Math Week 1-4 Notes
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This 31 page Class Notes was uploaded by unknown1998 on Tuesday September 13, 2016. The Class Notes belongs to MATH 132 at Ball State University taught by Moore in Fall 2016. Since its upload, it has received 4 views.
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Date Created: 09/13/16
CLASS NOTES – 9.1 Limit of a Function What does the word “limit” mean to you? Can you give situations in which you’ve dealt with limits? We’re going to discuss limits in this class throughout the rest of the semester. The mathematical meaning of a limit may be very close to or very different from your perception of the word limit. 2 f (0) Let’s consider the function,f (x) x 3x 4. What does the graph of this function look like? What is ? f (2) ? f (2) ? Each of these represents a number which is, in fact, the ycoordinate which corresponds to the x value given in the parentheses. Since this is a function, there is only one ycoordinate for each x value. Now let’s look at the ycoordinates not at a particular xvalue, but for values close to an xvalue. For instance, let’s look at the yvalues corresponding to xvalues which are close to 1. Values that are close to 1 might be 0.9 or 0.99 or 0.999 or 1.1 or 1.01 or 1.001. All of these values are close to 1. Some are closer than others. Let’s list these xvalues from smallest to largest. 0.9, 0.99, 0.9991 , 1.001, 1.01, 1.1 (The closer the numbers are physically to th1 , the closer their value is to 1 also.) Now let’s find the corresponding yvalues. x 0.9 0.99 0.999 1.001 1.01 1.1 y Does there seem to be a pattern to the yvalues? What conclusion would you draw about the yvalues as x gets closer to 1? Is this true for both sides1o? We can write this in a much more concise manner. Instead of writing “as the x values get closer to 1”, we write lim f (x) x 1 . Instead of writing “the yvalues corresponding to the xvalues as x gets closer to 1”, we wx1e , or in this specific case: 2 lim(x 3x4) x1 In general lim f (x) L means that the yvalues get closer to L as the xvalues get closer to a from both sides. xa Example Evaluate: lim f (x) when (x) x 3 x1 x 9 This is read “the limit as x approaches ofx 3 divided by 2 ”. You must learn how to read this correctly. 1 x 9 This means “what happens to the ycoordinates of the points on this curve as the xcoordinates get closer and closer to 1?” x 1.1 1.01 1.001 0.999 0.99 0.9 x 3 2 x 9 x 3 f (1) x1 x 9 Example Evaluate: im f (x) when f (x) x 3 x3 x 9 x 2.9 2.99 2.999 3.001 3.01 3.1 x 3 2 x 9 x 3 f (3) lx3x 9 Example x 3 Evaluate: x 3f (x when f (x) 2 x 9 x 2.9 2.99 2.999 3.001 3.01 3.1 x 3 x 9 x3 lim 2 f (3) x 3x 9 Complete the chart where x a, using the graph below. 2 3 4 5 6 4 —3 2 lim lim lim a xaf(x) xa f(x) xa f(x) 6 _________________________________________________________________________________ 2 ______________________________________________________________________________ 1 ________________________________________________________________________________ 1 _________________________________________________________________________________ 4 — - 7 When evaluating limits of the form lim f (x), the first thing to do is to replace all x’s with c’s. xc To simplify these limits, we need to use the following Properties of Limits: If k is a constant, m f L and lim g M , then the following are true. xc xc I. im k k IV. lim f x LM xc xc II. im x c V. lim f L , if M 0 xc xc g M lim f x g x LM lim n f n lim f x n L III.xc VI. xc xc Example 2 2 lim 2x 1 2(2) 1 9 3 x2 x 6x 4 (2) 6(2) 4 12 4 EXAMPLE 2 limx Evaluate: x 2 6x1 Evaluate: x7 2 2 3 Evaluate: lx4x 7 Evaluate: x 3x 1 2 2 2 x 1 (3) 1 8 4 x3 2 = (3) 3(3)2 2 x 3x 2 When x 3, y 4 and asx 3 , y 4 . 3 What does this tell us about the graph of the function? It tells us that whx 2 , y . This means that there 4 2, 3 is a point on the graph with coordinates 4 . We also know that as x2 , y 4 . 