Answer to Question #192512 in Calculus for Sarita bartwal

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 take some other value then the function is differentiable . The function |$x|$ is continuos at $x=0$ and not differentiable at $x$ = 0.

but the given function you see is 3-d function and if you 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.The function f(x,y,z) =$|x+2y+z|$ is continuos at the given point but not differentiable at the given point so by 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 , similarly for this same set if we put some other set of value then the given function is differentiable at all the other points .