Math150Sols4S11

Math150Sols4S11 - QUIZ ON CHAPTER 4 SOLUTIONS MATH 150...

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QUIZ ON CHAPTER 4 SOLUTIONS MATH 150 – SPRING 2011 – KUNIYUKI 105 POINTS TOTAL, BUT 100 POINTS = 100% Show all work, simplify as appropriate, and use “good form and procedure” (as in class). Box in your final answers! No notes or books allowed. A scientific calculator is allowed. 1) Consider fx () = x + 5 16 ± x 2 . (13 points total) a) Find and box in all critical numbers of f . If there are none, write “NONE.” Show all work, as we have done in class. (9 points) The domain of f is x ± ± x ² ± 4 {} , since those are the real values for which 16 ± x 2 ² 0 . ± = Lo ² DH i ³ Hi ² DLo Lo 2 = 16 ³ x 2 ´ µ · ² D x x + 5 ´ µ · ³ x + 5 ´ µ · ² D x 16 ³ x 2 ´ µ · 16 ³ x 2 2 = 16 ³ x 2 ´ µ · ² 1 ´ µ · ³ x + 5 ´ µ · ²³ 2 x ´ µ · 16 ³ x 2 2 = 16 ³ x 2 ³³ 2 x 2 ³ 10 x 16 ³ x 2 2 = 16 ³ x 2 + 2 x 2 + 10 x 16 ³ x 2 2 = x 2 + 10 x + 16 16 ³ x 2 2 = x + 8 x + 2 16 ³ x 2 2 Notice that the values of x that make ± f DNE (namely, ± 4) are not in the domain of f . Therefore, they cannot be critical numbers (CNs) of f .
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Now, solve ± fx () = 0 for real x in the domain of f . ± = 0 x + 8 x + 2 16 ² x 2 2 = 0 x + 8 x + 2 = 0, 16 ² x 2 ³ 0; i.e., x ³ ± 4 x = ± 8 or x = ± 2 Our two real solutions are in the domain of f , so they are critical numbers (CNs). The critical numbers (CNs) of f are: ± 8 and ± . b) Find the absolute maximum point and the absolute minimum point on the graph of f in the usual xy -plane, if x is restricted to the interval ± 3, 0 ² ³ ´ µ . (4 points) f is continuous everywhere on ± except at ± 4 and 4. In particular, f is continuous on ± ² ³ ´ µ , so the Extreme Value Theorem (EVT) applies, and there exist an absolute maximum and an absolute minimum on ± ² ³ ´ µ . Our candidates for x are ± 2 , which is the critical number (CN) of f in ± , and the endpoints of the interval ± ² ³ ´ µ . x Answers a = ± 3 f ± 3 = 2 7 ² 0.286 ± f ± 2 = 1 4 = 0.25 ± 2, 0.25 is the A.Min.Pt. on ± ² ³ ´ µ . b = 0 f 0 = 5 16 = 0.3125 0, 0.3125 is the A.Max.Pt. on ± ² ³ ´ µ . Graph (optional) :
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2) State Rolle’s Theorem - the hypotheses and the conclusion. (8 points) If a function f is continuous on a closed interval a , b [] and is differentiable on the open interval a , b () , and if fa = fb , then there exists a real number c in a , b such that ± fc = 0 (i.e., the tangent line there at c , is horizontal). We assume a < b . 3) Sketch the graph of fx = x 6 + 6 x 5 ± 3 in the usual xy -plane. (30 points) • Find all critical numbers of f and label them CNs. • Find all points at critical numbers. Indicate these points on your graph. • Find all inflection points (if any) and label them IPs. Indicate these points on your graph. • Classify all points at critical numbers as local maximum points, local minimum points, or neither.
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This note was uploaded on 09/08/2011 for the course MATH 150 taught by Professor Bart during the Spring '06 term at Mesa CC.

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Math150Sols4S11 - QUIZ ON CHAPTER 4 SOLUTIONS MATH 150...

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