Question

34
Maximum value of f(x) = x + sin 2x, x belongs to [0,2 pie) is​

Answer 1

$\huge\mathfrak\red{Answer}$

HOW TO FIND MAXIMUM

VALUE

$let \: any \: function \: f(x) \\ to \: find \: the \: maximum \: or \: minimum \\ value \: you \: need \: to \: differentiate \: it \\ and \: equate \: the \: result \: obtained \: with \: zero \\ \\ let \: the \: derivative \: of \: f(x) \: is \: g(x) \\ \\ g(x) = 0.......(for \: max \: or \: mini \: value) \\ \\ you \: get \: a \: solution \: of \: x \: but \: we \: are \\ not \: sure \: that \: the \: given \: solution \ \\ will \: give \: max \: or \: minimum \: value \\ \\ for \: checking \: the \: value \\ \\ differentiate \: g(x) \: than \: put \: the \: \\ value \: of \: x \: if \: the \: required \: answer \: \\ is \: less \: than \: zero \: than \: the \: given \: \\ value \: of \: x \: will \: give \: maximum \: value \\ \\ if \: answer \: obtained \: is \: positive \: than \\ this \: particular \: value \: give \: minimum \: \\ value \: of \: this \: function \: f(x).$

Your Solution.

$let \: y \: = x + \sin(2x) \\ differentiate \\ \\ dy = dx + d( \sin(2x) ) \\ dy = dx + 2 \cos(2x).dx \\ \\ for \: mxima \: dy = 0 \\ \\ 0 = dx(1 + 2. \cos(2x) ) \\ 0 = 1 + 2. \cos(2x) \\ \frac{ - 1}{2} = \cos(2x) \\ 2x = \frac{2\pi}{3} \\ x = \frac{\pi}{3} .....(1) \\ \\ again \: differentiate \: dy = dx + 2. \cos(2x) .dx \\ \\ \frac{dy}{dx} = 1 + 2. \cos(2x) \\ \frac{ {d}^{2}y }{d {x}^{2} } = 0 - 4 \sin(2x) \\ \\ \frac{ {d}^{2}y }{d {x}^{2} } = - 4 \sin(2x) \\ \\ put \: (1) \: here \\ \\ value \: will \: be \: positive \\ \\ maximum \: value \\ \\ f(x) = \frac{\pi}{3} + \sin( \frac{2\pi}{3} ) \\ f(x) = \frac{\pi}{3} + 0.8 \\ f(x) = 1.04 + 0.8 \\ f(x) = 1.84$