How do you find the inverse of #f(x) = x^2 + x# and is it a function?

1 Answer
May 16, 2018

inverse relation is #g(x) = \frac{-1\pm \sqrt{1+4x)}{2}#

Explanation:

let #y = f(x) = x^2 + x#
solve for x in terms of y using the quadratic formula:
#x^2+x-y = 0#,
use quadratic formula #x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}#

sub in # a=1, b=1, c = -y#
#x = \frac{-1\pm \sqrt{1^2-4(-y)}}{2}#
#x = \frac{-1\pm \sqrt{1+4y)}{2}#

Therefore the inverse relation is #y = \frac{-1\pm \sqrt{1+4x)}{2}#

Note that this is a relation and not a function because for each value of y, there are two values of x and functions cannot be multivalued