6 O ne Sided Limits Definition: Limit from the Right x cf x L the values of f (x) approach the value L as x c but x c Limit from the Left limf L the values of f (x) approach the value L as x c but x c x c lxcf (x) L if and only if ( ) xcf (x) L and xcf (x) L Example x x 0 Let f(x) 3 when 0 x 3 x 3 (x3 )2 Evaluate: lim (x) lim (x) x0 f (x) f (0) x0 x0 lim f (x) lim f (x) lim f (x) f (1) x1 x1 x1 x3f (x) x3 (x) x3 f (x) f (3) EXAMPLE 3x5 when x 3 Let f 2x when 3 x 2 x when x 2 Evaluate: x0f (x) x0 (x) x0 f (x) f (0) lim (x) limf (x) x 3 (x) f (3) x3 x3 7 lim f (x) lim f (x) lim f (x) x 2 x 2 x2 f (2) 8 As we know, there are times when the limit exists as c , but there is no ycoordinate when c. Can we evaluate these limits? Yes! x c There are rational functions which may have 0 in the denominator for a particular value . These forms of limits are called Indeterminate Form. There are 2 situations where this might occur. In one situation, both the numerator and denominator approach 0. In the other, only the denominator approaches 0. We can make some generalizations about each type. lim f lim g 0 Limits of the Form xc g where xc . lim f 0 lim g 0 0 Type I : Ixc and xc , then the fractional expression has 0he indeterminate form at x = c. We can factor x – c from (x) and g(x) , reduce the fraction, and then find the limit of the resulting expression, if it exists. f Type II : Ifxcf 0 and xc g 0 , then im does not exist. The line x = c is a vertical xc g asymptote. EXAMPLE: 2 lim x 4 When we replace x with 2, we get . This does NOT mean that the limit ax2 does not x2 x 2 0 exist. This means that there is no ycoordinate whenx 2 ; there is not a point on the graph of this function with an xcoordinate of 2. Simplify the expression before taking the limit. (x 2)(x 2) lx2 x2(x 2) NOTE: This is only true ix 2!!!. Otherwise we’d be dividing by 0! x 2 We can now evaluate lim(x2) . That limit is 4. Therefore, even though there is no pox 2, as x2 x 2, y 4. There would be a hole in the graph at the point (2, 4). EXAMPLE: x 6x8 Evaluate: lim x5 x5 9 Assignment: Pp. 553ff 1 11 all; 14; 1536 mult. of 3; 53, 56, 59 For 53, 56, and 59, evaluate and then explain what your answer means in the context of the problem. 10 CLASS NOTES 9.2 CONTINUITY AND LIMITS AT INFINITY Definition: A function, f, is continuous at x= c if all three of the following conditions are met: 1) im f exists, xc 2) exists, and 3) xc f f If any one or more of these conditions are not met, then we say f is discontinuous at x = c. Graphs EXAMPLE: x 3x2 f x1 is discontinuous at x = 1, because, when x = 1, the denominator is zero. Therefore there is no ycoordinate when x = 1, i. does not exist. However, because the numerator and denominator lim f both have a factor of x –x1 does exists and it is __________. Theorem: a) Any polynomial function is continuous for all real values of ¡ i.e. b) Any rational or radical function is continuous wherever it is defined, i.e. for its domain. EXAMPLE: x3 a) f x4 x The domain of f Rs except x = – 4 and x = 5. Therefore f is continuous for all real values of x except –4 and 5. Another way of writing this would be: f is continuous over the intervals ___________________________________________________. b) f x6 The domain is ______________________________. Therefore f is continuous for ____________ or over the interval ___________________. EXAMPLE: P. 564 # 8 P. 564 #16 2 Limits at Infinity So far we’ve looked at and evaluated limits as x approached a particular value. But we also need to evaluate limits as and x . This means that we’re going to look for patterns in the ycoordinates of the points at the extreme left and right ends of the graph .f EXAMPLE: Evaluate lim 2 x x c 10 100 1000 10,000 100,000 1,000,000 lim f xc 2 Evaluate xx c –10 –100 –1000 –10,000 –100,000 –1,000,000 lim f xc EXAMPLE: 2 Evaluate