※ 引述《huangko (淺水者)》之銘言：
: Suppose that f is continuous on ﹝a.b﹞ , differentiable in (a,b) , and
: f(a)=f(b)=0 . Prove that for each real number 阿法 there is some c 屬於(a,b)
: such that f'(c)=阿法f(c) .
: '幫忙一下 謝謝!!
Let g(x) = (e^(-αx))(f(x))
Then g is continuous on [a,b] and differentiable on (a,b)
g(a) = (e^(-αa))(f(a)) = 0 , g(b) = (e^(-αb))(f(b)) = 0
By Rolle's Theorem , there exists c belonging to (a,b) such that g'(c) = 0
g'(x) = (-α)(e^(-αx))(f(x)) + (e^(-αx))(f'(x))
g'(c) = (-α)(e^(-αc))(f(c)) + (e^(-αc))(f'(c)) = 0
Since e^(-αc) ≠ 0 , (-α)(f(c)) + f'(c) = 0 => f'(c) = (α)(f(c))
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