Find the domain of f(x)=1/√(x-x2).

Hi Menka,

To find the domain of functions, you need to find those values of X, where the function f(x) is well defined.

And hence for the above function to be well defined, the denominator cannot be zero, and the term inside the root must be positive.

So we get x-x² > 0,

which will give x(x-1) > 0..... which gives x > 1, or x<0.

So domain is (-∞,0)U(1,∞).

Hope that helps,

All the best,

Regards,

Ashwin (IIT Madras).

For =1/√(x-x²). to exist , x-x² should be greater > 0 , for this purpose

x(1-x)>0

This can be posiible only when x and (1-x) are either simultaneously > 0 or x and 1-x should be simultaneously < 0

Case I: x>0 and ((1-x) > 0 ==> x<1 )so function is defined between (0,1)

Case II: x<0 and (1-x) < 0 ==> x > 1 which is really not possible.

So the function is defined for (0,1)

Hi Menka,

Extremely sorry for my silly error in writing x-x^2 = x(x-1)

It should be x(1-x) > 0 (or) x(x-1)<0.

Hence 0<x<1 is the domain of this fuction.

ie x belongs to (0,1).

Hope it helps.

Best Regards,

Ashwin (IIT MadraS).