Question

Let $f( x )=\left(1+\frac{1}{ x }\right)^{ x }$ and $g ( x )=\left(1+\frac{1}{ x }\right)^{ x +1}$, both $f$ and $g$ being defined for $x >0$, then prove that $f(x)$ is increasing and $g(x)$ is decreasing.

• A. Both $f ( x )$ and $g ( x )$ are increasing
• B. $f(x)$ is increasing and $g(x)$ is decreasing
• C. $f(x)$ is decreasing and $g(x)$ is increasing
• D. both $f ( x )$ and $g ( x )$ are decreasing.

Answer 1

Answer: (B)

$f(x)$ is increasing
(discussed in class)
$g(x)=\left(1+\frac{1}{x}\right)^{x+1}>0$
$= e ^{( x +1)\left[\ln \frac{ x +1}{ x }\right]} $
$g^{\prime}(x)=\left(1+\frac{1}{x}\right)^{x+1}\left[(x+1)\left(\frac{1}{x+1}-\frac{1}{x}\right)+\ln \left(\frac{x+1}{x}\right)\right] $
$=\left(1+\frac{1}{ x }\right)^{ x +1}\left[\ln \left(\frac{ x +1}{ x }\right)+1-\frac{ x +1}{ x }\right] $
$\text { let } \frac{x+1}{x}=u ; \text { for } x>0, u \in(1, \infty) $
$\text { consider } h ( u )=\ln u +1- u$
$h^{\prime}(u)=\frac{1}{u}-1<0 \text { in }(1, \infty) $

$\text { hence } h(u) \text { is a decreasing function in }(1, \infty) $
$\text { h (u) }<\text { h (1) } $
$h ( u )<0 \text { [but } h (1)=0 \text { ] }$
$\therefore g ^{\prime}( x )<0 \Rightarrow \text { gis decreasing } $
$\therefore \text { fis increasing and g is decreasing } \Rightarrow (B)$