Inverse of #f(x)=(x^2+2)/(x-3)# ?

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Answer 1 · 2018-05-03T21:18:07

Depending upon which branch we seek, we have:

# f^(-1)(x) = x/2 +sqrt(x^2/4-3x-2) #

# f^(-1)(x) = x/2 -sqrt(x^2/4-3x-2) #

Let:

# y = (x^2+2)/(x-3) #

Then in order to find the inverse, #f^(-1)(x)#, we rearrange the above equations so that we have #x=x(y)#, that is #x# as a function of #y# alone.

So that:

# y(x-3) = x^2+2 #

# :. yx-3y = x^2+2 #

# :. x^2-yx+2+3y = 0 #

This is a quadratic in #x#, so we can complete the square:

# :. (x-y/2)^2-(y/2)^2+2+3y = 0 #

# :. (x-y/2)^2 = y^2/4-3y-2 #

# :. x-y/2 = +-sqrt(y^2/4-3y-2) #

# :. x = y/2 +-sqrt(y^2/4-3y-2) #

This is not a true function, as it is multi-valued, as a result of the quadratic term in #x# in the original function.

The graphs of the functions are:

and we clearly see that the inverse is a reflection of the original function in the line #y=x#

So, depending upon which branch we seek, we have:

# f^(-1)(x) = x/2 +sqrt(x^2/4-3x-2) #

# f^(-1)(x) = x/2 -sqrt(x^2/4-3x-2) #