Give the coordinates of the extreme point of the graph of each function, and the corresponding maximum or minimum values of the function.(a) $f (x) = 2 x^{2} + 4 x - 16$
(b) $f (x) = - x^{2} - 6 x - 8$

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Solution

at the point of extremum, f'(x) changes its sign
$o r s a y f^{'} (x) = 0 (a) f (x) = 2 x^{2} + 4 x - 16 f^{'} (x) = 4 x + 4 = 0 ∴ x = - 1 a s f^{'} (x) c h a n g e s f r o m - v e t o + v e ∴ a t x = - 1 i s a p o i n t o f m i n i m a a n d f (- 1) = 2 (- 1)^{2} + 4 (- 1) - 16 = 2 - 4 - 16 = - 18 (b) f (x) = - x^{2} - 6 x - 8 f^{'} (x) = - 2 x - 6 = 0 ∴ x = - 3 a s f^{'} (x) c h a n g e s s i g n f r o m + v e t o - v e,$
hence x=-3 is a point of maxima and maximum value $f (- 3) = - (- 3)^{2} - 6 (- 3) - 8 = - 9 + 18 - 8 = 1$

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