How do you find $f^-1(x)$ given $f(x)=3/(x^2+2x)$ ?

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Answer 1 · 2017-01-30T08:50:26

The answer is $=-1+-sqrt(x(x+3))/x$

Let $y=3/(x^2+2x)$

Then,

$y(x^2+2x)=3$

$yx^2+2yx-3=0$

Comparing this equation to

$ax^2+bx+c=0$

Calculating the discriminant

$Delta=b^2-4ac=(2y)^2-4*y+(-3)$

$=4y^2+12y=4y(y+3)$

So,

$x=(-b+-sqrtDelta)/(2a)$

$x=((-2y)+-sqrt(4y(y+3)))/(2y)$

$x=-1+-sqrt(y(y+3))/y$

Interchanging $x$ and $y$

$y=-1+-sqrt(x(x+3))/x$

Therefore,

$f^-1(x)=-1+-sqrt(x(x+3))/x$

How do you find f^-1(x)f^-1(x) given f(x)=3/(x^2+2x)f(x)=3/(x^2+2x)?