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# 2 e questions on logical thinking one says that a

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Unformatted text preview: r, = 1+ 1 and (1 1) = 2. Thus the answer is = 1 − ( − 1) + 2( − 1), that is, + − 2 = 0. 2 E ( Questions on logical thinking One says that a function ( ) is independent of the variable if ( ) for any . In a similar manner, one deﬁnes a function independent of . 2 Let ( ) be a function whose domain is all of R2 . Prove that (a) If ≡ 0, then is independent of the variable ; (b) If ≡ 0, then is independent of the variable ; 2 1 )= (c) If ≡ ≡ 0, then is constant. Notice that the converse to each of these statements is obvious: if a function doesn’t depend on , then of course its derivative with respect to is 0. (a) Given that ≡ 0, consider the function ( ) for a ﬁxed value of . For any 1 < 2 , by the Mean Value Theorem there is a point 0 ∈ ( 1 2 ) such that S ( )− ( 1 ( )= 2 ) = = which shows that ( )= ( 1 (b) In the same manner, if ( )=0 0 0 ). 2 ≡ 0, we show that ( 1) =( 2) for any 1 2 ∈ R. (c) Assume that ≡ ≡ 0 and let’s prove that the function is constant. In other words, we need to prove that its values at any two points are equal. Applying (a) and (b), we have () () ( 1 1) = ( 2 1) = ( 2 2) for any ( E at ( 1) 1 Let ( ∈ R2 and ( )=| 2) 2 ∈ R2 . |. Determine all points ( ). S Since | | is not diﬀerentiable at cases separately here: (a) = 0. Then since ( =2 so and tiable. (b) ) ∈ R2 such that is diﬀerentiable 2 2 1/2 )= 2 = 0, it’s natural to consider the following · 1 2 , we have 2 2 −1/2 2 = | | 2 = | | , being elementary functions, are continuous and hence is diﬀeren- = 0 and = 0 = 0. Then let’s ﬁx = 0 and consider the function ( 0) = | 0 | · | |. It is not diﬀerentiable at = 0 and hence (0 0 ) is not...
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## This document was uploaded on 02/10/2014.

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