# The value of $f\left( x \right) = {x^x}$ has stationary point at?

$

a.{\text{ }}x = e \\

b.{\text{ }}x = \dfrac{1}{e} \\

c.{\text{ }}x = 1 \\

d.{\text{ }}x = \sqrt e \\

$

Answer

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Hint: - Use $\dfrac{d}{{dx}}\left( {f\left( x \right)} \right) = 0$, to find out the stationary point.

To find out the stationary point differentiate the given function w.r.t the given variable and put that to zero.

$\therefore \dfrac{d}{{dx}}\left( {f\left( x \right)} \right) = 0$

So first simplify the function take log on both sides

$\therefore \log f\left( x \right) = \log {x^x}$

As we know $\log {a^b} = b\log a$ so, apply this property of logarithmic

$\therefore \log f\left( x \right) = x\log x$

Now differentiate above equation w.r.t.$x$

As we know differentiation of$\dfrac{d}{{dx}}\log f\left( x \right) = \dfrac{1}{{f\left( x \right)}}\left( {\dfrac{d}{{dx}}f\left( x \right)} \right)$, and in $x\log x$we use chain rule of differentiation.

$

\therefore \dfrac{1}{{f\left( x \right)}}\left( {\dfrac{d}{{dx}}f\left( x \right)} \right) = \dfrac{d}{{dx}}x\log x \\

\therefore \dfrac{1}{{f\left( x \right)}}\left( {\dfrac{d}{{dx}}f\left( x \right)} \right) = x\dfrac{d}{{dx}}\log x + \log x\dfrac{d}{{dx}}x \\

\therefore \dfrac{1}{{f\left( x \right)}}\left( {\dfrac{d}{{dx}}f\left( x \right)} \right) = \dfrac{x}{x} + \log x\left( 1 \right) \\

\therefore \dfrac{1}{{f\left( x \right)}}\left( {\dfrac{d}{{dx}}f\left( x \right)} \right) = 1 + \log x \\

\therefore \left( {\dfrac{d}{{dx}}f\left( x \right)} \right) = f\left( x \right)\left( {1 + \log x} \right) \\

$

Now substitute the value of $f\left( x \right) = {x^x}$in the above equation

$\therefore \left( {\dfrac{d}{{dx}}f\left( x \right)} \right) = {x^x}\left( {1 + \log x} \right)$

Now according to stationary point condition equate this value to zero.

$

\therefore {x^x}\left( {1 + \log x} \right) = 0 \\

\therefore \left( {1 + \log x} \right) = 0 \\

\therefore \log x = - 1 \\

$

Now take antilog

$\therefore x = {e^{ - 1}} = \dfrac{1}{e}$

So, the stationary point of the function $f\left( x \right) = {x^x}$is at $x = \dfrac{1}{e}$

Hence, option (b) is correct.

Note: - In such types of questions the key concept we have to remember is that always remember the condition of stationary point which is stated above, so differentiate the following function w.r.t. $x$ and equate the value to zero, then solve for $x$, which is the required stationary point.

To find out the stationary point differentiate the given function w.r.t the given variable and put that to zero.

$\therefore \dfrac{d}{{dx}}\left( {f\left( x \right)} \right) = 0$

So first simplify the function take log on both sides

$\therefore \log f\left( x \right) = \log {x^x}$

As we know $\log {a^b} = b\log a$ so, apply this property of logarithmic

$\therefore \log f\left( x \right) = x\log x$

Now differentiate above equation w.r.t.$x$

As we know differentiation of$\dfrac{d}{{dx}}\log f\left( x \right) = \dfrac{1}{{f\left( x \right)}}\left( {\dfrac{d}{{dx}}f\left( x \right)} \right)$, and in $x\log x$we use chain rule of differentiation.

$

\therefore \dfrac{1}{{f\left( x \right)}}\left( {\dfrac{d}{{dx}}f\left( x \right)} \right) = \dfrac{d}{{dx}}x\log x \\

\therefore \dfrac{1}{{f\left( x \right)}}\left( {\dfrac{d}{{dx}}f\left( x \right)} \right) = x\dfrac{d}{{dx}}\log x + \log x\dfrac{d}{{dx}}x \\

\therefore \dfrac{1}{{f\left( x \right)}}\left( {\dfrac{d}{{dx}}f\left( x \right)} \right) = \dfrac{x}{x} + \log x\left( 1 \right) \\

\therefore \dfrac{1}{{f\left( x \right)}}\left( {\dfrac{d}{{dx}}f\left( x \right)} \right) = 1 + \log x \\

\therefore \left( {\dfrac{d}{{dx}}f\left( x \right)} \right) = f\left( x \right)\left( {1 + \log x} \right) \\

$

Now substitute the value of $f\left( x \right) = {x^x}$in the above equation

$\therefore \left( {\dfrac{d}{{dx}}f\left( x \right)} \right) = {x^x}\left( {1 + \log x} \right)$

Now according to stationary point condition equate this value to zero.

$

\therefore {x^x}\left( {1 + \log x} \right) = 0 \\

\therefore \left( {1 + \log x} \right) = 0 \\

\therefore \log x = - 1 \\

$

Now take antilog

$\therefore x = {e^{ - 1}} = \dfrac{1}{e}$

So, the stationary point of the function $f\left( x \right) = {x^x}$is at $x = \dfrac{1}{e}$

Hence, option (b) is correct.

Note: - In such types of questions the key concept we have to remember is that always remember the condition of stationary point which is stated above, so differentiate the following function w.r.t. $x$ and equate the value to zero, then solve for $x$, which is the required stationary point.

Last updated date: 15th Sep 2023

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