2015 Math Camp Calculus Exam Solution

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1 015 Math Camp Calculus Exam Solution Problem 1: x = x x = 9 = 3 1. lim We also accepted ±3, even though it is not according to the prevailing convention 1. x x 4 x+4 =. lim = 4 0 = 4 0 = We also accepted the limit does not exist ; points were subtracted for answering + ; no points were given for answering not defined. 3. lim x 1 1 x 1 x Problem : 1. f(x) = x4 1 x = lim (1 x)(1+x) = lim x 1 1 x 1 x 1 (1+x) = = 1 f (x) = (4x3 )x (x 4 1) = 8x4 x 4 + = 6x4 + = 3x4 +1 (x) 4x 4x x. g(x) = (x + 1)e x g (x) = e x + e x 1 x (x + 1) = e x ( + x + 1 x ) 3. h(x) = ( 1 x + lnx) 1 3 h (x) = 1 3 ( 1 x + lnx) 4 3 ( 1 x + 1 x ) = 1 3 ( 1 x + lnx) 4 3 (x + 1 x ) Problem 3: 1 When you solve the equation x = 9, you are trying to find all possible values that might have been squared to get 9. But, when you are just simplifying the expression 9, the ONLY correct answer is 3. 1

2 1. Firstly, consider the interval 4 < x < 1. On this interval we have f (x) = > 0, so f(x) is increasing on ( 4, 1). For the endpoints, since f(x) is continuous on [ 4, 4], then although f (x) is not defined for x = 4, f(x) is increasing on [ 4, 1). Then consider the interval 1 x < 4. On this interval we have f (x) = 4 x. When the function is increasing we have f (x) > 0. i.e. 4 x > 0, which gives x <. So f(x) is increasing on [1, ). Again it is appropriate that we include the endpoint, i.e. f(x) is increasing on [1, ]. Since f(x) is continuous, we can combine the two intervals on which f(x) is increasing to get [ 4, ]. When f (x) < 0, i.e. x >, f(x) is decreasing. Including the endpoints we have f(x) is decreasing on [, 4]. So f(x) is increasing on [ 4, ] and decreasing on [, 4].. From (a) we know that f(x) is increasing on [ 4, ] and decreasing on [, 4]. To identify all local maxima and minima of f on [ 4, 4], we need to identify all the critical points on the interval. Firstly, f(x) has two type I critical points, i.e. endpoints at x = 4 and x = 4. Since f(x) is increasing on [ 4, ] and decreasing on [, 4], we know that both points are local minima. We do not need to recover the original function or to evaluate the function at x = 4 and x = 4 unless we are asked to identify the global minimum. Secondly, f(x) has one type II critical point where f (x) = 0, which is given in (a) as x =. Since f (x) is positive when x < and negative when x >, the first derivative test identifies x = as a local maximum, in fact the only local maximum. Lastly, f(x) has two type III critical points where f (x) does not exist, but they are exactly the endpoints. So there is no need for further work. So f(x) has one local maximum at x = and two local minima at x = 4 and x = 4. Problem 4: To find all global and local maxima and minima of f(x) = 3x 4 4x 3 6x + 1x + 7, we identify all its critical points. Firstly, f(x) is defined on [, ]. So there are two type I critical points, namely x = and x =. Their corresponding f(x) values are f( ) = 39 and f() = 3.

3 Secondly, type II critical points are values of x such that f (x) = 0. By applying the power rule, the constant multiple rule and the sum rule of differentiation, we have: f (x) = 1x 3 1x 1x + 1 Setting f (x) = 0, we have: 1x 3 1x 1x + 1 = 0 1(x 3 x x + 1) = 0 1[x (x 1) (x 1)] = 0 1(x 1)(x 1) = 0 1(x + 1)(x 1)(x 1) = 0 1(x + 1)(x 1) = 0 So there are two type II critical points: x = 1 and x = 1. Their corresponding f(x) values are f( 1) = 4 and f(1) = 1. Lastly there are no type III critical points because f (x) is defined for all x. We can already identify the global maximum and minimum of f(x) on [, ] by comparing the values of f(x) at the four critical points. Thus we have f(x) has a global maximum at x = with f( ) = 39 and a global minimum at x = 1 with f( 1) = 4. For local maxima and minima, clearly global maximum and minimum are local maxima and minima. We still need to decide whether x = 1 and x = are local maxima or minima. We do this by identifying the intervals on which f(x) is increasing and decreasing. Given f (x) = 1x 3 1x 1x + 1 = 1(x + 1)(x 1), we have the following: When x < 1, we have x + 1 < 0, x 1 < 0, (x 1) < 0. So f (x) < 0 and f(x) is decreasing on the interval [, 1]. When 1 < x < 1, we have x + 1 > 0, x 1 < 0, (x 1) > 0. So f (x) > 0 and f(x) is increasing on the interval [ 1, 1]. When x > 1, we have x + 1 > 0, x 1 > 0, (x 1) > 0. So f (x) > 0 and f(x) is increasing on the interval [1, ]. Since f(x) is continuous, we can combine the two intervals on which f(x) is increasing to get the interval [ 1, ]. So we have f(x) is decreasing on [, 1] and increasing on [ 1, ]. Since the first derivative is positive for both x < 1 and x > 1, the first derivative test gives that x = 1 is neither a local maximum nor a local minimum. Since f(x) is increasing on [ 1, ], the endpoint x = is a local maximum with f() = 3. 3

