PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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# Write the Minimum Value of F(X) = Xx . - PUC Karnataka Science Class 12 - Mathematics

#### Question

Write the minimum value of f(x) = xx .

#### Solution

$\text { Given: } \hspace{0.167em} f\left( x \right) = x^x$

$\text { Taking log on both sides, we get }$

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

$\text { Differentiating w . r . t . x, we get }$

$\frac{1}{f\left( x \right)} f'\left( x \right) = \log x + 1$

$\Rightarrow f'\left( x \right) = f\left( x \right) \left( \log x + 1 \right)$

$\Rightarrow f'\left( x \right) = x^x \left( \log x + 1 \right) .............. \left( 1 \right)$

$\text { For a local maxima or a local minima, we must have }$

$f'\left( x \right) = 0$

$\Rightarrow x^x \left( \log x + 1 \right) = 0$

$\Rightarrow \log x = - 1$

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

$\text { Now },$

$f''\left( x \right) = x^x \left( \log x + 1 \right)^2 + x^x \times \frac{1}{x} = x^x \left( \log x + 1 \right)^2 + x^{x - 1}$

$\text { At }x = \frac{1}{e}:$

$f''\left( \frac{1}{e} \right) = \frac{1}{e}^\frac{1}{e} \left( \log\frac{1}{e} + 1 \right)^2 + \frac{1}{e}^\frac{1}{e} - 1 = \frac{1}{e}^\frac{1}{e} - 1 > 0$

$\text { So,} x = \frac{1}{e}\text { is a point of local minimum .}$

$\text { Thus, the minimum value is given by }$

$f\left( \frac{1}{e} \right) = \frac{1}{e}^\frac{1}{e} = e^\frac{- 1}{e}$

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Solution Write the Minimum Value of F(X) = Xx . Concept: Graph of Maxima and Minima.
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