Question

Let a > 0 and f be continuous on [-a, a]. Suppose that f'(x) exists and f'(x)<= 1 for all x2 ㅌ (-a, a). If f(a) = a and f(-a) =-a. Show that f(0) = 0.

Hint: Consider the two cases f(0) < 0 and f(0) > 0. Use mean value theorem to prove that these are impossible cases.

Answer #1

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Suppose f is continuous for x is greater than or equal to 0,
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Let a < b, a, b, ∈ R, and let f : [a, b] → R be continuous
such that f is twice differentiable on (a, b), meaning f is
differentiable on (a, b), and f' is also differentiable on (a, b).
Suppose further that there exists c ∈ (a, b) such that f(a) >
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prove that there exists x ∈ (a, b) such that f'(x)=0.
then prove there exists z ∈ (a, b) such...

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