MIT18_024S11_soln-pset5

# MIT18_024S11_soln-pset5 - Solutions for PSet 5 1(8.14:10(a...

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Solutions for PSet 5 1. (8.14:10) (a) By hypothesis, we know ± f ( x )= 0 for all x B ( a ). By deﬁnition ± f ( x )=( D 1 f ( x ) ,...,D n f ( x )), thus D 1 f ( x ··· = D n f ( x ) = 0. By the 1-dimensional theorem, this means that f ( x ) is constant on every line x + t e 1 ,... x + t e n for any x B ( a ). Let f ( a c . Given y B ( a ), we will prove that f ( y c . Let y a = d 1 e 1 + + d n e n , then: c = f ( a f ( a + d 1 e 1 = f ( a + d 1 e 1 + + d n e n f ( y ) . (b) The condition means, that a is a local maximum of the function f .I n particular, because f ( x 0 for all x B ( a ), 0 is a maximum for all the functions f ( a + t e i ). By the 1-dimensional theorem D i ( f ( a )) = f ± ( a + t e i ) = 0, and thus ± f ( a D 1 f ( a ) n f ( a )) = 0 . 2. (8.17:6) f ( x, y | xy | = | x | | y | (a) f ( x, y ) = 0 on the lines ( x, 0) and (0 ,y ). Thus ∂f = = 0 at the origin. ∂x ∂y (b) For f to have a tangent plane at the origin, it must have a total derivative at the origin. If indeed f had a total derivative at the origin then we expect f ± (( x, y ); (1 , 1)) to be ( , ) · (1 , 1) = + =0+0=0 . But in reality, f ± (( x, y = x ); (1 , 1)) is the partial derivative along the line x = y . On this line, f ( x, x x ,thu s f ± ((

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MIT18_024S11_soln-pset5 - Solutions for PSet 5 1(8.14:10(a...

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