Question

Find the maximum value of ${\left( {\dfrac{1}{x}} \right)^x}$.
(A) \[{\left( e \right)^e}\]
(B) \[{\left( e \right)^{\dfrac{1}{e}}}\]
(C) \[{\left( e \right)^{ - e}}\]
(D) \[{\left( {\dfrac{1}{e}} \right)^e}\]

Answer 1

Hint: Given problem tests the concepts of derivatives and their applications. These type questions can be easily solved if we keep in mind the concepts of maxima and minima. In the problem, we are required to find the maximum value of the function in the variable x, ${\left( {\dfrac{1}{x}} \right)^x}$. We do so by differentiating the function with respect to x and finding the critical points at which the function attains maximum or minimum values. Then, we use the second derivative test to check whether the critical point represents the minimum or maximum value of the function.

Complete step by step answer:
We have the function in x as ${\left( {\dfrac{1}{x}} \right)^x}$.
Let us assume the function ${\left( {\dfrac{1}{x}} \right)^x}$ to be equal to y.
So, $y = {\left( {\dfrac{1}{x}} \right)^x}$
Taking natural logarithm on both sides of the equation, we get,
\[\ln \left( y \right) = \ln {\left( {\dfrac{1}{x}} \right)^x}\]
Now, we know the property of logarithms as $\ln {a^x} = x\ln a$. So, we get,
\[ \Rightarrow \ln \left( y \right) = x\ln \left( {\dfrac{1}{x}} \right)\]
Now, we differentiate both sides of the above equation. So, we get,
\[ \Rightarrow \dfrac{d}{{dx}}\left[ {\ln \left( y \right)} \right] = \dfrac{d}{{dx}}\left[ {x\ln \left( {\dfrac{1}{x}} \right)} \right]\]
Now, we know that the derivative of $\ln x$ is $\left( {\dfrac{1}{x}} \right)$. So, we get,
\[ \Rightarrow \left( {\dfrac{1}{y}} \right)\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left[ {x\ln \left( {\dfrac{1}{x}} \right)} \right]\]
We know the product rule of differentiation $\dfrac{d}{{dx}}\left[ {f\left( x \right) \times g\left( x \right)} \right] = f\left( x \right) \times g'\left( x \right) + f'\left( x \right) \times g\left( x \right)$. So, we get,
\[ \Rightarrow \left( {\dfrac{1}{y}} \right)\dfrac{{dy}}{{dx}} = x\dfrac{d}{{dx}}\left[ {\ln \left( {\dfrac{1}{x}} \right)} \right] + \ln \left( {\dfrac{1}{x}} \right)\dfrac{{d\left( x \right)}}{{dx}}\]
Now, we know that derivative of x with respect to x is $1$. So, we get,
\[ \Rightarrow \left( {\dfrac{1}{y}} \right)\dfrac{{dy}}{{dx}} = x \times \dfrac{1}{{\left( {\dfrac{1}{x}} \right)}} \times \dfrac{d}{{dx}}\left[ {\left( {\dfrac{1}{x}} \right)} \right] + \ln \left( {\dfrac{1}{x}} \right)\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = y\left[ {{x^2} \times \dfrac{{ - 1}}{{{x^2}}} + \ln \left( {\dfrac{1}{x}} \right)} \right]\]
We know the power rule of differentiation as $\dfrac{{d\left[ {{x^n}} \right]}}{{dx}} = n{x^{\left( {n - 1} \right)}}$. So, we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = y\left[ {\ln \left( {\dfrac{1}{x}} \right) - 1} \right] - - - - \left( 1 \right)\]
Now, we substitute the value of y and equate the first derivative of the function to zero to find the critical points of the function. So, we get,
\[ \Rightarrow {\left( {\dfrac{1}{x}} \right)^x}\left[ {\ln \left( {\dfrac{1}{x}} \right) - 1} \right] = 0\]
Shifting the terms to right side of the equation, we get,
\[ \Rightarrow \ln \left( {\dfrac{1}{x}} \right) = 1\]
Taking exponential functions on both sides of the equation. We also know that ${e^{\ln x}} = x$. So, we get the equation as,
\[ \Rightarrow \left( {\dfrac{1}{x}} \right) = e\]
\[ \Rightarrow x = \dfrac{1}{e}\]
So, we get \[x = \dfrac{1}{e}\] as a critical point of the function.
Now, we find the second derivative of the function. From equation $\left( 1 \right)$, we have,
\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{dy}}{{dx}}\left[ {y\left( {\ln \left( {\dfrac{1}{x}} \right) - 1} \right)} \right]\]
Again, using the product rule of differentiation,
\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \left( {\ln \left( {\dfrac{1}{x}} \right) - 1} \right)\dfrac{{dy}}{{dx}} + y\dfrac{d}{{dx}}\left( {\ln \left( {\dfrac{1}{x}} \right) - 1} \right)\]
We know the value of $\dfrac{{dy}}{{dx}}$. So, we get,
\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \left( {\ln \left( {\dfrac{1}{x}} \right) - 1} \right) \times y\left[ {\ln \left( {\dfrac{1}{x}} \right) - 1} \right] + y\dfrac{d}{{dx}}\left( {\ln \left( {\dfrac{1}{x}} \right) - 1} \right)\]
Simplifying the expression, we get,
\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = y\left[ {{{\left( {\ln \left( {\dfrac{1}{x}} \right) - 1} \right)}^2} + \dfrac{1}{{\left( {\dfrac{1}{x}} \right)}}\left( {\dfrac{{ - 1}}{{{x^2}}}} \right)} \right]\]
\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = y\left[ {{{\left( {\ln \left( {\dfrac{1}{x}} \right) - 1} \right)}^2} - \dfrac{1}{x}} \right]\]
Now, substituting the value of x as \[\dfrac{1}{e}\].
\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = {\left( {\dfrac{1}{{\left( {\dfrac{1}{e}} \right)}}} \right)^{\left( {\dfrac{1}{e}} \right)}}\left[ {{{\left( {\ln \left( {\dfrac{1}{{\left( {\dfrac{1}{e}} \right)}}} \right) - 1} \right)}^2} - \dfrac{1}{{\left( {\dfrac{1}{e}} \right)}}} \right]\]
Now, we know that \[\ln e = 1\]. So, we get,
\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = {e^{\dfrac{1}{e}}}\left[ {{{\left( {\ln e - 1} \right)}^2} - e} \right]\]
\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = {e^{\dfrac{1}{e}}}\left[ {{{\left( {1 - 1} \right)}^2} - e} \right]\]
\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = - e \times {e^{\dfrac{1}{e}}}\]
So, the second derivative of the function is negative for \[x = \dfrac{1}{e}\]. So, \[x = \dfrac{1}{e}\] is the point of local maxima. So, the maximum value of function is attained at \[x = \dfrac{1}{e}\].
So, $y\left( {\dfrac{1}{e}} \right) = {\left( {\dfrac{1}{{\left( {\dfrac{1}{e}} \right)}}} \right)^{\left( {\dfrac{1}{e}} \right)}}$
$ \Rightarrow y\left( {\dfrac{1}{e}} \right) = {e^{\left( {\dfrac{1}{e}} \right)}}$
Therefore, the maximum value of the function ${\left( {\dfrac{1}{x}} \right)^x}$ is ${e^{\left( {\dfrac{1}{e}} \right)}}$.Option (B) is the correct option.

Note:
We can solve the problems involving the maxima and minima concept by two methods: the first derivative test and the second derivative test. The first derivative test helps in finding the local extremum points of the function. The second derivative test involves finding the local extremum points and then finding out which of them is local minima and which is local maxima.