Calculus questions (again =( )?

Question

I totally got stuck at this calculus assignment again. :(
Need some help from you guys again. :)
So I have to find and explicitly express a function that is continuous and increasing in its all real number domain, it has to be bounded, but not differentiable. I thought about y = |x|, but it's not bounded, and half of its part is decreasing. Any idea guys?
another question is if f(x) = (x^3)cos(1/x) where x does not equal to 0, and f(x) is also equal to 0 (a point itself).
is this function f(x) differentiable, continuous, monotonic, bounded, and integrable on the interval [-1,1]?
f(x) = | (x^3)cos(1/x) x does not equal to 0
| 0
hope i made this question look clear enough. :(
Thanks in advance

Answer 1

f(x) = x^(1/3) is a good function that's not differentiable at x=0 b/c the tangent line is vertical, and it's always increasing b/c its derivative, 1/3x^(-2/3), is never negative b/c the number is squared.

f(x) = x^3cos(1/x)
NOT: monotonic, differentiable, integrable, continuous.
It is bounded b/c for some value M, the y-values of the function are within that M. That value M is:
f(-1) = -cos(-1), and f(1) = cos(1)
The reason is that if you had x^3 only, then it would be all of those things except monotonic. However, including the cos(1/x) makes it so that there is no finite value that the function approaches as x--> 0. That is, there are infinite oscillations as you get infinitely closer to zero for the function. Thus, the two ends of the function do not meet at zero and the function is not continuous.
Another way to think about it is that you would have 1/0, which is a non-fixable point of discontinuity. (vertical asymptote)

Answer 2

f´(x) = 3x^2cos 1/x +x^3*sin(1/x)(1/x^2)=cos(1/x)*x*(3x+1)
so if you define f´(0)=0 it is differentiable and so continuos.
But not monotonic as the sign in[-1,1] of the derivative is

-1++++++(-1/3)------(0)+++++ and (-1/3 is a local max and 0 a local min.
It is bounded and as continuos it is integrable(Cauchy)
I would like to add
As cos 1/x is bounded,as x=>0 lim x^3 cos 1/x =0 so with f(0)=0
the function is continuos for all x
the same holds for the derivative as lim f´(x) x=>0 =0
The sign around zero is wrong as cos 1/x changes it sign
infinite times
If you take x=1/npi cos 1/x= cos npi which is 1 or -1 so zero Is NOT a local minimum