find-the-relative-maximum-or-minimum-or-neither-at-the-given-critical-points-of-the-function-f-x-6x-x-2-4-4-x-2-1-2-8x-x-2-1-3-x-2-4-4-x-1-x-2-

Question Number 159669 by zakirullah last updated on 19/Nov/21

f i n d t h e r e l a t i v e m a x i m u m o r m i n i m u m

o r n e i t h e r a t t h e g i v e n c r i t i c a l

p o i n t s o f t h e f u n c t i o n ?

f^{^{'}} (x) = 6 x {(x^{2} - 4)}^{4} {(x^{2} - 1)}^{2} + 8 x {(x^{2} - 1)}^{3} {(x^{2} - 4)}^{4},

x = 1, x = 2

Answered by MohammadAzad last updated on 19/Nov/21

S o l u t i o n :

f^{^{'}} (x) = 2 x {(x^{2} - 4)}^{4} {(x^{2} - 1)}^{2} [3 + 4 (x^{2} - 1)]

f^{^{'}} (x) = 2 x {(x^{2} - 4)}^{4} {(x^{2} - 1)}^{2} (4 x^{2} - 1)

n o t i c e t h a t t h e {(x^{2} - 4)}^{4} {(x^{2} - 1)}^{2} i s ⩾ 0

s o w e o n l y n e e d t o w o r r y a b o u t 2 x (4 x^{2} - 1)

- \underset{- \frac{1}{2}}{∙} + \underset{0}{∙} - \underset{\frac{1}{2}}{∙} +

x = 1 a n d x = 2 a r e a f t e r \frac{1}{2} s o

t h e r e i s n o l o c a l e x t r e m a

a t x = 1 a n d x = 2

Commented by MohammadAzad last updated on 19/Nov/21

Commented by zakirullah last updated on 23/Nov/21

a b o u n d l e o f t h a n k s s i r G