find the maximum value of (2/(3cosh2x +2))


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Question Number 84220 by Rio Michael last updated on 10/Mar/20

$find the maximum value of \frac{2}{3 \cosh 2 x + 2}$
Commented by mr W last updated on 10/Mar/20

$\cosh 2 x ⩾ 1$ $3 \cosh 2 x + 2 ⩾ 3 \times 1 + 2 = 5$ $\frac{2}{3 \cosh 2 x + 2} ⩽ \frac{2}{5} = m a x i m u m$
	Answered by TANMAY PANACEA last updated on 10/Mar/20
	$(2/(3(((e^(2x) +e^(−2x) )/2))+2))=y y=(4/(3e^(2x) +3e^(−2x) +4))=(4/(3a+(3/a)+4))=(4/(3{((√a) −(1/(√a)))^2 +2}+4)) y_(max) =(4/(3(0+2)+4))=(4/(10))=(2/5)$
	$\frac{2}{3 (\frac{e^{2 x} + e^{- 2 x}}{2}) + 2} = y$ $y = \frac{4}{3 e^{2 x} + 3 e^{- 2 x} + 4} = \frac{4}{3 a + \frac{3}{a} + 4} = \frac{4}{3 {{(\sqrt{a} - \frac{1}{\sqrt{a}})}^{2} + 2} + 4}$ $y_{m a x} = \frac{4}{3 (0 + 2) + 4} = \frac{4}{10} = \frac{2}{5}$
	Commented by Rio Michael last updated on 10/Mar/20

	$t h a n k s$
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