The function f(x)=x^x decreases on the interval : - Vidyadip

MCQ

The function $f(x)=x^x$ decreases on the interval :

A
$(0, e)$
B
$(0, e)$
✓
$\Big(0,\frac{1}{\text{e}}\Big)$
D
None of these

✓

Answer

Correct option: C.

$\Big(0,\frac{1}{\text{e}}\Big)$

Given, $\text{f}(\text{x})=\text{x}^{\text{x}}$
Applying $\log$ with base $e$ on both sides, we get
$\log(\text{f}(\text{x}))=\text{x}\log_\text{e}\text{x}$
$\frac{\text{f}'(\text{x})}{\text{f}(\text{x})}=1+\log_{\text{e}}\text{x}$
$\text{f}\ '(\text{x})=\text{f}(\text{x})(1+\log_\text{e}\text{x})$
$=\text{x}^\text{x}(1+\log_\text{e}\text{x})$
For $f(x)$ to be decreasing, we must have
$\text{f}\ '(\text{x}) < 0$
$\Rightarrow\text{x}^\text{x}(1+\log_\text{e}\text{x})<0$
Here, logarithmic function is defined for positive values of $x.$
$\Rightarrow\text{x}^\text{x} > 0$
$\Rightarrow1+\log_\text{e}\text{x} < 0$
$[$Since $\text{x}^\text{x}(1+\log_\text{e}\text{x}) < 0$
$\Rightarrow1+\log_\text{e}\text{x} < 0]$
$\Rightarrow\log_\text{e}\text{x} < -1$
$\Rightarrow\text{x}<\text{e}^{-1}$
$\big[\because\ \log_\text{a}\text{x} < \text{N}\Rightarrow\text{a}^\text{N}\text{ for }\text{a}>1\big]$
Here,
$\text{e} > 1$
$\Rightarrow\log_\text{e}\text{x} < -1$
$\Rightarrow\text{x} < \text{e}^{-1}$
$\Rightarrow\text{x}\in\big(0,\text{e}^{-1}\big)$
So, $f(x)$ is decreasing on $\Big(0,\frac{1}{\text{e}}\Big).$

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