Solutions to Math 19 Practice Final A, Winter 2011

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1 Solutions to Math 9 Practice Final A, Winter 20. (0 points) Find each of the following limits, with justification (show steps). If there is an infinite limit, then explain whether it is or. (a) (5 points) lim x 4 2x 7 x 4 If we plug in, we get 0, so we multiply by the conjugate: 0 lim x 4 2x 7 x 4 = lim = ( 2x 7)( 2x + 7) x 4 (x 4)( 2x + = 7) lim = 2x 7 x 4 (x 4)( 2x + 7) = lim x 4 2x 8 (x 4)( 2x + 7) lim = 2(x 4) x 4 (x 4)( 2x + 7) = lim 2 = 2 x 4 2x = 7 ( x + (b) (5 points) lim x ) x x x + 2 For each fraction we get an indetermination, so let us divide by x the top and bottom of each fraction: ( x + lim x ) = x x x + 2 lim x ( + x ) + 2 = = 0 x

2 Math 9, Winter 20 Solutions to Practice Final A, Winter 20 Page 2 of?? 2. (0 points) Careful: These two questions are unrelated. (a) (5 points) Find the inverse function of f(x) = x + 2x + y = x + 2x + y(2x + ) = x + 2xy + y = x + 2xy x = y x(2y ) = y x = y 2y Now we switch x and y y = x 2x So f (x) = x 2x (b) (8 points) A glider s height at time t is given by h(t) = t 2 + ln(t + ) + 5 meters. When is the glider moving upward and when is it moving downward? The velocity of the glider is given by the derivative: v(t) = h (t) = 2t + t + To find where the derivative is positive and where it is negative, let us find first the critical numbers. 2t + t + = 0 2t(t + ) + = 0 2t 2 + 2t + = 0 t = 2 ± = 2 ± 4 4 No solution. The derivative is defined for all t 0, so there are no critical numbers. Therefore the derivative has always the same sign. v(0) = > 0 Plugging in t = 0 for example, we get

3 Math 9, Winter 20 Solutions to Practice Final A, Winter 20 Page 3 of?? So the glider is always moving upward.

4 Math 9, Winter 20 Solutions to Practice Final A, Winter 20 Page 4 of?? 3. (8 points) Mark each statement below as true or false by circling T or F. No justification is necessary. T F The function f(x) = x 4 3x 3 is concave down at x = False f (x) = 4x 3 9x 2 f (x) = 2x 2 8x f ( ) = = 30 > 0 so it is concave up. T False F d dx (ln(0)) = 0 ln(0) is a number, so its derivative is zero. T F If g(x) is continuous at x = 3, then it is differentiable at x = 3 False For example, the function f(x) = x 3 is continuous at x = 3, but it is not differentiable there. f(x) T F If lim f(x) = 6 and lim g(x) = 0, then lim x 5 x 5 x 5 g(x) = ± True T F If f() > 0 and f(3) < 0, then there exists a number c between and 3 such that f(c) = 0 False This would be true if the question said that f(x) is continuous on the interval [, 3]. But we could have a function that is constant equal to 8 up to x = 2 and constant equal to 5 when x > 2, so that there would be no crossings of the x-axis. T F If f (4) > 0, then f(x) is increasing at x = 4 False We can only conclude that f(x) is concave up at x = 4. f (4) > 0, but f (4) < 0, so it is decreasing at x = 4. For example, f(x) = (x 5) 2 satisfies T F [(x 2 + ) 3 ] = 3(x 2 + ) 2x

5 Math 9, Winter 20 Solutions to Practice Final A, Winter 20 Page 5 of?? False By the chain rule, the derivative is 3(x 2 + ) 2 2x. T F If a function is decreasing, then it is one-to-one True. Increasing and decreasing functions are one-to-one.

6 Math 9, Winter 20 Solutions to Practice Final A, Winter 20 Page 6 of?? 4. (8 points) Use the definition of the derivative as a limit to find the derivative of f(x) = marks will be awarded if you get the right answer using other methods. x 2x. No f f(x + h) f(x) (x) = lim = lim h 0 h h 0 x+h 2x 2h h x 2x = = lim h 0 (x+h)( 2x) x( 2x 2h) ( 2x 2h)( 2x) h = x + h 2x 2 2xh x + 2x 2 + 2hx = lim = h 0 ( 2x 2h)( 2x)h h = lim h 0 ( 2x 2h)( 2x)h = = lim h 0 ( 2x 2h)( 2x) = ( 2x) 2

7 Math 9, Winter 20 Solutions to Practice Final A, Winter 20 Page 7 of?? 5. (24 points) Consider the function f(x) = 3 ln(x + 5) x (a) (2 points) Find the domain of f(x). x needs to be greater than 5 in order for ln(x + 5) to be defined, so the domain is ( 5, ) (b) (5 points) Where is f(x) increasing? And decreasing? Find the x-coordinates of all the local maxima and minima of f(x). f (x) = 3 x + 5 f (x) = 0 when 3 = x + 5 x = 2 Our diagram only has the point 2 f ( 3) = 3 2 > 0 f (0) = 3 5 < 0 So f(x) is increasing on ( 5, 2) and it is decreasing on ( 2, ) So there is a local maximum at x = 2. Because it will be useful later, we compute f( 2) = 3 ln(3) + 2 (c) (5 points) Where is f(x) concave up? And concave down? Find the x-coordinates of all the inflection points of f(x). f (x) = 3 (x + 5) 2 This can never be zero, in fact the numerator is always negative and the denominator is always positive. Therefore f(x) is concave down on ( 5, ) and there are no inflection points.

