題目是這樣的
Suppose that f and g are differentiable functions and f(x)g'(x)-g(x)f'(x) has
no zeros on some intervals I. Assume that there are numbers a and b in I with
a<b for which f(a)=f(b)=0 and that f has no zeros in (a,b). Prove that if
g(a)≠0 and g(b)≠0, then g has exactly one zero in (a,b).
題目給的hint是：解 h=f/g, 再解 k=g/f
h=f/g 這部份我寫這樣
Suppose that g has no zeros in (a,b) and consider h(x)=f(x)/g(x).
Since g(x)≠0 in [a,b] => h(x) is conti. on [a,b] and diff. on (a,b) and
h(a)=h(b)=0. Therefore by Rolle's thm, ∃c in (a,b) s.t. h'(c)=0 -><-
=> g has at least one zero in (a,b).
可是第二部分就看不懂提示了， 令 k=g/f
因為 f(a)=f(b)=0, 所以 k 在 a,b 兩點就沒有定義了 要怎麼用 Rolle's thm ？
謝謝
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※ 編輯: bks 來自: 140.115.221.192 (03/23 14:22)
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