maths integration

2013-03-11 1:45 am
http://upload.lsforum.net/users/public/d7969Untitleds184.jpg


why this question dont have to consider both x>2 and x <-2 ??
just consider x>2 ???

回答 (1)

2013-03-16 3:22 am
✔ 最佳答案
x^3/(x^2 – 4)^0.5

If x < 2, (x^2 – 4) will be a negative number. You can’t find the square root of a negative number unless you go into complex number. This function is defined for real number only.

If x = 2, (x^2 – 4) will be zero. The square root of a zero is zero. Then the function involves division by zero. The function is not defined if you have a zero in the denominator.

This function is only valid for x > 2.


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