MA 151, Applied Calculus, Fall 2017, Final Exam Preview Identify each of the graphs below as one of the following functions:

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1 MA 5, Applie Calculus, Fall 207, Final Exam Preview Basic Functions. Ientify each of the graphs below as one of the following functions: x 3 4x 2 + x x 2 e x x 3 ln(x) (/2) x

2 2 x 4 + 4x x + 0 5x 0 Function notation, solving function equations 2. (This problem as a whole is obviously too long for the final: but I coul have a problem with a couple of parts like this.) (a) Let f(x) = 3x + 5. Solve f(x) = 3. 3x + 5 = 3 3x = 2 x = 2/3 (b) Let f(x) = x 2. Solve f(x) = 5. x 2 = 5 x = ± 5

3 MA 5, Applie Calculus, Fall 207, Final Exam Preview 3 (c) Let f(x) = 2x 3 5. Solve f(x) = 0. 2x 3 5 = 0 2x 3 = 5 x 3 = 5 3 x = = 3 5/3 () Let f(x) = x 2 2x 5. Solve f(x) = 0. x 2 2x 5 = 0 (x 5)(x + 3) = 0 (x 5) = 0 or (x + 3) = 0 x = 5 or 3 (e) Let f(x) = x 2 2x 4. Solve f(x) = 0. x 2 2x 4 = 0 x = 2 ± ( 2) 2 4()( 4) 2() = 2 ± = 2 ± 60 2 (f) Let f(x) = e x 2x 2 e x. Solve f(x) = 0. e x 2x 2 e x = 0 e x ( 2x 2 ) = 0 e x = 0 or ( 2x 2 ) = 0 e x = 0 is impossible 2x 2 = 0 2x 2 = x 2 = /2 x = ± /2

4 4 (g) Let f(x) = +. Solve f(x) = 0. x x + = 0 x = x = = x (cross multiply) x = (h) Let f(x) = 2 5. Solve f(x) = 0. x2 2 x 2 5 = 0 2 x 2 = 5 2 x 2 = 5 2 = 5x 2 (cross multiply) x 2 = 2/5 x = ± 2/5 (i) Let f(x) = 3e x 5. Solve f(x) = 0. 3e x 5 = 0 3e x = 5 e x = 5/3 ln(e x ) = ln(5/3) x = ln(5/3) (j) Let f(x) = 4e 2x 3. Solve f(x) = 0. 4e 2x 3 = 0 4e 2x = 3 e 2x = 3/4 ln(e 2x ) = ln(3/4)

5 MA 5, Applie Calculus, Fall 207, Final Exam Preview 5 2x = ln(3/4) x = 2 ln(3/4) (k) Let f(x) = ln(x) + 5. Solve f(x) = 0. ln(x) + 5 = 0 ln(x) = 5 e ln(x) = e 5 x = e 5 (l) Let f(x) = 2 ln(x) 5. Solve f(x) = 0. 2 ln(x) 5 = 0 2 ln(x) = 5 ln(x) = 5/2 e ln(x) = e 5/2 x = e 5/2 (m) Let f(x) = 3 ln(x + ) + 7. Solve f(x) = 0. 3 ln(x + ) + 7 = 0 3 ln(x + ) = 7 ln(x + ) = 7/3 e ln(x+) = e 7/3 (x + ) = e 7/3 x = e 7/3 3. Let f(x) = x 2 an g(x) = x +. (a) Fin f(x) + g(x). x 2 + x +. (b) Fin f(x)g(x). ( x ). 2 x +

6 6 (c) Fin f(x) g(x). x 2 x +. () Fin f(g(x)). ( ) f x + = (e) Fin g(f(x)). g(x 2 ) = x 2 +. ( ) 2 x +. Linear, Power Function, an Exponential Moelling 4. The two sections of Applie Calculus that I taught this semester starte with 6 stuents. 5 weeks later I ha 56 stuents. (a) Assuming that the stuents continue roppe the class at a linear rate, write an equation for S, the number of stuents, as a function of t, the number of weeks since the beginning of the semester. (b) Accoring to your function, how many stuents i I have in week 5? (a) (b) line through (0, 6) an (5, 56) y = m(x x 0 ) + y 0 x 0 = 0 y = (5) = y 0 = m = 5 0 = 5 5 = 3 = 6 ( ) = y = (x 0) = 3 x + 6

