Let f(x)=(x^n)∣x∣ , where n is a positive integer.
(a) Show , from the first principle , that f'(0) exists and find its value.
(b) Show that f'(x) = (n+1)[x^(n-1)]∣x∣
Hence find ∫(x^n)∣x∣dx
A pure Mathematics question
2007-11-01 9:52 am
回答 (1)
2007-11-01 5:01 pm
✔ 最佳答案
(a)lim (x^n|x| - 0)/(x - 0) = lim x^(n+1)/x = 0
x->0+
lim (x^n|x| - 0)/(x - 0) = lim -x^(n+1)/x = 0
x-> 0-
So f'(0) exist
(b) When x > 0
f'(x) = (n+1)x^n = (n+1)x^(n-1)*x
When x < 0
f'(x) = -(n+ 1) x^n = (n+1)x^(n-1)*(-x)
combining the results, we have f'(x) = (n+1)x^(n-1)*|x|
Let g(x) = x^(n+1) |x|, g'(x) = (n+2) x^n |x|
So ∫(x^n)∣x∣dx = x^(n+1) |x/(n+2) + constant
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