[微積] 找critical point
題目:f(x)= sqrt(x^2-4),求f(x)的critical point
書上定義critical point是導數為0或導數不存在的點
首先我分析這函數的定義域,得到Dom(f)=R\(-2,2)
x
顯然對x < -2 及 x > 2 ,f(x)均可導,導數為 f'(x)=------------
sqrt(x^2-4)
此時導數均存在
接著再看端點x=-2與x=2
f(x) - f(-2) sqrt(x^2-4) - 0
考慮左導數 f' (-2) = lim -------------- = lim ----------------
- x-> -2 x - (-2) x -> -2 x + 2
sqrt(x-2)
= lim ----------- = -infty
x -> -2 sqrt(x+2)
類似地,右導數 f' (2) = +infty
+
因此f在x=-2,2 均不可導,所以f的critical point有-2,2
不過書上給的答案是"沒有critical point",這讓我很疑惑,麻煩大家幫忙,謝謝
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