Explanation:
Let f(x)=y
y=2x/X-5
y(x-5)=2x
xy-5y=2x
x(y-2)=5y
x=5y/y-2
Apply x=5y/y-2, on f(x)
f(5y/y-2)=2(5y/y-2)/[(5y/y-2)-5]
=(10y/y-2)/[(5y-5y+10)/y-2]
=10y/10
=y
So f^-1(x)=5y/y-2
1.6m questions
2.0m answers