Question

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

Answer 1

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