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sol14-2

# sol14-2 - SOLUTIONS CHAPTER 14.2 MATH 132 WI01 2 f(x...

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SOLUTIONS CHAPTER 14.2 MATH 132 WI01 2. f ( x ) = - 2 x 2 - 6 x +5, and the interval of definition is [ - 2 , 3]. To find absolute extrema we are intersted in finding a few good numbers, as few as possible, and plug those numbers in f , and compare the results - by picking the largest result we want to get the absolute maximum , and by picking the smallest result we want to have the absolute minimum . How do we find those numbers? Two candidates are the definition interval’s endpoints - in our case - 2 and 3. The others are going to be the critical numbers , that is, zeroes and DNEs 1 of the derivative of f , f 0 . Let’s compute f 0 : f 0 ( x ) = - 2 * 2 x - 6 = - 4 x - 6. f 0 ( x ) = 0 → - 4 x - 6 = 0 - 4 x = 6 x = 6 ( - 4) x = - 3 2 . f 0 exists for every x , so there’s no case of DNE, so - 3 2 is the only critical number. The last thing we must make sure is whether the critical number is INSIDE the interval [ - 2 , 3] ... well, in our case it is! Let’s start pluging in numbers: f ( - 2) = - 2 * ( - 2) 2 - 6 * ( - 2) + 5 = - 2 * 4 + 12 + 5 = - 8 + 12 + 5 = 9 f (3) = - 2 * 3 2 - 6 * 3 + 5 = - 2 * 9 - 18 + 5 = - 18 - 18 + 5 = - 27 f ( - 3 2 ) = - 2 * ( - 3 2 ) 2 - 6 * ( - 3 2 )+5 = - 2 * 9 4 +3 * 3+5 = - 9 2 +9+5 = - 4 .

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