For how many integral values of x will |(x-1!)| + |(x-2!)| + |(x-3!)| + |(x-4!)| asuume a minimum value?

Question

Answer 1

Since there is a modulus function, all the values except 1,2,3,4 will not give minimum values. So, for x = 1, 4 . Value = 9. for x = 2,3 . Value = 4. There for for values, x = 2,3. We have minimum value for the expression. Please correct me if I'm wrong.

Answer 2

Since there is a modulus function, all the values except 1,2,3,4 will not give minimum values. So, for x = 1, 4 . Value = 9. for x = 2,3 . Value = 4. There for for values, x = 2,3. We have minimum value for the expression. Please correct me if I'm wrong.

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For how many integral values of x will |(x-1!)| + |(x-2!)| + |(x-3!)| + |(x-4!)| asuume a minimum value?

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