what is one possible value of x for which the function g above is undefined? g(x) = √(x-1) (x-2) note: √ iss the root sign

A possible value of x for which g(x) = √(x-1)(x-2) is undefined is x = 1.5

Step-by-step explanation:

The function g(x) = √(x-1)(x-2) is undefined when (x-1)(x-2) < 0.

So (x-1)(x-2) < 0

⇒ x - 1 < 0 or x -2 < 2

⇒ x < 1 or x < 2.

So we look for the interval for which (x - 1)(x - 2) < 0

when x < 1, e.g x = 0, (x - 1)(x - 2) = (0 - 1)(0 - 2) = (-1)(-2) =2 > 0. So (x - 1)(x - 2) > 0 for x < 1

when 1 < x < 2, e.g x = 1.5, (x - 1)(x - 2) = (1.5 - 1)(1.5 - 2) = (0.5)(-0.5) = -0.25 < 0. So (x - 1)(x - 2) < 0 for 1 < x < 2

when x > 2 e.g x = 3, (x - 1)(x - 2) = (3 - 1)(3 - 2) = (2)(1) = 2 > 0. So (x - 1)(x - 2)  > 0 for  x > 2

Since (x - 1)(x - 2) < 0 for 1 < x < 2, this is the required interval.

So, g(x) = √(x-1)(x-2) is undefined in the interval 1 < x < 2

A possible value for x in this interval is x = 1.5

So, a possible value of x for which g(x) = √(x-1)(x-2) is undefined is x = 1.5

the answer is c

hope this !

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