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Question Number 39921 by Raj Singh last updated on 13/Jul/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 13/Jul/18

	$i i i) m a x v a l u e o f s i n 2 x = 1$ $m i n v a l u e o f s i n 2 x = - 1$ $s 0 m a x v a l u e o f h (x) = 1 + 5 = 6$ $m i n v a l u e - 1 + 5 = 4$ $i) f (x) =∣ x + 2 ∣ - 1$ $c r i t i c a l v a l u e o f x = - 2$ $w h e n x = - 2 f (x) = - 1$ $w h e n x > - 2 t h e n ∣ x + 2 ∣ i s + v e a n d i t s v a l u e$ $i n c r e a s e s a s w e i n c r e a s e x . . . s o i t h a s n o m a x$ $v a l u e$ $w h e n x < - 2 ∣ x + 2 ∣ i s - v e i f w e d e c r e a s e$ $x k e e p i n g x < - 2 . . . t h e n v a l u e o f ∣ x + 2 ∣ c o n t i n u e$ $t o d e c r e a s e . . . s o ∣ x + 2 ∣ h a s n o m i n v a l u e$
	Answered by tanmay.chaudhury50@gmail.com last updated on 13/Jul/18

	$i i) g (x) = - ∣ x + 1 ∣ + 3$ $c r i t i c a l v a l u e o f x = - 1$ $a t x = - 1 g (x) = 3$ $n o w w h e n x > - 1 a n d w e i n c r e a s e x k e e p i n g$ $c o n d i t i o n x > - 1 v a l u e o f g (x) d e c r e a s e s$ $s o i t h a s n o m i n v a l u e . . .$ $w h e n x < - 1 w e c o n t i n u e d e c r e a s e t h e$ $v a l u e o f x . . . t h e n t h e v a l u e o f x + 1 < 0$ $s o - ∣ x + 1 ∣ b e c o m e + v e . . . t h u s t h e v a l u e$ $o f g (x) i n c r e a s e s . . . t h u s g (x) h a s n o m a x v a l u e$
	Answered by tanmay.chaudhury50@gmail.com last updated on 13/Jul/18

	$i v) f o r a n y v a l u e o f x, m i n v a l u e o f s i n 4 x = - 1$ $m a x v a l u e o f s i n 4 x = + 1$ $s o m i n v a l u e o f f (x) =∣ - 1 + 3 ∣= 2$ $m a x v a l u e f (x) =∣ 1 + 3 ∣= 4$ $i f t h e p r o b l e m i s f (x) =∣ s i n (4 x + 3) ∣$ $t h e n m i n v a l u e o f s i n (4 x + 3) = - 1$ $t h d n m a x v a l u e o f s i n (4 x + 3) = 1$ $b u t v a l u e o f f (x) =∣ - 1 ∣= 1$ $f (x) =∣ 1 ∣= 1$ $s o f (x) h a s n o m i n n o m a x v a l u e . . .$ $s o$
	Answered by tanmay.chaudhury50@gmail.com last updated on 13/Jul/18

	$v) h (x) = x + 1 a s x \to 1 h (x) \to 2$ $x \to - 1 h (x) \to 0$ $s o v a l u e o f h (x) l i e s b e t w e e n 0 a n d 2$ $h (x) (0, 2) . . . b u t t h e r a n g e o f h (x) d o e s n o t$ $i n c l u d e l o w e s v a l u e 0 h i g h e s t v a l u e 2$ $t h u s h (x) h a s n o m a x o r m i n v a l u e . . .$
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