Let U= {x such that IIxII cannot equal 1} Define f on U by setting f(x)= { 0, II

Question

Let U= {x such that IIxII cannot equal 1} Define f on U by setting

f(x)= { 0, IIxII < 1and 1, IIxII >1}

a) Note that the gradient f(x)=0 for all x in U, but f is not constant on U. Explain how this does not contradict the theorem that states : Let U be an open connected set and let f be a differentiable function on U. If gradient f(x)=0 for all x in U, then f is constant on U.

b) Define the function g on U different from f such that the gradient f(x)= gradient g(x) for all x in U and f-g is (i) constant on U. (ii) not constant on U.

Explanation / Answer

a)

theorem that states : Let U be an open connected set and let f be a differentiable function on U. If gradient f(x)=0 for all x in U, then f is constant on U

f(x)= { 0, IIxII < 1and 1, IIxII >1}   Given

Here Connected set will be

    U =U1 union U2   where U1= (-1,1) and U2=(-inf,-1) union (1,inf)

     so,

for U1 : f'(x) = 0 and f(x) =0 = constant

Hence for U1 theorem is vaild

for U2 :

f'(x) =0 and f(x) = 1 = constant  

hence for U2 also theorem is valid

Hence U don't contradict above theorem on piecewise domain it doesn't voailate the theorem

b)

Take g(x) ={-1, IIxII < 1and 0, IIxII >1}  

g'(x) = 0 for all U

f-g = 1 = const for all U

Take g(x) ={ 2, IIxII < 1and 1, IIxII >1}  

g'(x) = 0 for all U

f-g = { -2, IIxII < 1and 0, IIxII >1 } Not constant

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