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PracticeExam3Solutions

# PracticeExam3Solutions - MATH 191 Sections 1 and 3 Calculus...

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MATH 191, Sections 1 and 3 Calculus I Fall 2007 Practice Exam 3 This practice exam, like the actual exam will be, is worth a total of 100 points, and point values for each question are given below. It is similar in length, format, and tested content to the actual exam, but in the interest in saving paper I have omitted space for you to work your solutions; you should work these out on your own paper. Answer every question fully and clearly. Please show your work where appropriate, and circle or box your final answer. Please let me know if you have any questions! 1. (10 points) True or false?: if f ( c ) = 0, then f has a local extremum at x = c . (If this statement is true, explain why it is true. If it’s false, give an example of a function that shows that it’s false.) False. Although it’s possible for a function to have an extremum at such a point x = c , it’s not necessary. The example f ( x ) = x 3 , at the point x = 0, is the one you probably have in mind! 2. (10 points total; 5 points each) Suppose that f ( x ) = 5 - 4 x + 2 x 2 - 2 x 3 on the interval [ - 2 , 2]. (a) Explain why the Mean Value Theorem applies to this function on the given interval. As a polynomial, the function f is continuous and differentiable everywhere, and therefore is certainly so on the interval [ - 2 , 2]. (b) Find a value of c lying on ( - 2 , 2) satisfying the conclusion of the Mean Value Theorem. (You may use without computation the fact that the slope of the secant line between ( - 2 , f ( - 2)) and (2 , f (2)) is - 12. Differentiating, we obtain f ( x ) = - 6 x 2 + 4 x - 4. We need to find a value of c such that f ( c ) = - 12, so we compute: f ( c ) = - 12 ⇔ - 6 c 2 + 4 c - 4 + 12 = 0 ⇔ - 6 c 2 + 4 c + 8 = 0 . Using the quadratic formula here yields c = - 4 ± 16 - 4( - 6)(8) - 12 = 1 ± 13 3 . Both of these values, it turns out, are on the desired interval. 3. (12 points) On a set of axes, draw the graph of a single function f satisfying the following properties simultaneously: (a) f ( x ) > 0 for all x in [ - 3 , - 1) and ( - 1 , 2], (b) f is concave up on [ - 3 , - 1) and (1 , 2], and concave down on ( - 1 , 1), and (c) f ( - 1) does not exist.

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PracticeExam3Solutions - MATH 191 Sections 1 and 3 Calculus...

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