Math Help - Prove f is increasing on [0, infinity)

1. Prove f is increasing on [0, infinity)

Suppose that n > 1 is a positive integer and define f : R --> R by
$f(x) = x^n$. Prove that f is increasing on [0, infinity).

2. What does the derivative $f'(x)$ for $x \in [0, \infty)$ say about $f(x)$ for $x \in [0, \infty)$?

3. Originally Posted by lvleph
What does the derivative $f'(x)$ for $x \in [0, \infty)$ say about $f(x)$ for $x \in [0, \infty)$?
so can I just say this:

f '(x) = nx^(n-1)
this is always positive for x>0 and n>1

therefore f is increasing on [0, infinity)

it there anything I need to prove here?

4. There is nothing else to prove. Just be careful because f'(x) is not always positive on $[0,\infty)$ since it is zero at x=0.

5. Originally Posted by lvleph
There is nothing else to prove. Just be careful because f'(x) is not always positive on $[0,\infty)$ since it is zero at x=0.

so what do I need to say about the 0 possibility, and how do I know f is deffierentiable? because it's polynomial?

6. I think because of the problem statement it is okay to say the function is increasing on the entire interval. You know it is differentiable, because the function and its derivative are continuous.

7. Originally Posted by lvleph
I think because of the problem statement it is okay to say the function is increasing on the entire interval. You know it is differentiable, because the function and its derivative are continuous.

do we know f and f' is continuous?

8. Well, it would be beyond the scope of your class to prove it. So, since it is a polynomial you know that it is continuous.

9. Originally Posted by lvleph
Well, it would be beyond the scope of your class to prove it. So, since it is a polynomial you know that it is continuous.

so we can directly say it is defferentiable then, cause polynomialis defferentiable everywhere

10. Yes.

11. Originally Posted by lvleph
Yes.
ok I got it, tanks so much!! it's hard to believe it's that easy, lol, only 2 lines explanation?