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)#