Asked by Vanilla
Find the inverse function of f
f(x) = x^2 + x, x >(or equal to) (-1/2)
Thanks!
f(x) = x^2 + x, x >(or equal to) (-1/2)
Thanks!
Answers
Answered by
Reiny
let y = x^2 + x , x ≤ -1/2
the inverse is
x = y^2 + y
solving for y ...
y^2 + y - x = 0
y = (-1 ± √(1 - 4x))/2
f(x) = -1/2 ± √(1-4x)/2 where x ≤ 1/4
but in the original we are only considering the left "wing" of the parabola, so in the inverse we have to consider only
f(x) = -1/2 - √(1-4x)/2 , for x ≤ 1/4
testing:
in original, let x = -5
f(-5) = 25-5 = 20 , point is (-5,20)
in the inverse
f(20) = -1/2 - √81/2 = -1/ - 9/2 = -5
the inverse is
x = y^2 + y
solving for y ...
y^2 + y - x = 0
y = (-1 ± √(1 - 4x))/2
f(x) = -1/2 ± √(1-4x)/2 where x ≤ 1/4
but in the original we are only considering the left "wing" of the parabola, so in the inverse we have to consider only
f(x) = -1/2 - √(1-4x)/2 , for x ≤ 1/4
testing:
in original, let x = -5
f(-5) = 25-5 = 20 , point is (-5,20)
in the inverse
f(20) = -1/2 - √81/2 = -1/ - 9/2 = -5
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