Question

Let $f\left(x\right)=e^{(p+1)x}-e^x$ for real number p>0, then the value of $x=s_p$ for which $f\left(x\right)$ is minimum is,

• A. $-\frac{\ln (p+1)}{p}$
• B. $-\ln (p+1)$
• C. $-\ln p$
• D. $\ln \left(\frac{p+1}{p}\right)$

Answer 1

The correct option is A $-\frac{\ln (p+1)}{p}$

$f\left(x\right)=e^{(p+1)x}-e^x{f}^{\prime }\left(x\right)=e^{(p+1)x}(p+1)-e^x=p{e}^{(p+1)x}+e^{(p+1)x}-e^x=p{e}^{px}{e}^x+e^{px}{e}^x-e^x=e^x(p{e}^{px}+e^{px}-1)=0p{e}^{px}+e^{px}-1=0\Rightarrow p{e}^{px}+e^{px}=1\Rightarrow e^{px}(p+1)=1\Rightarrow e^{px}=\frac{1}{(p+1)}\Rightarrow ln{e}^{px}=ln\left(\frac{1}{(p+1)}\right)\Rightarrow px=ln\left(\frac{1}{(p+1)}\right)\Rightarrow x=-\frac{ln(p+1)}{p}{f}^{\prime \prime }\left(x\right)=(p+1)(p+1)e^{(p+1)x}-e^x>0$

$\therefore x=-\frac{ln(p+1)}{p}$ will give $f\left(x\right)$ as minimum.