integration doubt

The least value of \phi (x) = \int_{x}^{2}{log_{1/3}tdt}for x\epsilon \left(\frac{1}{10},4 \right) is at x = ........

ans--->1

1 Answers

Use Leibniz formula to differentiate the function \phi '(x) = -log_{1/3} x = log_{3}x..

now for minima we need to equate \phi '(x) = 0 ..so

log_{3}x = 0 ...

log_{3}x = \frac{log_{e}x}{log_{e}3} ...so ..\phi ''(x) = \frac{1}{xlog_{e}3} > 0

so x= 1 is a point of minima....

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