Analytic Geometry And Calculus I Exam 1 Practice Problems .

Analytic Geometry and Calculus IExam 1 Practice Problems Solutions 2/19/7Question 1Write the following as an integer: 2log4 (9) log2 (5)We have: 2log4 (9) log 2 (5) 2log4 (9) 2log2 (5) 5 2log4 (9) 1 log4 (9) 5 42 5 4log4 (9)1 5(9) 2 12 5(3) 15.Here we used the properties:ab c ab ac ,aloga (b) b,(ab )c abc (ac )b .

Question 2Eliminate the parameter t to find a Cartesian equation of the curve givenparametrically by the relations: x 1 3t, y 2 t2 and sketch the curve.Also sketch the part of the curve for which t 0.We have:x 1 3t,x 1 3t,1t (x 1),3y 2 t2 21 2 (x 1)31 2 (x 1)2 ,99y 18 (x 1)2 18 (x2 2x 1) 17 2x x2 .So the curve has the equation x2 2x 9y 17.This is the equation of a concave down parabola.1The equation y 2 (x 1)2 shows that the parabola has vertex at (1, 2)9and is symmetrical around the vertical line x 1.The point (1, 2) gives the global maximum for y.As t varies x ranges once through all possible real values, so the wholeparabola is traced. 117The parabola intersects the y-axis at t , so at the point 0,.39 The parabola intersects the x-axis at t 2, so at the points 1 3 2, 0 .When t 0, we are at the vertex (1, 2).As t increases from zero, the whole right half of the parabola is traced out.2

Question 3Find the following limits: lim x 2 x 2 x2 4When x 2, we have x 2 0, so x 2 (x 2).So we have:lim x 2 (x 2) x 2 lim 2x 4 x 2 (x 2)(x 2) lim x 2 lim x 2 11 1 lim .x 2 x 2 2 24 x 2 x2 4When x 2, we have x 2 0, so x 2 x 2.So we have:lim x 2 x 2 x 2 lim 2x 4 x 2 (x 2)(x 2) lim x 2 limx 2 x 2 x2 4Since lim x 2111 lim .x 2 x 2 2 24 x 2 x 2 x 2 6 lim 2, the limit lim 2does not exist.2x 2 x 4x 4 x 2 x 43

Question 4Suppose that:f (1) 2,f (3) 4,g(1) 3,g(2) 1,f 0 (1) 4,f 0 (3) 2,g 0 (1) 5,g 0 (2) 7.f (x).g(x)Find the equations of the tangent and normal lines to the curves y h(x),y k(x) and y m(x) at the points with x 1.Let h(x) f (g(x)), k(x) f (x)g(x) and m(x) We have h(1) f (g(1)) f (3) 4 and h0 (x) f 0 (g(x))g 0 (x), soh0 (1) f 0 (g(1))g 0(1) f 0 (3)(5) 2(5) 10.So the tangent line to y h(x) has slope 10 and goes through thepoint (1, 4), so has the equation:y 4 10(x 1) 10x 10,Also the normal line to y h(x) has slopepoint (1, 4), so has the equation:y 4 1(x 1),10y 10x 14.1and goes through the1010y 40 x 1, x 10y 39. We have k(1) f (1)g(1) 2(3) 6 and k 0 (x) f 0 (x)g(x) f (x)g 0 (x),so k 0 (1) f 0 (1))g(1) f (1)g 0 (1) 4(3) 2(5) 12 10 22.So the tangent line to y k(x) has slope 22 and goes through the point(1, 6), so has the equation:y 6 22(x 1) 22x 22, y 22x 16.Also the normal line to y k(x) has slope point (1, 6), so has the equation:y 6 1(x 1),221and goes through the2222y 132 (x 1) x 1,4x 22y 133.

f (1)2f 0 (x)g(x) f (x)g 0 (x) and m0 (x) ,g(1)3g 2 (x)4(3) 2(5)12 102f 0 (1)g(1) f (1)g 0 (1)0 .so m (1) 22g (1)3992So the tangent line to y m(x) has slope and goes through the point9 21,, so has the equation:3 We have m(1) y 22 (x 1), 9y 6 2(x 1) 2x 2,392x 9y 4.9Also the normal line to y m(x) has slope and goes through the2 2, so has the equation:point 1,3y 92 (x 1),326y 4 27(x 1) 27x 27,527x 6y 40.