7 MA 5, Applie Calculus, Fall 207, Final Exam Preview 7 5. The arctic ice caps appears to be shrinking at a rate of 4% per ecae. (a) Assuming that the rate of ecrease of ice remains the same, write an equation for I, the percent of the current ice that will be left, as a function of t, the number of ecaes from now. (b) Accoring to your equation, what percent of the current amount ice will be left in 50 years? (a) I = Ca t C = % of current ice, t=ecaes, a = + r, r = % change = ( 0.04) t I = (0.96) t (b) I = (0.96) 5 = 0.85 = 8.5% of current ice Functions in Economics 6. (Hughes-Hallet, 4e, #9) A company that makes Aironack chairs has fixe costs of $5000 an variable costs of $30 per chair. The company sells the chairs for $50 each. (a) Fin formulas for the cost an revenue functions. (b) Fin the marginal cost an marginal revenue. (c) Graph the cost an revenue functions on the same axes. () Fin the break-even point. (a) C(q) = q R(q) = 50q (b) MC = 30 MR50 From

8 8 (c) () C(q) = R(q) q = 50q 20q = 5000 q = q = (Hughes-Hallett, 4e, #28) The eman curve for a prouct is given by q = 20, p an the supply curve is given by q = 000p for 0 q 20, 000, where price is in ollars. (a) At a price of $00, what quantity are consumers willing to buy an what quantity are proucers willing to supply? Will the market push prices up or own? (b) Fin the equilibrium price of quantity. Does your answer to part (a) support the observation that market forces ten to push prices closer to the equilibrium price? (a) consumers willing to buy: q20, p = 20, (00) = 70, 000 suppliers willing to make: q = 000p = 000(00) = 00, 000 Market forces will move prices own, because there are too many items being mae. (b) supply = eman 000p = p

9 MA 5, Applie Calculus, Fall 207, Final Exam Preview 9 500p = p = = 80 This agrees with (a) because from p = 00 the price will move own to p = 80. Approximating the Derivative 8. Let f(x) = e x2. Fin an approximation of f (2) by filling in the following table using the same rules as on the quiz 2. x f(x) f(2) f(x) 2 x ERR The numbers we calculate for x =.999 an 2.00 are our best approximations. From them we can say f (2) Graphing Derivatives 9. Shown below is the graph of a cubic function, i.e. a function of the form y = ax 3 + bx 2 + cx +. Sketch a graph of the erivative; make sure that your sketch is positive/negative/zero in the right places, an that it has the right shape. 2 You shoul plug numbers in for x that are close to 2. They shoul be close enough that you get two approximations of f (2) that are the same to the secon ecimal place. You shoul always use at least 8 igits in every step of the calculation.

10 0 To see how to make the secon graph, start with the points where f is flat: graph of f(x) : m = 0 at x =.5 an x =.5

11 MA 5, Applie Calculus, Fall 207, Final Exam Preview This means that the graph of f (x) shoul have graph of f (x) : y = 0 at x =.5 an x =.5 Similarly, see where f is going up: graph of f(x) : m 2 at x = 2 an x = 2 This means that the graph of f (x) shoul have graph of f (x) : y = 2 at x = 2 an x = 2 Finally, see where f is going own: graph of f(x) : m 3 at x = 0 This means that the graph of f (x) shoul have graph of f (x) : y = 3 at x = 0 Marginal Cost an Revenue 0. The graph below shows a total cost function. Estimate the marginal cost at q = 700. We start by graphing a tangent line (simply move a straight ege towars the graph of C(q) until it just barely touches the graph at q = 700) Now we estimate the slope of the straight line. The best way to o this is to use points that are as far apart as possible, or to see if the line intersects exactly at one of the gri points. I ll use points far apart: (0, 20) an (000, 80)

