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Sample Exam 3 Fall 2011

Sample Exam 3 Fall 2011 - Math 2123.2 — Fall 2011 —...

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Unformatted text preview: Math 2123.2 — Fall 2011 — Sample Exam 2 Solution To receive credit, you must show your work. 1. [15 points] Complete the values in the table and use them to guess 2 6—1 MEL")— m—>O [1: to as much accuracy as the table allows. (2+x)x~1 x as The table allows us to guess 0.01 0.70057 m lim Old—1 z 0.69 0.001 0.69388 9590 1; 0 ? or even —0.001 0.69240 ac _ 1 0.6924 < lim ————<2 + m) < 0.6939. —0.01 0.68577 :r—>0 r 2. [12 points] Find an equation of the tangent line to the curve y = x5 — 5232 at the point Where :6 = —1. When a: = —1, y‘w:_1 = (—1)5 — 5(—1)2 = —6 and so the point is (—1,—6). The derivative of y is y’ = 5:54 — 5(2m) = 5.734 — 10.6 and so the slope of the tangent line is y’lxfl = 5(—1)4 —10(—1) = 15. By the point-slope form, an equation of the tangent line is y——6=15(m——1) y+6=15(a:+1). This may also be written as y : 15:13 + 9. 5. [15 points] Use appropriate derivative rules to calculate the derivative of SE2 563—1' y: We use the quotient rule with the Choices u = :32 and v = x3 — 1. Then we have du d1} 2% = 2m and do; = 3x2 and so 91g _ d<§> d3: .1 dx vii—2: - ”23—: 1)2 (333 - 1mm) - (9658332) _ ——(i:”——1)2— _ 2x4 — 2:13 — 3x4 _ W __ 7:134 ~ 23: 6. [15 points] Use appropriate derivative rules to calculate the derivative of 5 = 1 _ _ y ( x) We use the generalized power rule with the choices to = 1 — i = 1 — 33—1 and n = 5. d Then we have _u_ z 0 — —x“2 = I” and so dzc dy _ (“a”) dzv — dm du n—1_ — nu dac 1 4 72 = 1 _ _ 5( 3:) x 5 1 4 = 7(1 — —) 7. Let y = 3:54 — 4:113. (a) [7 points] Find the critical values of x. (b) [6 points] Make a sign chart for the derivative. ) (c [6 points] For each critical point, state whether it is a maximum, a minimum, or neither. (a) We have 3/ = 3(4903) — 4(3132) : 1231:3 — 12232. To find the critical :5 values we have to solve the equation y’ = O for ac: 12$3—12$2=0 12$2(9:—1)=0 m2=00r(m—1):O x=Oorzr=1. The critical :1: values are :c = 0 and :c = 1. OD) cc \ Test Value 3/ = 12:103 — 12x2 Sign (—oo,0) —1 #24 — 0 (0,1) 0.5 —1.5 — 1 (1,00) 2 48 + (0) Since the derivative is negative before and after the critical point that has $ = 0, this one is neither a maximum nor a minimum. Since the derivative is negative before and positive after the critical point that has x = 1, this one is a (relative) minimum. 1 3. [12 points] Use the four—step process to find the derivative of y = E 1 w + Am 1 1 1 ac + A55 — _Z:_y $+A92 — % A51: A30 y+Ay= Ay— ll H H H and so d—y— lim Ay lim '1 — #1 ‘1 dg: _ Ago—>0 A3: Ana—>0 35(30 + Ax) Mm + 0) m2 ' Note: You know that the derivative of y = l = :E‘1 is going to turn out to be 1‘ y’ = —x#2 = —fi because of stuff we have done since we discussed the four-step process. This allows you to know whether or not you have made an error, but doesn’t necessarily help to find it if you have. 4. [12 points] The function y = Zfi — :1: has a critical point at (1,1). What conclusion results from applying the second derivative test to this critical point? We start by finding the second derivative: 1 y/ : 2(ECE71/2) _ 1 : x—l/Q _ 1 1 I/ = __ —3/2 y 253 . At the critical point (1,1)7 Lt : l and 1 1 // _ __ —3/2 _ _ By the second derivative test, this critical point is a (relative) maximum. ...
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