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) = &quot;f (tx + (1 % t) y) &amp;x % y&amp; . &quot;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 ) &amp;x % y&amp; &quot;x!y&quot; % f (t0 x + (1 % t0 ) y) x!y = &amp;x % y&amp; (t % t0 ) &amp;x % y&amp; f (t0 x + (1 % t0 ) y+sv) % f (t0 x + (1 % t0 ) y) = &amp;x % y&amp; , s 12 where s := (t % t0 ) &amp;x % y&amp;. 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 &amp;x % y&amp; s\$0 t % t0 s &quot;f (t0 x + (1 % t0 ) y) &amp;x % y&amp; . &quot;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) = &quot;f (!x + (1 % !) y) &amp;x % y&amp; . &quot;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 &quot;f (!x + (1 % !) y) = &quot;v N &amp; i=i &quot;f (xi % yi ) (!x + (1 % !) y) , &quot; xi &amp;x % y &amp; if f is differentiable at all points of S and so we can rewrite (5) in the form f (x) % f (y) = N &amp; &quot;f (!x...
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## This document was uploaded on 03/31/2014.

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