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Department of Mathematical Sciences Instructor: Daiva Pucinskaite Calculus I July 13, 2016 Quiz 12 Functions from derivatives. Use the derivative f ′ given by f ′ (x) = x2 (x + 2)(x − 1) to determine the x-coordinates of the local maxima and minima of f , and the intervals of increase and decrease. Sketch a possible graph of f (f is not unique). Since f ′ (x) does not fail to exist for all x, the critical points of f are x = −2, x = 0, x = 1 because at those three points f ′ is 0. • f ′ (−3) = (−3)2 (−3 + 2) (−3 − 1) > 0 ⇒ f increases on (−∞, −2), | {z } | {z } | {z } >0 <0 <0 • f ′ (−1) = (−1)2 (−1 + 2) (−1 − 1) < 0 ⇒ f decreases on (−2, 0), | {z } | {z } | {z } >0 >0 <0 • f ′ (0.5) = (0.5)2 (0.5 + 2) (0.5 − 1) < 0 ⇒ f decreases on (0, 1), | {z } | {z } | {z } >0 >0 <0 • f ′ (5) = (5)2 (5 + 2) (5 − 1) > 0 ⇒ f increases on (−1, ∞), |{z} | {z } | {z } >0 >0 >0 Recall: Suppose c is a critical point of f , then ∗ f has a local maximum at c if f ′ changes sign form positive to negative, as x increases through. ∗ f has a local minimum at c if f ′ changes sign form negative to positive, as x increases through. • Since f increases on (−∞, −2), and decreases on (−2, 0), i.e. f ′ changes sign form positive to negative, the function f has a local maximum at the critical point c = −2. • Since f decreases on (0, 1), and increases on (−1, ∞), i.e. f ′ changes sign form negative to positive, the function f has a local minimum at the critical point c = 1. • Since f decreases on (−2, 1), thus f ′ does not changes sign at the critical point c = 0. The function f has neither a local minimum nor a local maximum at c = 0. A possible graph of f . 3 2 1 -3 -2 -1 0 -1 1 2 3 Department of Mathematical Sciences Instructor: Daiva Pucinskaite Calculus I July 13, 2016 Quiz 12 Maximum product. Find numbers x and y satisfying the equation 3x + y = 12 such that the product of x and y is as large as possible. Two numbers x and y satisfy the equation 3x + y = 12 if y = 12 − 3x. Thus we are searching for a number x such that x · y = x · (12 − 3x) is as large as possible. If the function f (x) = x(12 − 3x) = 12x − 3x2 takes the absolute maximum at a, i.e. f (a) ≥ f (x) for all x, |{z} |{z} a(12−3a) x(12−3x) then for x = a and y = 12 − 3a we have • 3x + y = 12, and • xy ≥ xy for all x and y with 3x + y = 12. Since f ′ (x) = (12x − 3x2 )′ = 12 − 6x implies that f ′ (c) = 12 − 6c = 0 for c = 2, the function f has only one critical point c = 2. Because f ′ (1) = 12 − 6 · 1 > 0 and f ′ (3) = 12 − 6 · 3 < 0, the function f has a local maximum at 2. Since 2 is the only critical point, the function f takes at the local maximum the absolute maximum. We have f (2) ≥ f (x) for all x. Therefore for x = 2 and y = 12 − 3 · 2 = 6 we have that the product 2 · 6 = 12 is larges number among xy with 3x + y = 12.