31_new - HW01 gilbert (55035) 1 This print-out should have...

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HW01 – gilbert – (55035) 1 This print-out should have 22 questions. Multiple-choice questions may continue on D A the next column or page find all choices before answering. Welcome to Quest. It’s a good idea to download this assignment as a PDF B E file and print it o f . You can then bring the assignment to class and discussion group. JEG C 001 10.0 points Determine x lim 0 ë x 1 x 2 x 1o . 1. limit = 1 correct 2. limit = 1 2 3. limit = 1 3 4. limit = 1 3 5. limit = 1 2 6. limit = −1 Explanation: After simplification we see that 1 1 = 1 − (1 − x) = 1 x − x 2 x x(1 − x) 1 − x for all x 6= 0. Thus limit = lim 1 = 1 . 002 10.0 points Consider the slope of the given curve at the five points shown. List the five slopes in decreasing order. 1. A, D, B, C, E 2. E, A, C, D, B correct 3. B, E, D, A, C 4. B, D, C, A, E 5. A, E, D, C, B 6. E, A, B, D, C Explanation: The order will be the one from most positive slope to most negative slope. Inspection of the graph shows that this is E, A, C, D, B . 003 10.0 points The derivative, f , of f has graph a b c graph of f
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HW01 – gilbert – (55035) 2 Use it to locate the critical point(s) x 0 at which f has a local maximum? 3. local min at x = −6 1. x 0 = c , a 2. x 0 = b , c 3. x 0 = c 4. x 0 = a 5. x 0 = b correct 6. none of a , b , c 7. x 0 = a , b 8. x 0 = a , b , c Explanation: Since the graph of f (x) has no ‘holes’, the only critical points of f occur at the x-intercepts of the graph of f , i.e. , at x 0 = a, b, and c. Now by the first derivative test, f will have (i) a local maximum at x 0 if f (x) changes from positive to negative as x passes through x 0 ; (ii) a local minimum at x 0 if f (x) changes from negative to positive as x passes through x 0 . Consequently, by looking at the sign of f (x) near each of x 0 = a, b, and c we see that f has a local maximum only at x 0 = b . 004 10.0 points Which one of the following properties does 4. local max at x = 6 correct 5. local max at x = 2 6. local max at x = −6 Explanation: By the Quotient rule, f (x) = x 2 + 12 2x(x 2) (x 2 + 12) 2 12 + 4x − x 2 = (x 2 + 12) 2 . The critical points of f occur when f (x) = 0, i.e. , at the solutions of f (x) = (2 + x)(6 − x) = 0 . (x 2 + 12) 2 Thus the critical points of f are x = −2 and x = 6. To classify these critical points we use the First Derivative Test. But the sign of f de-pends only on the numerator, so it is enough, therefore, to look only at a sign chart for (2 + x)(6 − x): + −2 6 From this it follows that f is decreasing on (−∞, −2), increasing on (−2, 6), and de- creasing on (6, ∞). Consequently, f has a local maximum at x = 6 . keywords: local maximum, local minimum,
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This note was uploaded on 04/04/2012 for the course M 408d taught by Professor Sadler during the Spring '07 term at University of Texas at Austin.

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31_new - HW01 gilbert (55035) 1 This print-out should have...

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