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Question: The value of \(f\left( x \right) = {x^x}\) has stationary point at? \( a.{\text{ }}x = e \\\ ...

The value of f(x)=xxf\left( x \right) = {x^x} has stationary point at?
a. x=e b. x=1e c. x=1 d. x=e  a.{\text{ }}x = e \\\ b.{\text{ }}x = \dfrac{1}{e} \\\ c.{\text{ }}x = 1 \\\ d.{\text{ }}x = \sqrt e \\\

Explanation

Solution

Hint: - Use ddx(f(x))=0\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.
∴ddx(f(x))=0\therefore \dfrac{d}{{dx}}\left( {f\left( x \right)} \right) = 0
So first simplify the function take log on both sides
∴log⁡f(x)=log⁡xx\therefore \log f\left( x \right) = \log {x^x}
As we know log⁡ab=blog⁡a\log {a^b} = b\log a so, apply this property of logarithmic
∴log⁡f(x)=xlog⁡x\therefore \log f\left( x \right) = x\log x
Now differentiate above equation w.r.t.xx
As we know differentiation ofddxlog⁡f(x)=1f(x)(ddxf(x))\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 xlog⁡xx\log xwe use chain rule of differentiation.
∴1f(x)(ddxf(x))=ddxxlog⁡x ∴1f(x)(ddxf(x))=xddxlog⁡x+log⁡xddxx ∴1f(x)(ddxf(x))=xx+log⁡x(1) ∴1f(x)(ddxf(x))=1+log⁡x ∴(ddxf(x))=f(x)(1+log⁡x)  \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(x)=xxf\left( x \right) = {x^x}in the above equation
∴(ddxf(x))=xx(1+log⁡x)\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.
∴xx(1+log⁡x)=0 ∴(1+log⁡x)=0 ∴log⁡x=−1  \therefore {x^x}\left( {1 + \log x} \right) = 0 \\\ \therefore \left( {1 + \log x} \right) = 0 \\\ \therefore \log x = - 1 \\\
Now take antilog
∴x=e−1=1e\therefore x = {e^{ - 1}} = \dfrac{1}{e}
So, the stationary point of the function f(x)=xxf\left( x \right) = {x^x}is at x=1ex = \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. xx and equate the value to zero, then solve for xx, which is the required stationary point.