¹ is a strictly decreasing continuous function on J, the image f(I).
For more on inverse functions and continuity:
http://people.reed.edu/~mayer/math111.html/chap14.pdf (page 298)
https://math.stackexchange.com/questions/672174/continuity-of-an-inverse-function
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