Inverse Functions

[track12]

Find the inverse of y= x+(x^-1)

I tried this so far...

y= (x^2+1)/x
y= x^2+1= xy
1= yx+x^2
1= (y-x)x
1/y-x= x

I can't seem to solve for x, can anyone help?

 

[Deleted member 4993]

Find the inverse of y= x+(x^-1)

I tried this so far...

y= (x^2+1)/x
y= x^2+1= xy


x^2 - xy + 1 = 0

solve for 'x' using quadratic equation(treating 'y' as constant).


1= yx+x^2
1= (y-x)x
1/y-x= x

I can't seem to solve for x, can anyone help?

 

[track12]

Ok thanks that helped a lot! But I'm still not sure I solved it correctly?

y=x+(x^-1)
y= (x^2+1)/x
x^2+1= yx
x^2-xy+1= 0
y (plus or minus) the square root of ((-y)^2 - 4) divided by 2
y (plus or minus) (y-16)/2
y (plus or minus) (y/2)-8
x=(3y/2)-8
y=(3x/2)-8

The last equation is what I think should be the answer?

 

[stapel]

y=x+(x^-1)
y= (x^2+1)/x
x^2+1= yx
x^2-xy+1= 0
y (plus or minus) the square root of ((-y)^2 - 4) divided by 2
y (plus or minus) (y-16)/2 <== How did this happen?
y (plus or minus) (y/2)-8 <== How did this happen?
x=(3y/2)-8

You have the following:

. . . . .\(\displaystyle y\, =\, x\, +\, \frac{1}{x}\)

You converted to the common denominator:

. . . . .\(\displaystyle y\, =\, \frac{x^2\, +\, 1}{x}\)

You then multiplied through by "x":

. . . . .\(\displaystyle xy\, =\, x^2\, +\, 1\)

Then you moved everything over together onto one side of the "equals" sign:

. . . . .\(\displaystyle 0\, =\, x^2\, -\, xy\, +\, 1\)

Now apply the Quadratic Formula to solve for x in terms of y:

. . . . .\(\displaystyle x\, =\, \frac{-(-y)\, \pm \, \sqrt{(-y)^2\, -\, 4(1)(1)}}{2(1)}\)

. . . . .\(\displaystyle x\, =\, \frac{y\, \pm \, \sqrt{y^2\, -\, 4}}{2}\)

But I don't understand what you did after this...? How did y[sup:j7wcc6s9]2[/sup:j7wcc6s9] - 4 become y - 16...?

(The "x=" is the inverse, though obviously it's not a function.)

Eliz.