AP Calc. BC 1998 MC # 91 and # 92 HELP??!!

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<p>t |0|2|4|6|
a(t)|5|2|8|3|
91) the date for the acceleration a(t) of a car from 0 to 6 seconds are given in the table above. If the velocity at t=0 is 11 feet per second, the approximate value of the velocity at t=6, computed using a left-hand reimann sum with three subintervals of equal length, is </p>

<p>the anwer is 41ft/sec... but I keep getting 30ft/sec....</p>

<p>92) Let f be the function given by f(x) = x^2-2x+3. the tangent line to the graph of f at x=2 is used to approximate values of f(x). Which of the following is the greatest value of x for which the error resulting from this tangent line approximation is less than 0.5? </p>

<p>the answer is 2.7, and this is supposedly a linear approx. problem, but I never really understood linear approx. :p... </p>

<p>Thanks for any help.</p>

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<p>hmmm… i just did 91 and i’m getting 30ft/sec too… i dunno maybe the key is wrong??</p>

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<h1>91: You’re forgetting that the left-hand Reimann sum accumulates velocity, so there are 30 ft./sec. accumulated, but at time 0, you started out with a velocity of 11 ft./sec. already, so at 6 seconds, you have 11 + 30 = 41 ft./sec. as the velocity.</h1>

<h1>92: The tangent line at x = 2 for this particular equation is y - 3 = 2(x - 2) or y = 2x - 1. Since f is concave up for all x, the tangent line will always underapproximate the value of the curve. Thus, we want to find where (x^2 - 2x + 3) - (2x - 1) < 0.5.</h1>

<p>This is equivalent to x^2 - 4x + 4 < 0.5, which can be easily solved using the graphing utility on the calculator, since this is calculator active.</p>