Question

The function f : R^3 - R. given by f(x, y, z)=(x + y + Z is differemtiable at
(2,3,-1).​

Answer 1

Answer:

The given function

f:\R^3\rightarrow \Rf:R

3

→R is defined by f(x,y,z)=|x|+|y|+|z|. f(x,y,z)=∣x∣+∣y∣+∣z∣.

Consider,the following maps.

f_1:\R^3\rightarrow \R f

1

:R

3

→R defined by f_1(x,y,z)=|x|f

1

(x,y,z)=∣x∣ ,

f_2:\R^3\rightarrow \Rf

2

:R

3

→R defined by f_2(x,y,z)=|y|f

2

(x,y,z)=∣y∣ ,

f_3:\R^3\rightarrow \Rf

3

:R

3

→R defined by f_3(x,y,z)=|z|f

3

(x,y,z)=∣z∣ .

Clearly,f_1,f_2,f_3f

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:\R^3\rightarrow \Rf:R

3

→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|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)(3,2,−1) .