Name: Local Maxima and Minima of a Function
Uploaded: 2019-07-19
Duration: 14 min 17 s
Description: Do you know what is minima or maxima of a function? This video also explains how to find local maxima and minima by differentiation in a detailed manner.

Question

Find the maximum value of

f (x) = {(\frac{1}{x})}^{2 x}

.

Accepted Answer

Given function: $f (x) = {(\frac{1}{x})}^{2 x}$

Given function is of the type variable power to the variable, so we cannot differentiate directly. So taking $\log$ on both sides will help us to differentiate the function.

$∴ \log_{e} y = \log_{e} {(\frac{1}{x})}^{2 x}$

$\Rightarrow \log_{e} y = 2 x \log_{e} \frac{1}{x} [∵ \log a^{m} = m \log a]$

$\Rightarrow \log_{e} y = - 2 x \log_{e} x \to (1)$

Now, differentiate $(1)$ with respect to $x$ , so we get

$\Rightarrow \frac{1}{y} \frac{d y}{d x} = - (2 x \cdot \frac{1}{x} + 2 \log_{e} x) [∵ \frac{d}{d x} (u v) = u . \frac{d v}{d x} + v \cdot \frac{d u}{d x}]$

$\Rightarrow \frac{1}{y} \frac{d y}{d x} = - (2 + 2 \log_{e} x)$

$\Rightarrow \frac{d y}{d x} = - y (2 + 2 \log_{e} x) \to (2)$

To find the critical points, the first derivative of the given function should be zero.

$∴ - {(\frac{1}{x})}^{2 x} (2 + 2 \log_{e} x) = 0$

$\Rightarrow (2 + 2 \log_{e} x) = 0$ [Since ${(\frac{1}{x})}^{2 x} \neq 0$ ]

$\Rightarrow \log_{e} x = - 1$

$\Rightarrow x = \frac{1}{e}$

To find whether $x = \frac{1}{e}$ gives maximum value or minimum value, we must differentiate again $(2)$ w.r.t $x$ ,

$\Rightarrow \frac{d^{2} y}{d x^{2}} = - \frac{d}{d x} (2 y) - \frac{d}{d x} (2 y \log_{e} x)$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = - 2 \frac{d y}{d x} - 2 y \times \frac{1}{x} - 2 \log_{e} x \frac{d y}{d x}$ (Applied product rule in differentiation)

$\Rightarrow \frac{d^{2} y}{d x^{2}} = - 2 [\frac{d y}{d x} + y \times \frac{1}{x} + \log_{e} x \frac{d y}{d x}]$

Now at $x = \frac{1}{e}$ , then $\frac{d^{2} y}{d x^{2}} = - 2 [\frac{d y}{d x} + y \times e + \log_{e} \frac{1}{e} \frac{d y}{d x}]$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = - 2 [0 + y \times e + \log_{e} \frac{1}{e} \times 0] [∵ \frac{d y}{d x} = 0]$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = - 2 y e < 0$

Since, $\frac{d^{2} y}{d x^{2}}$ is negative, so at $x = \frac{1}{e}$ , the function has a maximum value.

Hence, the maximum value of the function at $x = \frac{1}{e}$ is $e^{\frac{2}{e}}$ .

EXERCISE

Parthasarathi Mukhopadhyay Mathematics Solutions for EXERCISE

Questions from EXERCISE with Hints & Solutions

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