if-f-x-sin-1-cos-x-find-Df-and-Rf-the-function-notice-is-floor-

Question Number 87625 by M±th+et£s last updated on 05/Apr/20

i f f (x) = {s i n}^{- 1} (c o s [x])

f i n d D f a n d R f t h e f u n c t i o n

n o t i c e / [\dots] i s f l o o r

Answered by mind is power last updated on 05/Apr/20

{s i n}^{-} i s d e f i n d o v e r / [- 1, 1]

v o s [x] \in [- 1, 1] f i s d e f i n d o v e r R

c o s (n) i s d e n s e i n [- 1, 1]

\Rightarrow R f =] - \frac{π}{2}; \frac{π}{2} [

Commented by M±th+et£s last updated on 05/Apr/20

t h a n k y o u s i r

Commented by mind is power last updated on 05/Apr/20

c a n y o u s i r p l e a s r r e p o s t ? \int \frac{d x}{{((x + 1) \dots . (x + n))}^{2}}

Answered by mr W last updated on 06/Apr/20

i^{'} l l t r y t o a n s w e r t h e s e c o n d p a r t o f t h e

q u e s t i o n : r a n g e o f t h e f u n c t i o n = ?

\sin^{- 1} (\cos [x]) = y

a c c . t o d e f i n i t i o n : - \frac{π}{2} ⩽ y ⩽ \frac{π}{2}

\Rightarrow \cos [x] = \sin y = \cos (\frac{π}{2} - y)

\Rightarrow [x] = 2 n π \pm (\frac{π}{2} - y) = m

\Rightarrow y = \frac{π}{2} \pm (m - 2 n π)

\Rightarrow - \frac{π}{2} ⩽ \frac{π}{2} \pm (m - 2 n π) ⩽ \frac{π}{2}

\Rightarrow - π ⩽ \pm (m - 2 n π) ⩽ 0

\Rightarrow (2 n - 1) π ⩽ m ⩽ 2 n π

\Rightarrow ⌈ (2 n - 1) π ⌉ ⩽ m ⩽ ⌊ 2 n π ⌋

\Rightarrow 0 ⩽ m - 2 n π ⩽ π

\Rightarrow 2 n π ⩽ m ⩽ (2 n + 1) π

\Rightarrow ⌈ 2 n π ⌉ ⩽ m ⩽ ⌊ (2 n + 1) π ⌋

\Rightarrow y = \frac{π}{2} + m - 2 n π w i t h ⌈ (2 n - 1) π ⌉ ⩽ m ⩽ ⌊ 2 n π ⌋

\Rightarrow y = \frac{π}{2} - m + 2 n π w i t h ⌈ 2 n π ⌉ ⩽ m ⩽ ⌊ (2 n + 1) π ⌋

Commented by M±th+et£s last updated on 06/Apr/20

g o d b l e s s y o u s i r