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Unformatted text preview: jiwani (amj566) – Homework09 – Fouli – (58395) 1 This printout should have 23 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points If f is the function whose graph is given by 2 4 6 2 4 6 which of the following properties does f NOT have? 1. lim x → 2 f ( x ) = 2 2. local maximum at x = 4 correct 3. critical point at x = 2 4. lim x → 4 + f ( x ) = lim x → 4 f ( x ) 5. f ′ ( x ) < 0 on ( − 1 , 2) Explanation: The given graph has a removable disconti nuity at x = 4. On the other hand, recall that f has a local maximum at a point c when f ( x ) ≤ f ( c ) for all x near c . Thus f could have a local maximum even if the graph of f has a removable discontinuity at c ; simi larly, the definition of local minimum allows the graph of f to have a local minimum at a removable disconituity. So it makes sense to ask if f has a local extremum at x = 4. Inspection of the graph now shows that the only property f does not have is local maximum at x = 4 . 002 10.0 points Find all the critical numbers, x , of the function g ( x ) = 6 x + sin 6 x in (0 , ∞ ). 1. x = n + 1 6 π, n = 0 , 1 , 2 , . . . 2. x = 4 n + 1 6 π, n = 0 , 1 , 2 , . . . 3. x = 2 n + 1 6 π, n = 0 , 1 , 2 , . . . correct 4. x = 3 n + 1 6 π, n = 0 , 1 , 2 , . . . 5. x = n 6 π, n = 0 , 1 , 2 , . . . Explanation: After differentiation using the Chain Rule we see that g ′ ( x ) = 6 + 6 cos6 x = 6(1 + cos 6 x ) . Thus the critical points of g on (0 , ∞ ) are the solutions of 6(1 + cos 6 x ) = 0 lying in (0 , ∞ ), i.e. ., the positive values of x for which cos 6 x = − 1. But the positive values of x for which cos x = − 1 are x = π, 3 π, 5 π, . . . . Consequently, the critical points of g in (0 , ∞ ) are x = 2 n + 1 6 π, n = 0 , 1 , 2 , . . . . jiwani (amj566) – Homework09 – Fouli – (58395) 2 003 10.0 points Find all the critical points of the function f ( x ) = 2 sin x −  x  on the interval ( − π, π ). 1. x = − π 3 , 2 π 3 2. x = − 5 π 6 , − π 6 , , π 6 , 5 π 6 3. x = − π 6 , , π 6 4. x = 0 5. x = − π 3 , , 2 π 3 6. x = − 2 π 3 , π 3 7. x = − 2 π 3 , , π 3 correct 8. x = − 5 π 6 , , 5 π 6 Explanation: Since  x  is differentiable everywhere except at x = 0, while sin x is differentiable for all x , the point x = 0 is a critical point of f and all other critical points will be the solutions of f ′ ( x ) = 0. Now f ′ ( x ) = braceleftbigg 2 cos x − 1 , x > 0, 2 cos x + 1 , x < 0. But on (0 , π ) cos x = 1 2 = ⇒ x = π 3 , while on ( − π, 0) cos x = − 1 2 = ⇒ x = − 2 π 3 . Consequently, the only critical points of f on ( − π, π ) are x = − 2 π 3 , , π 3 ....
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 Spring '11
 Jensen
 Calculus, MVT

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