Asked by Anne
I need to find the inverse function of f(x)=sqrt(x^2+2x), where x>o
This is as far as I got:
x=sqrt(y^2+2y)
How in the world am I going to isolate for y? Should I complete the square underneath the square root symbol, and then use the quadratic equation? Does that work? :/
This is as far as I got:
x=sqrt(y^2+2y)
How in the world am I going to isolate for y? Should I complete the square underneath the square root symbol, and then use the quadratic equation? Does that work? :/
Answers
Answered by
helper
y = sqrt(x^2 + 2x)
square both sides
y^2 = x^2 + 2x
x^2 + 2x = y^2
complete the square
x^2 + 2x + 1 = y^2 + 1
(x + 1)^2 = y^2 + 1
take the square root of both sides
+ x + 1 = (sqrt(y^2 + 1))
- x - 1 = (sqrt(y^2 + 1))
x = -1 + - (sqrt(y^2 + 1))
interchange x and y
y = -1 +- (sqrt(x^2 + 1))
square both sides
y^2 = x^2 + 2x
x^2 + 2x = y^2
complete the square
x^2 + 2x + 1 = y^2 + 1
(x + 1)^2 = y^2 + 1
take the square root of both sides
+ x + 1 = (sqrt(y^2 + 1))
- x - 1 = (sqrt(y^2 + 1))
x = -1 + - (sqrt(y^2 + 1))
interchange x and y
y = -1 +- (sqrt(x^2 + 1))
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