Math 118, Summer 1999 Calculus for Students of Business and Economics Midterm

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1 Math 118, Summer 1999 Calculus for Students of Business and Economics Midterm Instructions: Try all the problems. Show all your work. Answers given with no indication of how they were obtained may receive no credit. Good luck! 1. Calculate the following limits. x a) lim x x 1 1 x b) lim x 4 1+x c) lim x 4 x 4 d) lim x a x a x a. Sketch the graph of a function, f(x), that has all of the following properties: (i) lim x f(x) = (ii) lim x + f(x) = (iii) lim x f(x) =1 (iv) lim x f(x) =. Consider the function x+1 x< f(x) = 5 x= x 1 x> a) Is lim x f(x) defined? If it is, find its value and clearly explain your reasoning. If it is not, explain clearly why not. b) Is f() defined? If it is, find its value. If it is not, explain why not. c) Is f continuous at x =? Explain carefully. 4. Find the derivatives of the following functions. You do not need to simplify your answers. a) f(x) =7x 14x b) f(x) = x + x 1 c) f(x) =(x+1) 1/ (x 1) 4 d) f(x) =(x 7) 4/ e) f(x) =( x 1) /5 5. Shown below is the graph of a function f(x). Use the graph to answer the questions about this function. (Sorry, no graph available in electronic version!) a) Which is bigger, f(4) f() or f (4)? Explain. 4 b) Which is bigger, f() or f ()? Explain. 6. A company s revenue from car sales, C (measured in thousands of dollars), depends on how much it spends on advertising, x (also measured in thousands of dollars). In other words, C = f (x). Suppose the company is currently spending 100 thousand dollars on advertising. What does the statement f (100) = 0.5 mean in practical terms? If this statement is true, should the company spend more or less on advertising? 1

2 7. A disease is spreading across the nation. Let P (t) denote the number of people that have been infected by the disease by time t. A vaccination has been found and as more and more people get vaccinated, the disease spreads less and less rapidly. What do we know about the signs of P and of P? Explain carefully. 8. A tropical storm is 40 miles offshore and it is headed directly towards the shore at 4 mph. Meteorologists stuing the behaviour of the storm from a van, want to stay exactly 50 miles from the storm and remain on the shoreline. How fast are they driving? 9. Consider the curve x + xy 1 =1. ThepointP =(1,1)liesonthiscurve. y a) Find at the point P. dx b) Estimate the y-coordinate of the nearby point on the curve, Q, whosex-coordinate is Shown below is the graph of f (x), the derivative of f(x). Answer the following questions about f. (Sorry, no graph available in electronic version!) a) On which intervals is f increasing? b) On which intervals is f concave up? c) Find all the critical points of f and classify them as local maxima, local minima, or neither. d) At what value of x does f attain its global maximum on the interval [0, ]? 11. Consider the function f(x) =x / (x 1). Its derivative is f (x) = 5x (you do not x 1/ need to show this). a) Find all critical points of the function and classify them as local maxima, local minima or neither. b) Find the global maximum and minimum of f on the interval ( 1, ], if they exist. Explain your work carefully. 1. For what value of x in the interval 1 x 4 is the graph of the function f(x) = x (1/)x the steepest? Explain carefully.

