Question

if f(x, y) = x2 – 3xy + 2y2 then using by mean value theorem to express the difference f(1,2) – f(2, – 1) by partial derivatives, compute θ and check that it is between 0 and 1.​

Answer 1

Answer:

) (1) Write with reason, which of the following are finite or infinite: A = {xlx is a multiple of 3) (ii) B=(yly is a factor of 13) (iii) C = {..., -3, -2, -1,0) (iv) D-xlx=2", n EN)​

Step-by-step explanation:

Answer 2

Answer:

Step-by-step explanation:

f(x,y)= x^2-3xy+2y^2                          …(1)

∂f/∂x=f_x=2x-3y                                    ….(2)

∂f/∂y=f_y=-3x+4y                                ….(3)

Cleary, f_x   and f_y  both are continuous function of x and y  

Therefore, Mean Value theorem is applicable  

By the Mean Value theorem,  

If (a,b) and (a+h,b+k) are points of domain of F , there exist θ∈(0,1) such that  

f(a+h ,b+k)- F(a,b)=hf_x  (a+θh ,b+θk)+kf_y (a+θh,b+θk)   ….(4)

a=2  ,b=-1  ,h=-1,k=3  

LHS of (4)  

f(1,2)-f(2,-1)=-9                             (from 1)

RHS of (4)  

hf_x  (a+θh ,b+θk)+kf_y (a+θh,b+θk)=(-1){2(2-θ)-3(-1+3θ) }+3 {-3(2-θ)+4(-1+3θ)}

=56θ-37                                                                      

Now ,  

-9=56θ-37  

θ=1/2   ∈(0,1)

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