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Question 1167057: What is the inverse function of f(x)=x^2+x, x ≥ -1/2?
Found 2 solutions by Theo, solver91311:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! actually, x needs to be greater than or equal to -1/4, not -1/2, as you will see later.
if you let y = f(x), then your function of f(x) = x^2 + x becomes:
y = x^2 + x
since x^2 + x = (x + 1/2)^2 - 1/4, then your function becomes:
y = (x + 1/2)^2 - 1/4
add 1/4 to both sides of this function to get:
y + 1/4 = (x + 1/2)^2
take the square root of both sides of this function to get:
sqrt(y + 1/4) = x + 1/2
subtract 1/2 from both sides of the function to get:
sqrt(y + 1/4) - 1/2 = x
replace x with y and y with x to get:
sqrt(x + 1/4) - 1/2 = y
switch sides to get:
y = sqrt(x + 1/4) - 1/2
that's your inverse function.
if it is truly the inverse function, then it must be a reflection about the line y = x and (x,y) in the normal function must be the same distance from the line y = x as (y,x) is in the inverse function.
i graphed both the normal function and the inverse function and the line y = x to show you that the inverse function is a reflection about the line y = x.
i also graphed the line y = -x + 6 to show you that the point (x,y) in the normal function is the same distance from the line y = x as the point (y,x) in the inverse function.
here's what the graph looks like.
the distance between the point (1.646,4.354) on the normal graph and the point (3,3) on the line y = x is the same distance between the point (4.354,1.646) on the inverse graph and the same point of (3,3) on the line y = x.
to confirm, use the distance formula of d = sqrt((x2-x1)^2 + (y2-y1)^2)
when (x1,y1) from the normal function = (1.646,4.354) and (x2,y2) from the line (y = x) = (3,3), then:
d = sqrt((3 - 1.646)^2 + (3 - 4.354)^2) = 1.914845163
when (x1,y1) from the inverse function = (4.354,1.646)) and (x2,y2) from the line (y = x) = (3,3), then:
d = sqrt(3 - 4.354)^2 + (3 - 1.646)^2) = 1.914845163
the distances are the same, confirming that the inverse function for y = x^2 + x is the function y = sqrt(x + 1/4) - 1/2.
Answer by solver91311(24713)  (Show Source):
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