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 xintercepts 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 03 1  3x2 3 x3 = lim = . 6 2 = x+ 2x3  6x + 2 x+ 2  2 + 3 200 2 x x lim 1 3 1  3x2 03 3 x3 = lim = . 6 2 = 3  6x + 2 x 2x x 2  2 + 3 200 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 pointslope equation y  12 = 48(x  1/2)
or slopeintercept equation y = 48x + 36.
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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.
 Fall '07
 Vorel
 Math, Calculus

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