Answered: a) Find and locate local extreme values…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

(a) Find and locate local extreme values off(x) = (x+ 1)^4−4x−2,

(b) Then find the intervals of increase, decrease and concave upward, concavedownward and inflection points.

Expert Solution

Step 1

$G i v e n, f (x) = (x + 1)^{4} - 4 x - 2$

Step 2

$N o w, f (x) = (x + 1)^{4} - 4 x - 2 D i f f e r e n t i a t i n g o n b o t h s i d e s, w e g e t \Rightarrow f' (x) = \frac{d ((x + 1)^{4} - 4 x - 2)}{d x} \Rightarrow f' (x) = \frac{d ((x + 1)^{4})}{d x} - \frac{d (4 x)}{d x} - \frac{d (2)}{d x} \Rightarrow f' (x) = 4 (x + 1)^{3} \frac{d (x + 1)}{d x} - 4 \frac{d (x)}{d x} - 0 \Rightarrow f' (x) = 4 (x + 1)^{3} (1 + 0) - 4 (1) - 0 \Rightarrow f' (x) = 4 (x + 1)^{3} - 4 F i n d i n g t h e c r i t i c a l p o i n t s b y e q u a t i n g f' (x) = 0, w e g e t \Rightarrow f' (x) = 0 \Rightarrow 4 (x + 1)^{3} - 4 = 0 \Rightarrow (x + 1)^{3} - 1 = 0 \Rightarrow (x + 1)^{3} = 1 \Rightarrow (x + 1) = {(1)}^{\frac{1}{3}} \Rightarrow (x + 1) = 1 \Rightarrow x = 0 S i g n a n d b e h a v i o u r :$

	$- \infty < x < 0$	$x = 0$	$0 < x < \infty$
Sign	-	0	+
Behavior	Decreasing	Minimum	Increasing

$F r o m t h e t a b l e w e h a v e : I n c r e a s i n g i n t e r v a l = (0, \infty) D e c r e a s i n g i n t e r v a l = (- \infty, 0) P o i n t o f m i n i m a i s a t x = 0 . f (0) = {(0 + 1)}^{4} - 4 (0) - 2 = 1 - 0 - 2 = - 1 S o, t h e l o c a l e x t r e m e v a l u e i s - 1 a t x = 0 G r a p h o f f (x) s h o w n b e l o w w i t h t h e l o c a l e x t r e m e v a l u e s :$

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