3 1. a) lim x x x 1 = lim Solutions x(x 1) (x 1)(x +1) = lim x x +1 = 1 1 x b) lim x 4 1+x = 5 c) lim = since, as x goes to 4 from below, the top is always and the x 4 x 4 bottom is negative and gets closer and closer to zero. d) lim x a x a x a x a = lim x a (x a)( x + a) = 1 a. a) lim f(x) = lim x x (x + 1) =. On the other hand, lim f(x) = lim 1) =. x + x +(x Since these two are both defined and equal it follows that lim f(x) =. x b) f() = 5 c) No, the function is not continuous at x = since lim f(x) f(). x 4. a) f (x) =1x 14 b) f (x) = (x 1)(x) (x + )() = x x 9 (x 1) (x 1) c) f (x) = 1 (x+1) 1/ (1)(x 1) 4 +(x+1) 1/ 4(x 1) (1) = (x 1) (9x +7) (x +1) 1/ d) f (x) = 4 (x 7) 1/ (x) = 8 x(x 7) 1/ e) f (x) = 5 ( x 1) /5 1 (x) 1/ () = 5 x( x 1) /5 6. f (100) = 0.5 means that if the amount spent on advertising were increased from its present value of 100 thousand dollars, then the revenue in car sales would increase at the rate of 50 cents for every one extra dollar spent on advertising. If this statement were true it would not be smart for the company to spend more on advertising since the extra sales would only recoup half of the extra money spent on advertising. 7. As time goes on, people will continue to get infected, so P is an increasing function of t. Thus, P is positive. However, the disease is spreading less and less rapidly. In other words, the rate at which P is increasing is itself decreasing. Thus P is negative. 8. Let x denote the distance (in miles) of the storm from the shoreline. Notice that, although the storm is presently 40 miles from the shore, it is heading towards the shore, so this distance will change as time progresses. For this reason we label it with avariable, x, rather than simply labeling it 40. The storm is heading towards the shore at 40 mph. In terms of derivatives this means that dx = 40. Let y denote the dt distance (also in miles) of the van from the point on the shoreline where the storm will hit. We want to find how fast the meteorologists are driving (right now). In other words we want to find dt / x=40. Since the meteorologists want to stay 50 miles from the storm, we know that at all points in time x + y = 500. This is an identity of functions of time since it will be true at all points in time. Differentiating both sides

4 4 with respect to time gives us x dx dt +y = 0. Evaluating everything at that point dt in time when x = 40 gives (40)( 40) + y/ x=40 =0. Butx +y = 500 so dt x=40 when x = 40, y = i.e. y = 0. Thus (40)( 40) + (0) =0and dt x=40 dt = 160 =5. mph. 9. a) Differentiating with respect to x gives x + y +xy dx + 1 y dx =0. Thus, x+y +(xy + 1 y ) x + y =0and = dx dx xy + 1. Thus, at the point (1, 1), y dx = = 4. b) Let y denote the amount by which the y-coordinate changes when we move along the curve from (1, 1) to the nearby point whose x-coordinate is 1.1. Using the marginal analysis approximation formula we get ( ) y dx / (1,1) ( x) = (.1) = Thus the y-coordinate of the nearby point is approximately 1+( 0.075) = a) When f is increasing, f is positive. So look for all those values of x on the graph where the graph lies above the x-axis. b) When f is concave up, f is positive which means that f is increasing. So look for all those intervals on the graph where the graph is increasing. c) Critical points occur at values of x where f is either equal to 0 or where it is undefined. So look for values of x on the graph where the function touches or crosses the x-axis or where the function is undefined. Use the first derivative to classify each of these points as a local max, a local min, or neither. For instance, if as you go from left to right, the graph crosses the x-axis going from below the axis to above the axis, then this means that f is changing from negative to positive so f has a local minimum at that point. 11. a) The function is defined at all values of x, so critical points occur where f is either equal to 0 or is undefined. Thus, f has critical points at x =0andat x=/5. The sign graph of f reveals that f is positive in the intervals (, 0) and (.5, ) and is negative in (0,.5). This means that the function has a local maximum at x = 0 and a local minimum at x =/5. b) We evaluate the function at all the critical points and at the one endpoint x =. The other endpoint is not contained in the interval so we evaluate a limit there instead. f(0)=0;f(.5) = (/5) / (/5); f() = / ; lim f(x) =. The x 1 + largest value here is /. This is the value of the function at x =sothis is the global maximum of the function. The smallest of these numbers is. This is the limit as x approaches the left endpoint of the interval. However, this endpoint is not contained in the interval so the function does not have a global minimum on the interval.

5 1. The derivative f (x) =4x x =x(4 x) measures the slope of the graph. The graph is steep when its slope is either large and positive or large and negative. The graph of f is an upside down parabola with zeros at x =0andx= 4. So, the largest positive value that f attains occurs at x = (the vertex of the parabola). Here, f () = (4 ) = 4. In the interval [ 1, 4] the largest negative value that f attains occurs at x = 1. Here f ( 1)=( 1)(4 + 1) = 5. Thus the graph is steepest at x = 1. 5