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

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