Question #106745
State whether the following statements are true or false. Give reasons for your answers.
a) The function R^3 → R. f : , given by f (x, y,z) = |x| + |y| + |z| is differentiable at
3,2(-1)

b) The function
f(x,y)=max{y/x,x} is a homogeneous function on R^2

c) The domain of the function f / g where f (x, y) = 2xy and g(x, y) = x^2 + y^2 is R^2 .
1
Expert's answer
2020-03-30T04:38:27-0400

a) True

The given function

f:R3Rf:\R^3\rightarrow \R is defined by f(x,y,z)=x+y+z.f(x,y,z)=|x|+|y|+|z|.

Consider,the following maps.

f1:R3Rf_1:\R^3\rightarrow \R defined by f1(x,y,z)=xf_1(x,y,z)=|x| ,

f2:R3Rf_2:\R^3\rightarrow \R defined by f2(x,y,z)=yf_2(x,y,z)=|y| ,

f3:R3Rf_3:\R^3\rightarrow \R defined by f3(x,y,z)=zf_3(x,y,z)=|z| .

Clearly,f1,f2,f3f_1,f_2,f_3 are differentiable at (3,2,1)(3,2,-1) ,since the absolute value functions are differentiable everywhere except (0,0,0)(0,0,0).

Again, we known that sum of differentiable functions is again differentiable.

Therefore,f:R3Rf:\R^3\rightarrow \R defined by

f(x,y,z)=f1(x,y,z)+f2(x,y,z)+f3(x,y,z)=x+y+zf(x,y,z)=f_1(x,y,z)+f_2(x,y,z)+f_3(x,y,z)=|x|+|y|+|z|

is differentiable at (3,2,1)(3,2,-1) .


b) False.

A homogeneous real valued function of two variable x and y is a real valued function that satisfies the condition

f(rx,ry)=rkf(x,y)f(rx,ry)=r^kf(x,y) for some constant kk and all real number r.r.

Now, the given real valued function of two variable is

f(x,y)=max{yx,x}f(x,y)=max\{ \frac{y}{x},x \} .

Putting,x=rx and y=ryx=rx \ and \ y=ry ,we get

f(rx,ry)=max(ryrx,rx)=max(yx,rx)rkf(x,y).f(rx,ry)=max(\frac{ry}{rx},rx)=max(\frac{y}{x},rx)\neq r^kf(x,y).

for some k and all real number r.

Hence,ff is not a homogeneous function.



c) False.

Given function are

f:R2: Rf:\R^2:\rightarrow \ \R defined by f(x,y)=2xy.f(x,y)=2xy.

g:R2:Rg:\R^2:\rightarrow\R defined by g(x,y)=x2+y2g(x,y)=x^2+y^2 .

Now ,

fg(x,y)=2xyx2+y2\frac{f}{g}(x,y)=\frac{2xy}{x^2+y^2}

but at (0,0),(0,0),

fg(0,0)=00\frac{f}{g}(0,0)=\frac{0}{0}


which is not defined.

Hence , R2\R^2 is not a domain of fg\frac{f}{g} .


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