(Pure Mathematics) About differentiable function

KGPPRA

2007-12-01 9:58 am

Let f(x) be a differentiable function with f(0) = 0 and f"(x) < 0 for x∈(0,1]

(a) Prove taht g(x) is monotonic decreasing function for x∈(0,1]
where g(x)=f(x)/x
(b) Show that f(1) ≤ g(x) ≤ f'(0) for x∈(0,1]
THX~

回答 (1)

myisland8132

2007-12-01 10:17 pm

✔ 最佳答案

ANSWER
(a)
g'(x)
=[xf'(x)-f(x)]/x^2
Since for any x∈(0,1], there is exist a c∈(0,x] such that
f(x)-f(0)=(x-0)f'(c)
f(x)=xf'(c)
So
g'(x)
=[xf'(x)-xf'(c)]/x^2
=[f'(x)-f'(c)]/x
<0 [because f'(x) is monotonic decreasing ]
We prove taht g(x) is monotonic decreasing function for x∈(0,1]
(b)
Because g(1)=f(1) and g(x) is monotonic decreasing, so
g(x)>=g(1)=f(1) for x∈(0,1]
On the other hand
g(0)=f(x)-f(0)/(x-0)=f'(0)
So g(x)<=g(0)=f'(0) for x∈(0,1]
Combine f(1) ≤ g(x) ≤ f'(0) for x∈(0,1]

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