Inverse Fucntions

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