How do you find the inverse of #f(x) = root5((3 x - 5) / (x - 5)) #?

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Answer 1 · 2015-09-17T12:33:48

Let #y = f(x)# and solve for #x# to find:

#f^(-1)(y) = 10/(y^5-3) + 5#

Let #y = f(x) = root(5)((3x-5)/(x-5)#

Then:

#y^5 = (3x-5)/(x-5) = (3x-15+10)/(x-5) = 3+10/(x-5)#

Subtract #3# from both ends to get:

#y^5 - 3 = 10/(x-5)#

Multiply both sides by #(x-5)/(y^5 - 3)# to get:

#x - 5 = 10/(y^5-3)#

Add #5# to both sides to get:

#x = 10/(y^5-3) + 5#

So:

#f^(-1)(y) = 10/(y^5-3) + 5#

Answer 2 · 2015-12-18T16:01:47

#f^-1(x)=(5(x^5-1))/(x^5-3)#

Write as

#y=root5((3x-5)/(x-5))#

Switch #x# and #y# and solve for #y#.

#x=root5((3y-5)/(y-5))#

#x^5=(3y-5)/(y-5)#

#x^5(y-5)=3y-5#

#x^5y-5x^5=3y-5#

#x^5y-3y=5x^5-5#

#y(x^5-3)=5(x^5-1)#

#y=(5(x^5-1))/(x^5-3)#