Answer to Question #106745 in Calculus for Nikesh gautam pandit ji

Question

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 .

Answer 1

a) True

The given function

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

Consider,the following maps.

$f_1:\R^3\rightarrow \R$ defined by $f_1(x,y,z)=|x|$ ,

$f_2:\R^3\rightarrow \R$ defined by $f_2(x,y,z)=|y|$ ,

$f_3:\R^3\rightarrow \R$ defined by $f_3(x,y,z)=|z|$ .

Clearly,$f_1,f_2,f_3$ are differentiable at $(3,2,-1)$ ,since the absolute value functions are differentiable everywhere except $(0,0,0)$.

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

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

$f(x,y,z)=f_1(x,y,z)+f_2(x,y,z)+f_3(x,y,z)=|x|+|y|+|z|$

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

b) False.

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

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

Now, the given real valued function of two variable is

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

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

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

for some k and all real number r.

Hence,$f$ is not a homogeneous function.

c) False.

Given function are

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

$g:\R^2:\rightarrow\R$ defined by $g(x,y)=x^2+y^2$ .

Now ,

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

but at $(0,0),$

$$\frac{f}{g}(0,0)=\frac{0}{0}$$

which is not defined.

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