How do you find the inverse of $f(x) = x^2 + x - 2$ ?

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Answer 1 · 2015-11-21T09:42:18

Since it is not one to one, this function has no inverse function unless you restrict its domain.

Let $y = f(x) = x^2+x-2 = (x+1/2)^2-9/4$

Add $9/4$ to both ends to get:

$(x+1/2)^2 = y + 9/4$

So:

$x+1/2 = +-sqrt(y+9/4)$

Hence:

$x = -1/2+-sqrt(y+9/4)$

So if $y > -9/4$ then there are two Real values of $x$ such that $f(x) = y$ .

If we restrict the domain of $f(x)$ to $[-1/2, oo)$ , then we can define:

$f^(-1)(y) = -1/2+sqrt(y+9/4)$

Alternatively, if we restrict the domain of $f(x)$ to $(-oo, -1/2]$ then we can define:

$f^(-1)(y) = -1/2-sqrt(y+9/4)$

How do you find the inverse of f(x) = x^2 + x - 2f(x) = x^2 + x - 2?