First slide

Maxima and minima

Question

$Let f (x) = (x - 3)^{5} (x + 1)^{4} then$

Moderate

A

x = 7/9 is a point of maxima
B

x = 3 is a point of minimum
C

x = -1 is a point of maxima
D

f has no point of maximum or minimum

Solution

$\begin{array}{l} f^{'} (x) = (x - 3)^{4} (x + 1)^{3} (9 x - 7) \\ and \begin{aligned} f^{''} (x) & = 8 (x - 3)^{3} (x + 1)^{2} (9 x^{2} + 14 x + 1) \\ f^{''} (x) & = 24 (x - 3)^{2} (x + 1) (21 x^{3} - 49 x^{2} + + 7 x + 13) \\ f^{i v} (x) & = 24 (x - 3) (3 x - 1) (21 x^{3} - 49 x^{2} + 7 x + 13) + 168 (x + 3)^{2} (x + 1) (9 x^{2} - 14 x + 1) \end{aligned} \end{array}$
$f^{v} (x) = 48 (3 x - 5) (21 x^{3} - 49 x^{2} + 7 x + 13) + 336 (x - 3) (9 x^{2} - 14 x + 1) (9 x - 7)$
$f (3) = f^{'} (- 1) = f^{'} (7 / 9) = 0$
$\begin{aligned} At x = - 1, f^{i v} (x) < 0 so x = - 1 is a point of maxima \\ At x = 3, f^{''} (x) = f^{'''} (x) = f^{i v} (x) = 0 but f^{v} (x) \neq 0 \end{aligned}$
$\underline{So} f has neither maximum nor minimum at x = 3$
$At x = 7 / 9, f^{''} (x) > 0 and therefore is a point of minima$

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