Suppose that x(t) is a continuously differentiable function satisfying x\'(t) le

Question

Suppose that x(t) is a continuously differentiable function satisfying x'(t) less than or equal to x(t), x(t) greater than or equal to 0 for t in [0,1] and x(0) = 0. Show that x(t) = 0 for t in [0,1]. I'm assuming this a proof by contradiction problem where you assume x(0) > 0 for some x in (0,1], but I am a little rusty on proofs and can't figure out how to precisely show that if this is the case,x'(t) will also be > 0 at some point and if this is the case x'(t) will be greater than x(t) at some point in (0,1]. (i.e. if 0.5,0.5 is a point, x'(t)= 1 but x(t)= .5 Any assistance would be appreciated, thanks in advance!

Explanation / Answer

Since x'(t) 0 x'(a) = Lt (h->0) (x(a+h) - x(a))/h = Lt (h->0) (x(a+h) - 0)/h = Lt (h->0) (x(a+h))/h > 0 [as x(a+h) > 0 and h > 0] But x(a) = 0, therefore, x'(a) > x(a) But this is a contradiction. Therefore, x(t) = 0 for all t in (0, 1]

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