12 2 appear to be on the line. Thus MC = m = = 3/5. A company is making complicate things with a cost function given by C(q) = 7e q 5 q + 000q 2 an revenue function given by R(q) = 25000q. (a) Fin the marginal cost an marginal revenue. (b) At q = 300 is the profit increasing or ecreasing? (a) (b) MC = q C(q) = 7e q 5 2 q / q MR = q R(q) = MC(300) = 7e (300) / (300) = MR(300) = Profic is ecreasing since MC > MR. Basic erivatives 2. Fin the following erivatives (a) x 3x6 8x 5 (b) x 5 3 x (c) 5 3 x 2/3 (note that 3 x = x /3 ) 0 x x 5 50x 6 (note that 0 x 5 = 0x 5 )

13 () x 7ex 7e x (e) 7 ln(x) x 7 x Combinations of functions 3. Fin the following erivatives (a) 3(2x 5)6 x ( x 3 2x 5 MA 5, Applie Calculus, Fall 207, Final Exam Preview 3 ) 6 ( ) 5 = 8 2x 5 2x 5 = 8(2x 5) 5 2 (b) x 5 3 7x + x 5 3 7x + = 5 3 ( ) 2/3 7x + 7x + = 5 3 ( 7x + ) 2/3 ( 7) (c) 0 x (4x 5) 5 0 ( ) 6 x ( 4x 5 ) = 50 4x 5 4x 5 5 = 50 (4x 5) 6 4 () x 7e x+0 x 7e x+0 7e x + 0 x + 0 7e x+0 ( )

14 4 (e) 7 ln(3x + 5) x ( ) x 7 ln 3x + 5 = 7 3x + 5 3x Fin the following erivatives (a) x (3x x)( 0 x 5 7ex ) x (3x x) }{{} f f = 8x x 2/3 ( 0 x 5 7ex ) } {{ } g g = 50x 6 7e x ( f g + fg = 8x ) ( ) 0 3 x 2/3 x 5 7ex + (3x x) ( 50x 6 7e x) (b) 2x 2 x x e x + x 2 2x 2 x f e x + x 2 g f = 4x f g fg g = e x + 2x Derivatives an Tangent Lines g 2 = (4x )(ex + x 2 ) (2x 2 x)(e x + 2x) (e x + x 2 ) 2 5. (a) Fin the Equation of tangent line at x = 5 for f(x) = ln(x). ym(x x 0 ) + y 0 x 0 = 5 y 0 = f(5) = ln(5) f (x) = x m = f (5) = 5 y = (x 5) + ln(5) 5

15 MA 5, Applie Calculus, Fall 207, Final Exam Preview 5 (b) Fin the Equation of tangent line at x = 9 for f(x) = 5 x. ym(x x 0 ) + y 0 x 0 = 9 y 0 = f(9) = 5 9 = 5 f (x) = 5 2 x /2 m = f (9) = 5 2 (9) /2 = = = 5 6 y = 5 (x 9) Local Max/Mins 6. Let f(x) = 3x 4 4x 3. (a) Using your calculator, graph f(x). Use a winow that shows all the interesting features (in particular it shoul show the local max/mins). Note: your winow has to be small enough that I can see two things: () that the point at (0, 0) is neither a max nor min, (2) that the point (, ) is a local min.

16 6 (b) Describe the critical points, the local max an mins, an where the function is increasing/ecreasing. critical points: x = 0, x = is l. min at x =. x = 0 is neither f to right of x =. f to left of x =. (c) Take the first erivative of f(x), fin the critical points algebraically. f (x) = 2x 3 2x 2 f (x) = 0 2x 3 2x 2 = 0 2x 2 (x ) = 0 x 2 = 0 or (x ) = 0 x = 0, () Summarize your work in a D# table (st Derivative Number Line Table 3 ) table that shows the first erivative test an the conclusions that it gives you. neither l.min x = 0 x = f f f f < 0 f = 0 f < 0 f = 0 f > 0 Global Max/Mins 7. Let f(x) = x 0 0x. (a) Fin the critical points of f(x). f (x) = 0x 9 0 f (x) = 0 0x 9 0 = 0 0(x 9 ) = 0 x 9 = 0 x 9 = x = 9 x = 3 This shoul be a number line with the following information: you shoul label on the number line each critical point. Above each critical point you shoul inicate whether that point is a local max/min/neither. On top of the number line an between the critical points you shoul inicate whether f is increasing or ecreasing. On the bottom of the number line, between the critical points an at each critical point, you shoul inicate whether f is +, or 0.

