Suppose n > 1 is a positive integer. Prove for all x > 1.

(hint: suppose we know f and g are differentiable on the interval (-c,c) and f(0) = g(0), if f'(x) > g'(x) for all , then f(x) > g(x) for all .

Printable View

- Nov 20th 2009, 05:54 PM450081592Prove x^n > x for all x > 1.
Suppose n > 1 is a positive integer. Prove for all x > 1.

(hint: suppose we know f and g are differentiable on the interval (-c,c) and f(0) = g(0), if f'(x) > g'(x) for all , then f(x) > g(x) for all . - Nov 20th 2009, 06:38 PMlvleph
Do you know how to prove something by induction? If not I am unsure how to do this for general n.

- Nov 20th 2009, 06:46 PM450081592
- Nov 20th 2009, 06:49 PMlvleph
What you want to do is use the hint on a base case (n=2). Then you will make the assumption that the problem hypothesis holds for n=k. Use this assumption to prove that it is true for n=k+1.

- Nov 20th 2009, 06:52 PM450081592
- Nov 20th 2009, 06:54 PMlvleph
Because it is easy and the problem statement says n > 1.

- Nov 20th 2009, 06:59 PM450081592
- Nov 20th 2009, 07:03 PMDrexel28
- Nov 20th 2009, 07:05 PMlvleph
Okay, I will show you the base case.

Assume n=2.

Now let's show that these are equal at

Now we differentiate the two

For

Therefore we see that for . - Nov 20th 2009, 07:07 PMlvleph
- Nov 20th 2009, 08:30 PM450081592
- Nov 20th 2009, 09:26 PM450081592
- Nov 20th 2009, 09:33 PMlvleph
We know because you are trying to prove and the hint was suggesting you

Wouldn't you agree for and ? But we know that and . - Nov 20th 2009, 09:35 PM450081592
- Nov 20th 2009, 09:37 PMlvleph
Yes, because the hint was to prove and we wanted to prove so it makes sense to assume .