Question 5 Let f (x) (x 1)2x 1 x2 .f (a h) f (a)to find f 0 (1).h 0h We have f (1) (1 1)21 1 12 0(2) 2) 0.Use the definition f 0 (a) limThen we get:f (1 h) f (1)f (1 h) 0 limh 0h 0hhp(1 h 1)21 h 1 (1 h)2f (1 h) lim limh 0h 0hhp1 h1 (1 h)2(h)2 limh 0hp1 h lim 21 (1 h)2h 0p 21 0 1 (1 0)2 21 1 12 2 1 1 2 2.f 0 (1) lim6

Question 6x2. Let f (x) xe 2 . Find f 0 (x) and f 00 (x).f 0 (x) 1e x22x2x2 xe 2 ( 2x)2x2 e 2 x 2 e 2x2 (1 x2 )e 2 ,f 00 (x) 2xe 2xe x22x22x2 (1 x2 )e 2 ( 2x)2x2 x(1 x2 )e 2x2 e 2 ( 2x x x3 )x2 e 2 (x3 3x)x2 x(x2 3)e 2 . Find the intervals where f is increasing and the intervals where f isdecreasing.When x 1, we have f 0 0, so f is decreasing on [1, ).When x 1, we have f 0 0, so f is decreasing on ( , 1].When 1 x 1, we have f 0 0, so f is increasing on [ 1, 1]. Find intervals where f is concave up and concave down respectively. When x 3, we have f 00 0, so f is concave up on [ 3, ). 00When 0 0, so f is concave down on [0, 3]. x 3, we have f 00When 3 x 0, we have f 0, so f is concave up on [ 3, 0].When x 3, we have f 00 0, so f is concave down on ( , 3].7

Sketch the curve y f (x).As x , the curve approaches the x-axis from below.As x increases, the curve decreases and is initiallydown, switch concave 3ing to concave up at the inflection point ( 3, 3e 2 ).1It then decreases further to its absolute minimum at ( 1, e 2 ).Then it increases, crossing the axes at the origin, where it inflects toconcave down.1It increases further to its absolute maximum at the point (1, e 2 ).Thenswitching to concave up at the inflection point it decreases,3( 3, 3e 2 ).Then it decreases further, approaching the y-axis from above as x . Give the domain and range of the function f .The domain is all the real numbers.The range is from its absolute minimum to its absolute maximum, so11is the closed interval [ e 2 , e 2 ]. Is f one-to-one?Explain your answer.f is not one-to-one because it has an absolute maximum and in theneighorhood of the absolute maximum it decreases on both sides.1Alternatively, we may say that y takes all values in the interval [0, e 2 ] asx ranges over the interval [0, 1], by the intermediate value theorem and1all values in the interval (0, e 2 ] over the interval [1, ), again by the1intermediate value theorem, so takes all values in the interval (0, e 2 )exactly twice, once in the interval (0, 1) and once in the interval (1, ).In particular f is not one-to-one.8

Analytic Geometry and Calculus I Exam 1 Practice Problems Solutions 2/19/7 Question 1 Write the following as an integer: 2log4(9) log2(5) We have: 2log4(9) log2(5) 2log4(9) 2log2(5) 5 2log4(9) 5 41 2 log 4(9) 5 4log4(9) 1 2 5(9)12 5(3) 15: Here we used the properties: a b c a ac; aloga(b) b; (ab) c abc (a )b: Question 2 Eliminate the parameter t to nd a Cartesian equation .