Question

The function $f\left( x \right)={{x}^{\dfrac{1}{x}}}$ has stationary point at
A. $x=e$
B. $x=1$
C. $x=\sqrt{e}$
D. $x=\dfrac{1}{2}$

Answer 1

Hint: At first, we take log on both sides of the equation. After that, we differentiate both sides with respect to x. Then, we multiply $f\left( x \right)$ on both sides to get an equation for $f'\left( x \right)$ . We equate $f'\left( x \right)$ to zero. Then solving, we get the stationary point.

Complete step by step solution:
The term stationary point means the point on the x-y plane where the value of the first derivative of a function becomes zero. In other words, if $f\left( x \right)$ is the function then, at the stationary point of $f\left( x \right)$ , $f'\left( x \right)$ must be zero, as that is the definition of a stationary point.
The given function that we have at our disposal is $f\left( x \right)={{x}^{\dfrac{1}{x}}}$ . We can see that unfortunately we cannot directly differentiate the function with respect to x. This is because both the base and the power contain the variable x. so, we need to think of an alternative. We can take natural logarithm on both sides of the function. The function then becomes,
$\Rightarrow \ln \left( f\left( x \right) \right)=\dfrac{1}{x}\ln x$
Now, we differentiate both sides of the equation with respect to x.
$\Rightarrow \dfrac{1}{f\left( x \right)}\times f'\left( x \right)=\dfrac{1}{x}\left( \dfrac{1}{x} \right)-\dfrac{1}{{{x}^{2}}}\ln x$
Now, we simplify the above equation by opening up the brackets and get,
$\Rightarrow \dfrac{f'\left( x \right)}{f\left( x \right)}=\dfrac{1}{{{x}^{2}}}-\dfrac{1}{{{x}^{2}}}\ln x$
Taking $\dfrac{1}{{{x}^{2}}}$ common in the right-hand side of the equation, we get,
 $\Rightarrow \dfrac{f'\left( x \right)}{f\left( x \right)}=\dfrac{1}{{{x}^{2}}}\left( 1-\ln x \right)$
Multiplying both sides of the equation with $f\left( x \right)$ , we get,
$\Rightarrow f'\left( x \right)=\dfrac{{{x}^{\dfrac{1}{x}}}}{{{x}^{2}}}\left( 1-\ln x \right)$
Replacing the denominator with indices, we get,
$\Rightarrow f'\left( x \right)={{x}^{\dfrac{1}{x}-2}}\left( 1-\ln x \right)$
Now, for the stationary point, $f'\left( x \right)$ must be zero. This means that the right-hand side of the equation must be zero.
${{x}^{\dfrac{1}{x}-2}}\left( 1-\ln x \right)=0$
We know that exponential functions can never be zero. Thus, we are left with only one option and that is,
$\begin{align}
  & \left( 1-\ln x \right)=0 \\
 & \Rightarrow \ln x=1 \\
 & \Rightarrow x=e \\
\end{align}$

So, the correct answer is “Option A”.

Note: Students may get confused after seeing such functions which are directly differentiated, becomes a tedious job. We should always remember to take natural logarithms on both sides in such cases. We should also remember to multiply $f\left( x \right)$ on both sides towards the end so that we get an equation for $f'\left( x \right)$ .