Find-the-supremum-and-the-infimum-of-x-sin-x-on-the-interval-0-pi-2-

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 = \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 \dots