repost to rational functions question?

Question

this function I am trying to figure out is y= 4x OVER or divided by x squared + 1 and having to determine the inverse I believe by exchanging x and y, but I can't put it together.

Alyssa

Answer 1

Not positive about answer:
y=(4x/x)squared + 1
y=(16xsquared/xsquared) + 1
y=(16)+1
y=17

Answer 2

You are right - you exchange x and y.

x = (4y)/(y^2 + 1)

(y^2 + 1) = 4(y/x)

y^2 - (4/x)y + 1 = 0

which is a quadratic equation in y. You could either factor it, or cheat and use the quadratic formula.

y = (4/x) +/- \sqrt[(16/x^2) - 4]/2 = (1/x)[2 +/- \sqrt(4 - x^2)]

SO there are two possible "inverses" (which should be expected, since the original equation has a square in it).

y = (1/x)[2 + \sqrt(4 - x^2)] or y = (1/x)[2 - \sqrt(4 - x^2)]

Answer 3

This is not the easiest inverse in the world to determine:
y = 4x/(x^2 + 1)
yx^2 + y = 4x
yx^2 - 4x + y = 0
x = (4 ± √(16 - 4y^2))/2y
x = (2 ± √(4 - y^2))/y

interchanging variables now, you have

y = (2 ± √(4 - x^2))/x

Answer 4

You switch x and y,
x = 4y/(y^2+1)
y^2+1-4y/x=0
y^2-(4/x)y+1=0

Solve for y,
y
= {(4/x)±√[16/x^2-4]}/2
= {2±√(4-x^2)}/x