Is the Function y=x^3 Monotonic at x=0?

[player1_1_1]

Homework Statement

y=x^3 is it monotonic in 0?

The Attempt at a Solution

if i try to solve y'>0 it will be 0 in point x=0 so that function is not strictly decreasing in 0 but in other way we have x_1<x_2\Rightarrow f(x_1)<f(x_2) so it is scrictly decreasing, its paradox, why?

 

[CompuChip]

Why is y' = 0 not allowed for a stricly increasing function?
Isn't x_1 < x_2 \implies f(x_1) \le f(x_2) the definition of monotonic (or strictly monotonic, if you replace the equality by a strict equality)?

 

[gb7nash]

Homework Statement

y=x^3 is it monotonic in 0?

The Attempt at a Solution

if i try to solve y'>0 it will be 0 in point x=0 so that function is not strictly decreasing in 0 but in other way we have x_1<x_2\Rightarrow f(x_1)<f(x_2) so it is scrictly decreasing, its paradox, why?

Plug in a value before 0 and after 0 in y'=3x^2. You'll see that y' is strictly positive for all real values except 0, where the slope is zero. So it follows that this is neither a minimum nor maximum, and 0 is the only point that has slope zero. We also know that y is a continuous function, so y must be strictly increasing, i.e. monotone.

 

[player1_1_1]

so generally if x_0 is isolated point where derivative is 0 and its positive around this function can be strictly monotonic there yeah? and what about this theorem what i found? if function is scrictly monotonic in every point then function then inverse function has derivative in any point, but function y=\sqrt[3]{x} doesn't have derivative in 0, why?

 

[Char. Limit]

Because the derivative of the function is not defined at x=0. The best you can hope for is a limit.

 

[AlephZero]

and what about this theorem what i found? if function is scrictly monotonic in every point then function then inverse function has derivative in any point

Either you have misunderstood something about the statement of the theorem, or it is just wrong.

Counter example:
f(x) = -1 + x when x < 0
f(x) = 0 when x = 0
f(x) = 1 + x when x > 0
f(x) is strictly increasing, but you can't even define a limit of f^-1(0), let alone a derivative, because f^-1(x) is not defined when -1 < x < 0 and 0 < x < 1

 

[HallsofIvy]

a^3- b^3= (a- b)(a^2+ ab+ b^2)= (a- b)((a^2+ ab+ b^2/4)+ (3/4)b^2)= (a-b)((a+b/2)^2+ 3/4 b^2)

Since a^2+ ab+ b^2 is the sum of two squares it is never negative and a^3- b^3 is positive if and only if a- b is.