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# Exam2_3 4 - 3 1 abs min value = 2 2 abs min value = 4 3 abs...

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Version 208 – Exam 2 – Radin – (58305) 4 with T (20) = lim t 20 T ( t ) T (20) t 20 . But T (20) T (10) T (20) 10 20 = 166 160 10 = 3 5 degrees/min . Consequently, temp = T (30) 154 F . 008 10.0 points Use differentials to estimate the amount of paint needed to apply a 1 / 4 cms thick coat of paint to a sphere having diameter 4 cms. 1. amount 2 cm 3 2. amount 2 π cm 3 3. amount 4 cm 3 4. amount 4 3 π cm 3 5. amount 4 3 cm 3 6. amount 4 π cm 3 correct Explanation: A sphere of radius r has volume, V = 4 3 πr 3 . Thus dV = 4 πr 2 dr , Δ V 4 πr 2 Δ r . When r = 1 2 (4) = 2 , Δ r = 1 4 , therefore, differentials estimate the amount of paint needed as amount = Δ V 4 π cm 3 . keywords: differentials, application of differ- entials, surface area sphere, volume sphere 009 10.0 points Find the absolute minimum value of f ( x ) = 1 3 x 3 2 x 2 + 3 x + 3 on the interval [0
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Unformatted text preview: , 3]. 1. abs. min. value = 2 2. abs. min. value = 4 3. abs. min. value = 3 correct 4. none oF the other answers 5. abs. min. value = 1 Explanation: The absolute minimum value oF f on [0 , 3] occurs either at an endpoint oF [0 , 3] or at a critical point oF f in (0 , 3). Now f ′ ( x ) = x 2 − 4 x + 3 = ( x − 1)( x − 3) , so the critical points oF f occur at x = 1 , 3. But only x = 1 lies in (0 , 3). On the other hand, f (0) = 3 , f (1) = 13 3 , f (3) = 3 . Consequently, abs. min. value = 3 . 010 10.0 points ±ind all the critical values oF f ( x ) = x (5 − x ) 6 . 1. x = 5 , 5 7 correct 2. x = − 5 , 1...
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