21-356 Lecture 6

Erentiable for all t 0 1 with g t f tx 1

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Unformatted text preview: := f (tx + (1 % t) y), t $ [0, 1], is di!erentiable for all t $ (0, 1), with g # (t) = "f (tx + (1 % t) y) &x % y& . "v Proof. Fix t0 $ (0, 1) and consider g (t) % g (t0 ) f (tx + (1 % t) y) % f (t0 x + (1 % t0 ) y) = t % t0 t % t0 $ % f t0 x + (1 % t0 ) y+ (t % t0 ) &x % y& "x!y" % f (t0 x + (1 % t0 ) y) x!y = &x % y& (t % t0 ) &x % y& f (t0 x + (1 % t0 ) y+sv) % f (t0 x + (1 % t0 ) y) = &x % y& , s 12 where s := (t % t0 ) &x % y&. Since s # 0 as t # t0 , it follows that there exists g # (t0 ) = lim t$t0 = g (t) % g (t0 ) f (t0 x + (1 % t0 ) y+sv) % f (t0 x + (1 % t0 ) y) = lim &x % y& s$0 t % t0 s "f (t0 x + (1 % t0 ) y) &x % y& . "v Proof of the Mean Value Theorem. Consider the function g (t) := f (tx + (1 % t) y), t $ [0, 1]. Since compositions of continuous functions is continuous, we have that g is continuous. By the previous lemma, we are in a position to apply the mean value theorem to the function g to find ! $ [0, 1] such that g (1) % g (0) = dg (!) (1 % 0) , dt that is, f (x) % f (y) = "f (!x + (1 % !) y) &x % y& . "v Remark 22 If f is defined on a larger domain E and all points of S except at most x and y are interior points of E , then by (1), we have that "f (!x + (1 % !) y) = "v N & i=i "f (xi % yi ) (!x + (1 % !) y) , " xi &x % y & if f is differentiable at all points of S and so we can rewrite (5) in the form f (x) % f (y) = N & "f (!x...
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This document was uploaded on 03/31/2014.

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