Answer to Question #199231 in Calculus for Nikhil Rawat

Question

Let f: R^3→R be function defined by :

F(x,y,z)= |x+2y+z| . Show that f is not differentiable at the point (1,-1,1).

Answer 1

If we take example of $|x|$ 

Then, it is clearly not differentiable at $x=0$ but if we choose some other value then the given function is differentiable .

So, the function |$x|$ is continuous at $x=0$ but not differentiable at $x = 0$ .

but the given function in the problem is 3D Function and if we see the given points which is

 $R^{3}\to {R}$ $f(x,y,z) =$ |$x+2y+z|$ 

and if we put this point where we have to prove that given function is differentiable or not then the given $f(x,y,z)=0$ which shows that at this given point clearly the function is not differentiable.

So, the function $f(x,y,z) =$ $|x+2y+z|$ is continuous at the point (1,-1,1) but not differentiable at that given point so from the given example of |x| we prove for this given function

$f(x,y,z)$ $=|x+2y+z|$ and the function is differentiable at all the other point rather than (1,-1,1) which is given in the question.

And if we choose any other point rather than (1,-1,1), the the given function is continuous and differentiable both at that point.