y »x 2» x 1. Find x if a = be 2x, lna = 7, and ln b = 3 HAL ln 7 HBL 2 HCL 7 HDL 4 HEL e 3

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1 . Find if a = be, lna =, and ln b = HAL ln HBL HCL HDL HEL e a = be and taing the natural log of both sides, we have ln a = ln b + ln e ln a = ln b + = + = B. lim b b b = HAL b HBL b HCL b HDL b HEL b lim b lim b b b b b = lim b I b MI b + M b = lim b = b A I b + M so. Find f HL if f HL = +»» HAL HBL HCL HDL HEL Does not eist Consider the graph f f HL, +»» This graph has a cner at =, so the derivative does not eist at =, E

2 . Consider the function below. From the three points given, where is f HL < f HL?. fhl b c a HAL a onl HBL b onl HCL c onl HDL a and b HEL a and c F the point at a, f HL < 0 and f HL > 0 so a onl A. Find the equations of all hizontal and vertical asmptotes f the curve = H 9L HAL =, =, and = HBL = 0, =, = 0, and = HCL = 0, =, and = HDL =, =, = 0, and = HEL =, and = Consider lim H 9L = 0 and lim H 9L = 0 and at = 0 we have a removable discontinuit. so there are asmptotes at = 0, =, and = C. If a b f HL d = 8, a =, f is continuous, and the average value of f bd is, then b = HAL 0 HBL HCL HDL HEL f av = b b a f HL d = a H8L b = so b b = D. i d dt Ht + t L { z dt = HAL HBL 0 HCL HDL HEL i d dt Ht + t L { z dt = Ht + L dt = A t + t D HH8 + 8L H + LL = = 0 B

3 8. Find Hf HLLD at = if f HL and g HL are defined b the following graphs : ghl fhl HAL 0 HBL HCL HDL HEL Does not eist Hf HLLD = g Hf HLL f HL so at =, g Hf HLL f HL = g HL f HL = i { z HL = B 9. Find the intercept f the tangent line to the curve of = cos H L, at = HAL HBL HCL HDL HEL 8 d = sin H L so at = d, d d 0 = i j { z = + 0. lim H + L» +» = = and = 0 so so H0L = D HAL HBL HCL HDL HEL Does not eist Consider that lim H + L» +» the denominat is positive, so lim D = because the numerat is negative and H + L» +» = H L H L =

4 . Given that f HL =, f HL =, g HL =, and g HL =, then find D i f HL z at = g HL { HAL 8 HBL 8 HCL 8 HDL 8 HEL D i f HL z = g HL f HL f HL g HL and at =, g HL { Hg HLL g HL f HL f HL g HL HL H L = = Hg HLL = 8 A. If the graph of f HL = + then the value of is has a point of inflection at =, HAL HBL HCL 0 HDL HEL f HL = and f HL = + and we now that f H L = 0 DNE so f H L = + = 0 so = and = E. lim h 0 tan H H + hll tan H L h HAL 0 HBL cot H L HCL sec H L HDL sec HL HEL Does not eist = f HL = lim h 0 so lim h 0 f H + hl f HL and h tan H H + hll tan H L h D Htan H LL = lim h 0 tan H H + hll tan HL h = sec H L D. Find d if = d HAL + HBL HCL HDL HEL None of these = Implicitl differentiating, we have i d z = 0 d { = i d z d { I M = B d d = and = H L H d L I M = d H L =. f HL = 0 H L has a local maimum at HAL = HBL = HCL = and = HDL = HEL None of these

5 f HL = 0 H + L = so f HL = = 0 H 8 + L = 0 H L H L so there are critical points at =, and so the sign line... + f HL + and there is a local maimum at =, A. Given the graph of f shown below, suppose that Newton s Method is used to approimate the zero between and. If we choose =, what are the most plausible values f and? = fhl HAL =.9 and =.0 HBL =. and =. HCL =. and =. HDL =. and =. HEL =. and = 8. follow the tangent lines to the ais, and the most plausible pair are =. and =. D. Find the area bounded b the curves = and = 0 HAL HBL HCL HDL 0 HEL 9 First, find the points of intersection, = + = 0 H L H + L = 0 so =, and A = HH + L L d A = A + D = i j i j + 8 z i { + z z = + {{ 9 = E 8. Find f H0L if f HL = ln HH L H + L L HAL HBL HCL 0 HDL HEL

6 f HL = ln HH L H + L L = lnh L + ln H + L and f HL = HL + + HL = 0 + so + f H0L = 0 + = 0 + = A 9. Find the area of the rectangle of maimum area having two vertices on the ais and two vertices above the ais on the graph of = = HAL HBL 9 HCL 8 HDL HEL 8 Area = H L = 8 so A = 8 so there is a critical value = and the maimum area is where 8 = = A = i j { z i j z = { i 8 z = { 9 B

7 0. Consider h HL in the graph below. If HL = h HL Hln L, find HeL Hwhere e.8l hhl HAL HBL + e HCL HDL e HEL + e HL = h HL Hln L + h HL i z and so HeL = h HeL Hln el + h HeL i { z e { HeL = HL HL + e i z HeL = C e {. ln 0 + d = HAL ln HBL ln HCL Hln L Hln L HDL ln ln HEL ln ln ln 0 + d, u = +, du = Hln L d ln du = d so u du = A ln u D = ln ln = ln A. Suppose that f HL = ln, and f HL denotes the inverse of f. Then f HL d = HAL e + C HBL e + C HCL + C HDL + C HEL e + C First, find the inverse function, so = ln e = e ln H L = e = e so f HL d = e d = e + C B

8 . If the curve below represents the complete polnomial graph of f HL, then the graph of f HL will cross the ais in how man points? 8 fhl HAL 0 HBL HCL HDL HEL Loo f hizontal tangents. It loos lie there are distinct hizontal tangents, and so the answer is E. ln d = HAL Hln L HDL Hln L HBL Hln L HEL Hln L + HCL Hln L ln d using Integration B Parts, u = ln dv = d ln d = ln ln d = A ln D du = d v = so d = ln so B = i jh ln L i j0 { z = ln. Which of the following statements is ê are true? I. If f is continuous everwhere, then f is differentiable everwhere. II. If f is differentiable everwhere, then f is continuous everwhere. III. If f is continuous and f HL > f ever D, then f HL d > 9 HAL I onl HBL II onl HCL III onl HDL I and III onl HEL II and III onl Continuit is a consequence of differentiabilit, so II must be true, and III is a consequence of continuit, so the answer is E

9 . Find the values f where the graph of = is concave downward HAL No values of HBL < HCL > HDL < HEL > = so = H L H L = and = H L H L H L = H L so the second derivative is negative when > E Consider the following curve of f HL H not f HLL f problems and 8 c d e b f HL a. Of the five points shown, the least value f f HL occurs at : HAL a HBL b HCL c HDL d HEL e We are looing f the minimum accumulated area, and this occurs at c, C 8. Of the five points shown, the greatest value f f HL occurs at : HAL a HBL b HCL c HDL d HEL e We are looing f the maimum slope on the curve, and this occurs at b, B 9. Find f HL if f HL = ln Htan H LL HAL HBL HCL HDL HEL 8 f HL = i j tan H L + { z H L so f HL = i z HL = { D

10 0. + sin d = cos HAL HDL + HBL HCL + HEL + + sin d = cos sec d + tan cos d + sin cos d = sec tan d = A tan D tan + sec sec + A + A sec D = +