17 MA 5, Applie Calculus, Fall 207, Final Exam Preview 7 (b) Apply the Global Max/Min test to fin the absolute max/min on the interval 0 x 2. x y G. max: x = 2, y = 004 G. min: x =, y = 9 Optimizing Cost an Revenue 8. Shown below is a graph of cost an revenue for a certain company: (a) If the prouction level is q = 400, shoul the company increase prouction to q = 40? Why or why not? Here s what the graphs look like with a small part of the tangent lines shown:

18 8 It shoul be clear that the slope of C is less than the slope of R. In other wors: so they shoul increase prouction. MR > MC profit increases with q (b) Fin two critical points of profit. Which one has maximum profit? Why? We fin two points where the slopes are the same for C an R: these are at q = 26 an q = 730: The one that is maximum profit is q = 730. The best way to tell this is just to the left of 730 we have MR > MC an just to the right we have MR < MC. 9. A company has a cost function given by C(q) = 0q + an revenue function given by R(q) = 5 q.

19 MA 5, Applie Calculus, Fall 207, Final Exam Preview 9 (a) Fin the critical point of profit. MC = 0 MR = 5 2 q /2 MC = MR 0 = 5 2 q /2 0 = = 5 2 q q 20 q = 5 (cross multiply) q = 5 20 q = 3/4 q = (3/4) 2 q = 9/ (b) Is the critical point a maximium for profit? Why? The critical point is a maximum for profit. The best way to see this is to note that just to the left we have MR > MC an just to the right we have MR < MC: Average Cost 20. A company is making simple things with a cost function given by C(q) = 5q 2 +. (a) Fin the average cost at q = 3.

20 20 a(q) = C(q) q = 5q2 + q a(3) = 5(3)2 + 3 = (b) Is the average cost increasing or ecreasing at q = 3? MC = 0q MC(3) = 30 The average cost is increasing because MC(3) > a(3). (c) Repeat the previous two parts graphically, using the graph of 5q 2 + : The average cost equals the slope of the straight line shown below: a(3) = slope of line

21 MA 5, Applie Calculus, Fall 207, Final Exam Preview 2 = rise run = 45 3 If we look at q = 3 where the straight line intersects the curve, it s clear that the curve is steeper at that point. Thus, MC(3) > a(3) an so the average cost is increasing. Interpreting Integrals as Velocity an Area 2. Set up an integral to fin the area below y = e x2, above the x-axis, an between 0 an. e x2 x Set up an integral to fin the net istance travelle by a falling object with velocity given by v(t) = 9.8t + 2, from t = to t = t + 2 t Calculating using tables, graphs an calculators 23. Calculate 0 x 2 x with a Left Han Riemann Sum an n = 9. We start with the number line 0 Now we ivie into 9 equal pieces: x So the x-values are, 2, 3,..., 0. For the left han rule we use, 2,..., 9: 0 x 2 x = f() + f(2) + f(3) + f(4) + f(5) + f(6) + f(7) () + f(8) + f(9) = () 2 + (2) 2 + (3) 2 + (4) 2 + (5) 2 + (6) 2 + (7) 2 () + (8) 2 + (9) 2 = 285

22 22 For the right han rule we use 2, 2,..., 0: 0 x 2 x = f(2) + f(3) + f(4) + f(5) + f(6) + f(7) () + f(8) + f(9) + f(0) = (2) 2 + (3) 2 + (4) 2 + (5) 2 + (6) 2 + (7) 2 () + (8) 2 + (9) 2 + (0) 2 = 384 Note: it is require that you first write out the sum in at least one of the ways I ve one (i.e. either with f( ) or with ( ) 2 ) before you calculate the total.