Exam2_3 6 - Version 208 – Exam 2 – Radin – (58305) 5....

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Unformatted text preview: Version 208 – Exam 2 – Radin – (58305) 5. ππ −, 33 4. limit = − π π , , π correct 3 3 Explanation: After differentiation, 6. −π , − 6 1 3 5 5. limit = − correct 2 Explanation: Dividing by x3 in the numerator and denominator we see that f ′ (x) = 2 cos x − 1 . Now f will be decreasing on an interval [a, b] if f ′ (x) < 0 on (a, b). But for x in (−π, π ), f ′ (x) = 2 cos x − 1 = 0 when x = −π/3, π/3. Thus the graph 2 x − 5 x3 1+ 3 + 2 x3 With s = 1 1 2 + 2 −5 3 x =x . 3 +2 x3 1 , therefore, x 1 + 2 x − 5 x3 x → −∞ 3 + 2 x3 lim π −π − 34 − π − π 2 4 π 4 -1 π 2 3π 4 π = -2 limit = − of f ′ on (−π, π ) shows that f ′ (x) < 0 only on π 3 , π ,π . 3 014 Consequently, f will be decreasing only on the intervals −π , − s → 0− s 3 + 2s 2 − 5 . 3s 3 + 2 Consequently, the limit exists, and -3 −π , − lim π , 3 π ,π 3 . 5 . 2 10.0 points For which function f does its graph have vertical asymptotes x = − 1 , 5 and horizon5 tal asymptote y = 5. 1. f (x) = 5x2 − 24x − 5 25x2 + 1 2. f (x) = 5 x2 + 2 x − 5 5x2 + 24x − 5 3. f (x) = 25x2 − 2x + 5 correct 5x2 − 24x − 5 5 1. limit = 2 4. f (x) = 5x2 + 24x − 5 25x2 + 1 2. limit does not exist 5. f (x) = 25x2 − 1 5x2 + 24x − 5 1 3. limit = 3 6. f (x) = 5 x2 + 1 5x2 − 24x − 5 013 10.0 points Determine if the limit lim x → −∞ 1 + 2 x − 5 x3 3 + 2 x3 exists, and if it does, find its value. ...
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This note was uploaded on 02/14/2012 for the course MATH 408 K taught by Professor Clark,c.w./hoy,r.r during the Spring '08 term at University of Texas at Austin.

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