Math & Stat

It is because there are some continuous graphs that are not differentiable at certain point.

Take an easier example, the graph y = |x| (i.e. absolute value of x) is continuous everywhere, however, at x = 0, the function doesn't have derivative there.
More rigorously, we consider the derivative of |x|
{ -1 for x < 0
Note that d/dx (|x|) = { non exist for x = 0
{ 1 for x > 0

Since the left derivative is not equal to right derivative at x = 0, so this function is not differentiable there.

Back to your question, consider f(x) = 3/4(x^2 - 1)^2/3
we have f'(x) = 1/2 * 2x * (x^2 - 1)^(-1/3) = x / [x^2 - 1]^(1/3)
which is undefined at x = 1 and -1.
You may try to check the derivatives at x = 1, -1 to see whether their left and right
derivative are equal, if not, you may expect there are sharp point at these points.

此圖本人於Graphmatica畫了出來
這我想不算尖角
不過是頂點
於本人的觀念看來
某個function f(x)
在某一點尖角x=a
當lim(x→a+) f(x)不等於lim(x→a-) f(x)
例如f(x)=|x|
如以f(x)=3/4(x^2 - 1)^2/3計
d[f(x)]/dx=(1/x)(x^2-1)^(-1/3)
我在該軟件畫出,頂點位於x=-1,1
當lim(x→-1+) f(x)=lim(x→-1-) f(x)=1
lim (x→1+) f(x)=lim(x→1-) f(x)=1
該頂點不是尖角

2007-02-18 22:03:45 補充：
打漏了:於本人的觀念看來某個function f(x)在某一點尖角x=a當lim(x→a十) f(x)不等於lim(x→a-) f(x)例如f(x)=|x|,在x=0才算尖角