8 Math 9, Winter 20 Solutions to Practice Final A, Winter 20 Page 8 of?? (d) (5 points) Find all the asymptotes of f(x). (You can and should use that lim f(x) = ). x lim f(x) =, so there is no horizontal asymptote on the right side. x It does not make sense to do the limit at, because the domain is ( 5, ) So there are no horizontal asymptotes. x = 5 is a candidate for vertical asymptote. lim f(x) =, so x = 5 is a vertical asymptote. x 5 + There are no more vertical asymptotes because the function is continuous everywhere else. (e) ( point) Find the y-intercept of f(x) (if possible). Note: Do not try to find the x-intercepts. f(0) = 3 ln(5) The point is (0, 3 ln(5))

9 Math 9, Winter 20 Solutions to Practice Final A, Winter 20 Page 9 of?? (f) (6 points) Sketch the graph of f(x) in the axes provided below.

10 Math 9, Winter 20 Solutions to Practice Final A, Winter 20 Page 0 of?? 6. (4 points) Find the derivatives of the following functions using any method you like (that works). You do not need to simplify your answers. 3 (a) (4 points) f(x) = x 2 + f(x) = 2 ( 3 x 2 + ) /2 3 (x2 + ) 2/3 2x (b) (4 points) g(x) = cos(2 3 x ) g (x) = sin(2 3 x ) 2 3 x ln(3) ( ) (c) (6 points) Z(x) = (x 2 ) /x We need to use logarithmic differentiation ln(z(x)) = ln [ (x 2 ) /x] ln(z(x)) = x ln(x2 ) ln(z(x)) = ln(x2 ) x Z (x) Z(x) = 2x x 2 x ln(x2 ) x 2 ( 2x2 Z (x) = (x 2 ) /x x 2 ) ln(x2 ) x 2

11 Math 9, Winter 20 Solutions to Practice Final A, Winter 20 Page of?? 7. ( points) Linear approximation. (a) (6 points) Using linear approximation, estimate the value of We compare this with the value 2 + = 2 So we use f(x) = x 2, a = and x = x f (x) = 2x (x 2 + x) 2 f () = 3 4 Now linear approximation says f(x) f(a) + f (a)(x a) (0.9) = = (0.9) (b) (5 points) Is your estimate smaller or greater than the real value? Justify your answer. f (x) = ( 2)(x2 + x) 2 ( 2x ) 2(x 2 + x)(2x + ) (x 2 + x) 4 f () = = 28 > 0, so the function is concave up there. 6 Therefore, the estimate is smaller than the real value.

12 Math 9, Winter 20 Solutions to Practice Final A, Winter 20 Page 2 of?? 8. (8 points) A hemispherical bowl of radius 0 inches is filled with water to a depth of x inches. The volume of water in the bowl (in cubic inches) is given by the known formula V = π 3 (30x2 x 3 ). Suppose that you measure the depth of water in the bowl and you find it to be 5 inches with a maximum possible measurement error of 6 inches. Using differentials, estimate the maximum error in the calculated volume in the bowl. Recall dv = V (a) dx The question is giving us a = 5 (the original measure) and dx = 6 V (x) = π 3 (60x 3x2 ) V (5) = π 225π (300 75) = 3 3 So dv = 75π 6 = 75π 6 = 75π Therefore the maximum error in the volume is 75π 6 cubic inches. (the error)

13 Math 9, Winter 20 Solutions to Practice Final A, Winter 20 Page 3 of?? 9. (0 points) Find an equation of the tangent line to the curve x 3 + xy 2 + y 3 = 3 at the point (, 2) on the curve. Simplify your answer. We use implicit differentiation 3x 2 + y 2 + 2xyy + 3y 2 y = 0 y (2xy + 3y 2 ) = 3x 2 y 2 y = 3x2 y 2 2xy + 3y 2 y (, 2) = 7 6 So the equation of the tangent line is y 2 = 7 (x ) 6

14 Math 9, Winter 20 Solutions to Practice Final A, Winter 20 Page 4 of?? 0. (2 points) A farmer has 200 ft of fence to be used in constructing three sides of a rectangular pen (as in place to keep cows, not a writing instrument). An existing long straight wall is to be used for the fourth side of the pen. What dimensions will maximize the area of the pen? Justify your answer. We want to maximize the area A = xy We also know that x + 2y = 200 x = 200 2y Therefore A(y) = 200y 2y 2 The domain for this function is [0, 00] (the second number found from the fact that x 0) This function is continuous on [0, 00], so we can use the closed interval method. A (y) = 200 4y A (y) = 0 when y = 50 There are no more critical numbers because the function is differentiable everywhere. A(0) = 0 ft 2 A(00) = 0 ft 2 A(50) = = 5000 ft 2 So the side parallel to the wall should be 00 ft long and the other two sides should be 50 ft long.

15 Math 9, Winter 20 Solutions to Practice Final A, Winter 20 Page 5 of??. (2 points) The equation x 3 4x + = 0 has three different solutions. Find three intervals of length (that is, (0, ), ( 0, 9), etc.) where they are located. Justify your answer. We use the intermediate value theorem with f(x) = x 3 4x + and M = 0. We can do this because f(x) is continuous everywhere. f(0) = > 0 f() = 2 < 0 By the intermediate value theorem, there is c in (0, ) such that f(c) = 0 f(2) = > 0 By the intermediate value theorem, there is d in (, 2) such that f(d) = 0 f( ) = 4 > 0 f( 2) = > 0 f( 3) = 4 < 0 By the intermediate value theorem, there is a in ( 3, 2) such that f(a) = 0 They are located in the intervals ( 3, 2), (0, ) and (, 2).

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