f(x)={ (sgn x)^(sgn x) }^n ; n is an odd integer . Find whether f(x) is even or odd.

Hi Manuj,

sgnx = |x|/x

As |x|^2 = x^2, we have |x|/x = x/|x|.

Hence f(x) as mentioned above will be odd for odd integers " n", since f(-x) = -f(x).

Hope it helps.

Wish you all the best.

Regards,

Ashwin (IIT Madras).

It is clearly an odd function.

Solution:

Sgn(x) which stands for signum function of x which always return either 1 or -1 for positive and negative values respectively. It is  defined for  R-{0}

To find whether it is an odd or even function , lets take two values 1 and -1 for testing the function

If f(x)= f(-x) it is even otherwise odd.

for x=1, the function would be like this (1¹)ⁿ where n is an odd no, f(x)= 1

for x=-1, the function would be like this (-1^-1)ⁿ where n is an odd no, f(-x)= -1

since f(x) is not equal to f(-x). It is an odd function