Answer:
f(x) = (x ? 2)4
(x + 1)3
= (x ? 2)3
(x + 1)2 [3x ? 6 + 4x + 4]
= (x ? 2)3
(x + 1)2 (7x ? 2)
f?(x) = 0
Now f?(x) = (x ?
2)3 [(x + 1)2 × 7 +
(7x ?
2) × 2 (x + 1) ]
+ (x + 1)2
(7x ? 2) × 3 (x ? 2)2
f?(x)|x = 2
= 0 + 0 = 0
order
derivative test fails.
Hence we will use
1st order derivative test.
At x =2
When x isslightly
less than 2, then f?(x) < 0
When x is slightly
greater than 2, then f?(x) > 0
Therefore f(x) has
local minima at x = 2
At x = ? 1
When x is slightly
< ?1 then f?(x) > 0
When x is slightly
> ?1, then f?(x) > 0
Therefore f(x) has
neither maxima nor minima at x = ?1
is the
point of in flexion.
At
When x is slightly
< then
f?(x) > 0
When x is slightly
> then
f?(x) < 0
Therefore f(x) has
local maxima at .
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