Re: directional derivative,-

What about the well-known rectangular circular half-cone,
the graph of (x, y) -> z = sqrt(x^2 + y^2)?

At first sight the problem seems underspecified? incompletely posed? badly posed? provocative, i.e. aimed at putting in motion the reader's creativity?

Or did the author mean to ask:
"Prove that no real-valued function f ==that has a continuous derivative at a point C==
has a positive directional derivative at C for every possible direction u."

BTW, the paraboloid (x, y) -> z = x^2 + y^2 does not qualify because the slope at (0, 0) is zero in every direction.
This example shows that zero local slope is not in contradiction to positive average slope.

Ciao: Johan E. Mebius


chrizm7@xxxxxxxxx wrote:
A problem asks me to show that no real-valued function f has a
positive directional derivative at a fixed point c for every possible
direction u.

But wouldn't a paraboloid work? Fix c corresponding to the vertex.
Then for any direction u, the slope will always be positive
(increasing).

To be clear, the exact problem is worded: Prove that there is no real-
valued function f such that f'(c;u) > 0 for a fixed point c in R^n and
every nonzero vector u in R^n.

.