Find f^-1(x) the function is f(x)=2-(x+1/x)?

Question

Have tried to solve this several times, but can't figure out what I'm doing wrong. Could someone illustrate the problem in steps? Thank you!

Answer 1

f(x)=2-(x+1/x).    According to order of operations, this function should be f(x) = 2 - (x + (1/x)).   However this is a much more difficult problem.   

 

The function y = 2 - (x + 1/x)  is not a 1-1 function, and so has no inverse function.  You can see this by noticing that the function has the same value for x = a and x = 1/a so it can't be 1-1.

 

Here is the computation:

 

y = 2 - (x + 1/x)  = 2 - ((x² + 1)/x)

 

If x is not zero, then xy = 2x -x² -1,  and so x² + (y-2)x + 1 = 0

 

Using the quadratic formula, 

 

x = [-(y-2) ± √((y-2)² - 4)]/2

 

x = (2-y)/2 ± [√((y-4)y)]/2

 

So the inverse relation for this function is

 

y = (2-x)/2 ± [√((x-4)x)]/2

 

 

 

 

 

 

 

 

Answer 2

Kenneth is right, the answer is

 

 y = 1(1-x)

 

Here is the solution. If g is an inverse function to f, then we have f(g) =x

 

meaning that f and g are symmetric with respect to y=x line. Thus, we have

 

f(g) = 2- (g+1)/g=(g-1)/g = x or

 

1- (1/g) = x ⇒ g = 1/(1-x)

 

 

 

Answer 3

Does f(x) = 2 - (x+1/x) mean:

 

1.) f(x) = 2 - (x+1)/x ; or

 

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

 

For 1.:

 

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

 

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

 

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

 

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

 

check:

 

f^-1(f(x)) = 1/(1 - f(x)) = 1/(1 - (1 - 1/x)) = 1/(1/x)) = x √

f(f^-1(x)) = 1 - 1/(1/(1 - x)) = 1 - (1 - x) = x √

 

For 2.:

 

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

 

Let g = f^-1(x)

 

x = 2 - g - 1/g

 

g + 1/g = 2 - x

 

g^2 + (x - 2) g + 1 = 0

 

g = f^-1(x) = (-(x - 2) ± √((x - 2)^2 - 4(1)(1)))/(2(1))

 

f^-1(x) = (2 - x  ± √(x^2 - 4x + 4 - 4))/2

 

f^-1(x) = (2 - x ± √(x^2 - 4x))/2

 

f^-1(x) = (2 - x ± √(x(x - 4)))/2

 

Obviously the inverse is not a function because it has 2 y-values for each x-value.

 

The composition checks are very challenging, so here's a geometric solution from GeoGebra:

http://www.dropbox.com/s/7fjmaxqay06g7r8/sks23cu14013001.png

Answer 4

Gregory, I suspect you're having problems with the simplification step (step 3).  I think the key is to recognize that in your original equation, x divides both x and 1, so (x+1)/x = 1 + (1/x), 2 minus that is  equal to 1 - (1/x) [you have to distribute the negative sign across the parentheses].  In the steps below, I did this after the swap of x for y, but you can do it either before or after.  I hope this helps.  John

 

1st step:  Replace f(x) with y

 

      2        x + 1

y = --  -  --------

      1           x

 

2nd step:  Swap x every place it occurs for y and y every place it occurs for x

 

      2      y + 1
x = -- - --------
      1        y

 

3rd step:  Simplify equations to get y alone on one side of an equation

 

      2      y + 1       2     y     1                1           1
x = -- - -------- = --- - --- - --- = 2 - 1 - --- = 1 - ---
      1         y          1     y     y                y            y

 

x = 1 - (1/y) => (1/y) = 1 - x => y = 1/(1-x)

 

4th step: replace y with f^-1(x)

 

y = 1/(1-x) => f^-1(x) = 1/(1-x)

 

Note:  I've now edited the math, I lost a minus sign working through it the first time.  Ty to the tutors who caught my mistake.

 

Also, as other answers suggest, it is correct to note that the given function, strictly speaking does not have an inverse because it cannot be a 1-1 function.  Both the original function (at x=0), and the proposed inverse (at x=1) have holes in their respective domains because division by zero is not permitted.