4 So we have the following result: f(x) has a global and local maximum at x = with f( ) = 39, a global and local minimum at x = 1 with f( 1) = 4, and a local maximum at x = with f(x) = 3. Problem 5: Consider the equation xe y + x e y = 6. (a) Use implicit differentiation to find dy dx. Differentiating both sides gives e y + x e y dy dx + xe y x y dy e dx = 0 ( xe y x e y) dy dx = ey xe y dy dx = ey xe y xe y x e y. (b) Determine dy dx (, 0) and find the equation of the line tangent to the curve at this point. dy dx (, 0) = e0 e 0 e 0 e 0 = 5. Therefore the tangent line at (, 0) is y 0 = 5 (x ), or y = 5 x 5. Problem 6: f(x) = ln(x + 1) (a) f (x) = 4x x +1 (b) f (x) = 0 when x = 0 On the interval (, 0), f (x) < 0. So function is decreasing on (, 0). On the interval (0, ), f (x) > 0. So function is increasing on the interval (0, ). (c) By the First Derivative Test: local minima at x=0 since the function is decreasing on the interval (, 0) and increasing on the interval (0, ). 4

5 By the Second Derivative Test, f (0) = 4 (x +1) > 0 and hence a local minima. (d) f (x) = 4 8x (x +1) (e) To check for concavity. Check where the second derivative is equal to 0. where f (x) = 4 8x = 0 (x +1) This occurs at x = ± 1 Interval f (x) f (, 1 ) - Concave Down ( 1 1, ) + Concave Up ( 1, + ) - Concave Down (f) Inflection points can exist where f (x)=0 or where f (x) does not exist. f (x) exists everywhere. Therefore, the inflection points are x = ± 1. Problem 7: a) From the given information we know two equations: f(1) = 8 and f (1) = 0, the latter due to it being an inflection point. We also have two unknowns: a and b. We can therefore solve the two equations for the two unknowns. f (x) = a x 3 + b x f (x) = 6 x 4 b x 3 f (1) = 6a 1 4 b 1 3 = 0 b = 6a We can now plug (1) into the original function. b = 3a (1) 5

6 f(1) = a 1 b 1 = 8 a 3a = 8 a = 8 a = 4 () Plugging () back into (1), we end up with: a = 4 and b = 1. b) Looking at the first derivative, we see that there is a critical point at 0, where the first derivative does not exist. However, since 0 is not in the original domain of the function, we ignore it. The domain also is not on a closed interval. The only other critical points of the first derivative are where it equals zero. f (x) = 8 x x = x = 0 1x = 8 x = 8 1 x = 3 (3) The only possible local maximum or minimum is then at /3. When x < /3, f (x) < 0. When x > /3, f (x) > 0. Therefore, by the first derivative test, the function has no local maximum but has a local minimum at (/3, 9). c) 6

7 Question 8 (a) In order to find the level curves, equate the function to each value: x y 1 = 1 x y = 0 x y 1 = 0 x y = 1 y = 1 x x y 1 = 1 x y = y = x x y 1 = x y = 3 y = 3 x The first equation is satisfied when either y = 0 or when x = 0. Therefore the level curve at f(x, y) = 1 is the whole y-axis and x-axis. The other three level curves are just the graph of the corresponding equations. Notice: you must graph the function for both negative and positive values of x, hence the graph of each of them will be symmetric around the y-axis. (b) First compute the partial derivatives: f x = xy f x = x 7

8 Then put together the gradient vector: f = ( xy, x ) (c) Reading carefully the question, notice that we are asked for THE maximum of the function on the constraint, and we are explicitly asked to do so using the Lagrange multipliers technique. Therefore, all we need to do is to use the Lagrange multipliers technique to find one or more points, and one of these will be the maximum. In this case, g(x, y) = x + y 1. Hence, we have the following system of equations f = λ g ( xy, x ) = λ(1, 1) x + y = 1 The rest is just algebra. Let s proceed in the following way (many ways are valid). Suppose first that λ = 0. Then, it must be that x = 0. Plugging this into the constraint we have that y = 1. The point (0, 1) satisfies all three equations, hence is a possible solution. 8

9 Suppose now that λ 0. Then, for any value of λ 0 it must be true that xy = x, that is, that x(x y) = 0. This is satisfied in two ways: x = 0 or x = y. The first case, x = 0, is not possible in this case because that would contradict the fact that λ 0. So we have that the only solution in the case that λ 0 is that x = y. Plugging this in the constraint we have y + y = 1, that is, y = 1 3 and x = 3. We have a second valid point: ( 3, 1 3). From the statement of the question we know that one of the two points we found must be the maximum of the function at the constraint. In order to find which one is the maximum, we evaluate the function at each point, and the point giving the highest value of the function will be the answer. We have: f ( 3, 1 3 Since 3 7 > 1, the maximum is: f(0, 1) = (0) 1 1 = 1 ) ( ) ( 1 = 3 3 (x, y) = ) 1 = = 3 7 ( 3, 1 ) 3 Problem 9: Evaluate the following definite integral and sketch a graph of what this integral represents: 4 ( ) 3 + x dx. By the Fundamental Theorem of Calculus, we have ( 3 + x ) dx = F (4) F (1), where F (x) = (3 + x) dx = 3x x 3 + c, where c is a constant. Thus 4 ( ) 3 + x dx = F (4) F (1) 1 = ( ) ( ) =

10 The geometric interpretation of a definite integral is the area under the graph of the function, as illustrated by the shaded area below. 10

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