Answer to Question #160261 in Calculus for Priya

Question

The function f : R³ ->R ,given by f ( x ,y,z)=|x|+| Y |+|Z |is differentiable at (2,3,-1).is it true or false?give reasons for your answer

Answer 1

There exist neighbourhood U of (2, 3, -1) such that

$\forall(x,y,z)\isin U, f(x,y,z)=x+y-z$ by definition of the absolute value.So, if

$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=1$

$(x,y,z)\isin U: \frac{\partial f}{\partial z}=-1$

Since function $g(x,y,z)=const$ is continuous, all partial derivatives of f(x, y, z) are continuous at (2, 3, -1). Thus, f(x, y, z) is differentiable at (2, 3, -1).