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Given is a function f(x) = 4 x/ [(x- 2) (x- 3)]

The domain of the function is all the possible input values of ‘x’.

Here if we equate the dominator [(x- 2) (x- 3)] = 0

This gives the possible x = 2 and x = 3

When x = 2, 3 the function is undefined as denominator = 0 and hence those values are not included.

Given is a function f(x) = 5 x/ [(x- 1) (x- 5)]

The domain of the function is all the possible input values of ‘x’.

Here if we equate the dominator [(x- 1) (x- 5)] = 0

This gives the possible x = 1 and x = 5;

When x = 1, 5 the function is undefined as denominator = 0 and hence those values are not included.