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#