SOLUTION: I need to find the inverse function of m(x)= (1/x)-x I know the answer is =(-x+sqroot(x^2+4))/2 but I don't know how to get there

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Question 1132864: I need to find the inverse function of m(x)= (1/x)-x I know the answer is =(-x+sqroot(x^2+4))/2 but I don't know how to get there
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!


+m%28x%29=+1%2Fx-x
to find the inverse function of m%28x%29 (recall m%28x%29=y),
y=+1%2Fx-x..............swap x and y
x=+1%2Fy-y..............solve for y
x=+1%2Fy-y%5E2%2Fy
x=+%281-y%5E2%29%2Fy
xy=+1-y%5E2......move y%5E2 to the left
y%5E2%2Bxy=1----complete square on left side
%28y%5E2%2Bxy%2Bb%5E2%29-b%5E2=1........2ab=x =>since coefficient a=1, b=x%2F2
%28y%5E2%2Bxy%2B%28x%2F2%29%5E2%29-%28x%2F2%29%5E2=1
%28y%2Bx%2F2%29%5E2-x%5E2%2F4=1
%28y%2Bx%2F2%29%5E2=x%5E2%2F4%2B1
%28y%2Bx%2F2%29%5E2=x%5E2%2F4%2B4%2F4
%28y%2Bx%2F2%29%5E2=%28x%5E2%2B4%29%2F4
y%2Bx%2F2=sqrt%28%28x%5E2%2B4%29%2F4%29
y%2Bx%2F2=sqrt%28x%5E2%2B4%29%2Fsqrt%284%29
y%2Bx%2F2=sqrt%28x%5E2%2B4%29%2F2
y=+-x%2F2%2Bsqrt%28x%5E2%2B4%29%2F2
y=+%28-x%2Bsqrt%28x%5E2%2B4%29%29%2F2

=>+m%5E-1%28x%29=%28-x%2Bsqrt%28x%5E2%2B4%29%29%2F2