naren
2004-09-29 07:45:58 |
hellppppppp now!!!
prove that e^pie is greater than pie^e |
Federico Kereki
2004-09-29 08:56:50 |
Re: hellppppppp now!!!
Consider f(x)=x^(K-x), where K is a constant. (In our case, we'll take K=e+pi). We'll show that this function decreases as x increases from e to pi, so e^pi es greater.
We write f(x)=e^((k-x)ln(x)) so f'(x)=[e^((k-x)ln(x))][(k-x)ln(x)]' =x^(K-x) times (K/X-1-ln(x)). The first factor is always positive, so we can ignore it. The second factor is obviously decreasing as x grows, and for x=e and K=e+pi, it is negative -- so f(x) decreases from e to pi, and f(x) is higher at e.
HTH! |
