Math 1300 4.2 Optimization, part 2 Name: Solutions 1. Consider the ...
Math 1300 4.2 Optimization, part 2 Name: Solutions 1. Consider the ...
Math 1300 4.2 Optimization, part 2 Name: Solutions 1. Consider the ...
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<strong>Math</strong> <strong>1300</strong> <strong>4.2</strong> <strong>Optimization</strong>, <strong>part</strong> 2 <strong>Name</strong>: <strong>Solutions</strong>2. <strong>Consider</strong> <strong>the</strong> function f(x) = 2x 3 − 6x 2 − 144x.(a) Find <strong>the</strong> global maximum and minimum values of f(x) on [−6, 6].f ′ (x) = 6x 2 − 12x − 144 = 6(x 2 − 2x − 24) = 6(x − 6)(x + 4) so our critical pointsare x = −4, 6.f(−4) = 352 f(6) = −648 f(−6) = 216So <strong>the</strong> global max is 352 and occurs at x = −4 and <strong>the</strong> global min is −648 and occursat x = 6.(b) Find <strong>the</strong> global maximum and minimum values of f(x) on (−5, 5).From our critical points −4 and 6 we now only care about −4.f(−4) = 352lim f(x) = f(−5) = 320x→−5f(x) = f(5) = −620limx→5So f has no global maximum or global minimum on (−5, 5).