Proof that f(x) = ln(x) - x has a maximum, without using derivatives and taylor series expansion

Jun 2019
2
0
Israel
I need to proove that for any x in the range (x>0), f(x) <= f(1) = -1.
I tried using the limit definition of e, and the fact that e^x > x for any x>0, to no avail.
Can someone help me?
 

Prove It

MHF Helper
Aug 2008
12,897
5,001
You wouldn't know that $\displaystyle \begin{align*} \mathrm{e}^x > x \end{align*}$ without knowledge of Taylor Series, so I doubt that result could be used.

I really dislike when people put unnecessary restrictions on themselves. Always remember Occam's Razor - the most concise solution is the best one.
 
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Jun 2013
1,134
605
Lebanon
maybe we can use this

\(\displaystyle \ln x -x +1=\int_1^x \left(\frac{1}{t}-1\right) \, dt\)
 
Dec 2013
2,002
757
Colombia
You wouldn't know that $\displaystyle \begin{align*} \mathrm{e}^x > x \end{align*}$ without knowledge of Taylor Series
Why not? I can prove $\displaystyle \lim_{x \to \infty} \frac{\ln x}{x} = 0$ without recourse to Taylor series. On the other hand, I'd use the integral definition of the logarithm which rather suggests knowledge of derivatives.
 

Prove It

MHF Helper
Aug 2008
12,897
5,001
Why not? I can prove $\displaystyle \lim_{x \to \infty} \frac{\ln x}{x} = 0$ without recourse to Taylor series. On the other hand, I'd use the integral definition of the logarithm which rather suggests knowledge of derivatives.
Interesting, how would you do it?
 
Dec 2013
2,002
757
Colombia
\begin{align}
0 \lt \frac1t & \lt t^{c-1} &(t \gt 1, \, c \gt 0) \\
0 \lt \int_1^x \frac1t \,\mathrm dt & \lt \int_1^x t^{c-1} \,\mathrm dt \\
0 \lt \log t &\lt \frac1c ( x^c -1) \lt \frac{x^c}c \\
0 \lt \frac{\log^b x}{x^a} \lt \frac {x^{-a}}{c^b} &(a \gt 0, \, b \gt 0) \\
\text{(choose $c= \tfrac{a}{2b}$ and} \; 0 \lt \frac{\log^b x}{x^a} \lt \frac {x^{-\frac{a}2}}{c^b}
\end{align}
and take limits as $x \to \infty$.