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Math 141 Fall 1998 No Work-No Credit 1) Exam 4 Name SS# Use Newton's method to approximate the nearest root of f ( x) = 1 − tan x x2 using x1 = 3.8 as the initial guess. List each approximation until the approximation is repeated to 6 decimal places. X1=3.8 x2=3.369190 x5=3.236755 2) If 24 is x3=3.239422 x4=3.236756 x6=x5 A softball diamond has the shape of a square with sides 60 feet long. a player is running from second base to third base at a speed of ft/sec, at what rate is her distance from home plate changing when she 20 ft from third? Want dL dt at the instant y = 20. Also given L2 = x 2 + y 2 Then then dy ( 20)( −24) dL = − 7.6 = dt = 2 2 dt L ( 20) + (60) y dy = − 24 . dt differentiate to get: 3) For f ( x ) = sin x − cos x a) critical numbers determine: b) relative maximum 2 , when x = Set f’(x) = 0 f’(x) = cos x - sin x = 0 x= 3π 4 relative minimum 7π ,− 2 4 3π 7π , 4 4 c) intervals where f(x) is increasing: 3π 7π ,2π 0, ∪ 4 4 d) intervals where f(x) is concave up: π 5π 0, ∪ ,2π 4 4 decreasing: concave down: 3π 7π , 4 4 π 5π , 4 4 4) Determine the absolute maximum and absolute minimum values of f ( x ) = x 2 − 8 ln x over the interval [1,e] . 8 2 x2 − 8 f '( x ) = 2 x − = =0 x x Critical Number(s): So x = +/- 2, but only x = 2 is in the interval of interest. So we need: f(1) = 1, f(e) = -0.61, and f(2) = -1.545 So the absolute max is 1 when x = 1 and the absolute min is -1.545 when x = 2. 5) Evaluate: Rewrite lim (1 + 3x ) 1 2x x→ 0 (1 + 3x ) = e ln( 1+ 3 x ) Then (1 + 3x ) 1 2x =e ln(1+ 3 x ) 2 x so the original limit may be written as: lim (1 + 3x ) 1 2x x→ 0 = e lim ln( 1+ 3 x ) 2 x x→ 0 . Now consider just the exponent. ln(1 + 3x ) 3 3 LR lim = , x→ 0 = x → 0 2 (1 + 3 x ) 2x 2 lim So lim (1 + 3x ) 1 2x x→ 0 = e ( 32 )