I want to maximize this function $\log_2(1+p)+e^{-p} $.
The second derivative is $-\frac{1}{\ln2 (1+p)^2}+e^{-p}$
Is there any way to get a unique solution for p ?
Asked
Active
Viewed 131 times
0
-
Find critical point(s) $p_i$ and check with the $2$nd derivative. – farruhota Aug 03 '17 at 05:03
-
maximizing on where? on "R" function is unbouded – Red shoes Aug 03 '17 at 05:04
-
@farruhota The second derivative>0 which means it is not concave – user7341333 Aug 03 '17 at 05:15
-
@Red shoes $p=[0, p_{max}]$, p is bounded by a maximum value $p_{max}$ – user7341333 Aug 03 '17 at 05:17
1 Answers
0
There is no extreme (max/min) point.
Note that the function has the domain: $$1+p>0\Rightarrow p>-1.$$ Also note that the function is increasing in its domain: $$f'(p)=\frac{1}{(p+1)\ln{2}}-\frac{1}{e^p}>0, \ \ \ for \ \ \ p>-1.$$
Also note: $$\lim_\limits{p\to-1^{+}}\left[\frac{1}{\ln 2}\cdot\ln(1+p)+\frac{1}{e^p}\right]=-\infty$$ $$\lim_\limits{p\to+\infty}\left[\frac{1}{\ln 2}\cdot\ln(1+p)+\frac{1}{e^p}\right]=+\infty$$
farruhota
- 31,482