Find the supremum and the infimum of (x/(sin x)) on the interval (0, (π/2) ]


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Question Number 115516 by bobhans last updated on 26/Sep/20

$F i n d t h e s u p r e m u m a n d t h e i n f i m u m$ $o f \frac{x}{\sin x} o n t h e i n t e r v a l (0, \frac{π}{2}]$
Commented by john santu last updated on 26/Sep/20

$i n f i m u m y = \lim_{x \to 0} \frac{x}{\sin x} = 1$ $s u p r e m u m f (\frac{π}{2}) = \frac{π}{2}$
	Answered by TANMAY PANACEA last updated on 26/Sep/20
	$y=(x/(sinx)) (dy/dx)=((sinx−xcosx)/(sin^2 x)) for max\min (dy/dx)=0 so x=tanx (d^2 y/dx^2 )=((sin^2 x(cosx−cosx+xsinx)−(sinx−xcosx)(2sinxcosx))/(sin^4 x)) (d^2 y/dx^2 )=((xsin^3 x−2sin^2 xcosx+2xsinxcos^2 x)/(sin^4 x)) now putting x=tanx in (d^2 y/dx^2 ) ((d^2 y/dx^2 ))_(x=tanx) =((((sin^4 x)/(cosx))−2sin^2 xcosx+2sin^2 xcosx)/(sin^4 x)) ((d^2 y/dx^2 ))_(x=tanx) =secx>0 when x∈(0,(π/2)] wait...$
	$y = \frac{x}{s i n x}$ $\frac{d y}{d x} = \frac{s i n x - x c o s x}{{s i n}^{2} x}$ $f o r m a x ∖ m i n \frac{d y}{d x} = 0 s o x = t a n x$ $\frac{d^{2} y}{{d x}^{2}} = \frac{{s i n}^{2} x (c o s x - c o s x + x s i n x) - (s i n x - x c o s x) (2 s i n x c o s x)}{{s i n}^{4} x}$ $\frac{d^{2} y}{{d x}^{2}} = \frac{{x s i n}^{3} x - 2 {s i n}^{2} x c o s x + 2 {x s i n x c o s}^{2} x}{{s i n}^{4} x}$ $n o w p u t t i n g x = t a n x i n \frac{d^{2} y}{{d x}^{2}}$ ${(\frac{d^{2} y}{{d x}^{2}})}_{x = t a n x} = \frac{\frac{{s i n}^{4} x}{c o s x} - 2 {s i n}^{2} x c o s x + 2 {s i n}^{2} x c o s x}{{s i n}^{4} x}$ ${(\frac{d^{2} y}{{d x}^{2}})}_{x = t a n x} = s e c x > 0 w h e n x \in (0, \frac{π}{2}]$ $w a i t . . .$
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