hw1_sol_graded-1

# hw1_sol_graded-1 - Solutions explained for graded problems...

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Unformatted text preview: Solutions explained for graded problems in HW1, math118, 2008 1.2.32 (a) H(t) = -16t2 + 160t = 16t(-t + 10). parabola with x-intercepts at the points it will have a maximum value H(5), and Therefore H(t) is graphed as a (0, 0) and (10, 0), since -16 < 0 (5, H(5) = 400) is its vertex. (b) H(t) will be zero at times hitting the ground and the answer is feet. 1.5.32 t = 0 and t = 10 seconds; t = 10 corresponds to t = 0 corresponds to leaving the ground. Therefore t = 5 when the object as height H(5) = 400 t = 10 seconds. (c) The maximum height occurs at 1 -3 0-3 1 - 3x2 3 x3 = lim =- . 6 2 = x+ 2x3 - 6x + 2 x+ 2 - 2 + 3 2-0-0 2 x x lim 1 -3 1 - 3x2 0-3 3 x3 = lim =- . 6 2 = 3 - 6x + 2 x- 2x x- 2 - 2 + 3 2-0-0 2 x x lim 2.1.16 f (x) = 3 with domain x2 x = 0. f (x) = lim f (x + h) - f (x) =? h0 h Since, f (x + h) - f (x) = h = We conclude, 3 (x+h)2 - 3 x2 h = 3x2 -3(x+h)2 (x+h)2 x2 h = 3x2 - 3(x2 + 2xh + h2 ) h(x + h)2 x2 -6xh - h2 -6x - h 3x2 - 3x2 - 6xh - h2 = = . h(x + h)2 x2 h(x + h)2 x2 (x + h)2 x2 f (x) = lim for -6x - h -6x - 0 -6x -6 = = 2 2 = 3 2 x2 2 x2 h0 (x + h) (x + 0) (x )x x c = 1/2: f (1/2) = 12, -6 = -6(23 ) = -48, (1/2)3 x = 0. For the tangent line at m = f (1/2) = therefore the line has point-slope equation y - 12 = -48(x - 1/2) or slope-intercept equation y = -48x + 36. End. ...
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## This note was uploaded on 10/24/2008 for the course MATH 118x taught by Professor Vorel during the Fall '07 term at USC.

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