x3x 2x1 c –10 –100 –1000 10 1000 100,000 lim f x xc EXAMPLE: 3x 3 Evaluate: lim x x1 c –10 –100 –1000 10 1000 100,000 lim f xc 3 P. 564ff 1–15 all, 21–31 odd; 39, 41, 42, 46, 51, 52 4 CLASS NOTES 9.3 THE DERIVATIVE As stated earlier, the difference between calculus and algebra is that, using calculus, we can determine the rate at which a function is changing at an instant in time, even if the function is changing at different rates instant by instant. To work our way to this “instantaneous” rate of change, let’s begin by looking at the changes that occur over longer intervals of time. Driving Example Distance Time Average velocity distance travelled elapsed time Average velocity from parking lot to I69 ________________________________________________ Average velocity from McDonald’s to Exit 5 ______________________________________________ Average velocity from parking lot to Exit 5________________________________________________ Instantaneous velocity is when the elapsed time is 0! For instance, what is the instantaneous velocity when you pass the Nestle Rabbit sign? Let’s look at air pollution. Suppose it has been determined that the ozone level (in parts per billion) on a summer day in Muncie is given by P 8012t t 2 where t is time in hours and t = 0 is 9:00 am. We can find the ozone level at any time of day by substituting the number of hours past 9:00 in the morning for t. For instance the ozone level at 10:00 am is 2 . P 8012 1 91 The ozone level at 1:15 pm = P4.25 8012 4.5 5 2 113 . If we wanted to know how much the ozone level had changed from 10:00 to 1:15, we would just subtract 113 – 91 = 22 parts per billion. 2 (Since the amount of ozone increased, this answer should be positive.) The amount of time that elapsed between 10:00 and 1:15 is also found by subtracting; it’s 3.25 hours. Since the amount of ozone increased 22 parts per billion in 3.25 hours, we can find the average rate of change by dividing the change in the ozone level by the change in the time, i.e. 22 parts per billion 6.8 parts per hour . 3.25 hours So far we haven’t done anything mathematically different than we would have done in elementary school. The difference comes when we want to find how fast the ozone level is changing at exactly 10:00. We can not do this with our traditional math tools because the difference in time from 10:00 to 10:00 is 0 and we can’t divide by 0. Therefore we’ll have to approach this problem from a slightly different angle. We’re going to “sneak up” on the 10:00 hour by finding the average rate of change of the ozone levels for periods of time which are becoming shorter and shorter. For instance we could find the average rate of change of ozone levels from 10:00 to 10:30, then from 10:00 to 10:15 and then from 10:00 to 10:01. We still haven’t found the rate of change at exactly 10:00. To do this we’ll use our new math tool, limits. We will take the limit of this rate of change as the time interval approaches 0 hours. Since there are entirely too many words involved thus far, let’s discuss this same problem symbolically. We have been finding the rate of change in ozone levels when there’s a difference between the two times, which 2 we’ll refer to as h. So we could say that the ozone level at time t is P 8012t t and the ozone level h hours later is 2 . Therefore P h 012(t h)(t h) average rate of change = 2 P h t 8012(t h) t h 012t t 12h2thh 2 h 22t h =12 – 2t – h t h t h h h So the average rate of change between 10:00 and 10:30 could be found by substituting 1 for t and 0.5 hrs. for h. average rate of change = 12 – 2(1) – 0.5 = 9.5 parts per hour. The formula used to find the average rate of change is also called the difference quotient. However to find the instantaneous rate of change right at 10:00, we have to do the following: P h 1 P h t lim Instantaneous rate of change lim h0 h h0 h = h0 (12 – 2t – h) = 12 – 2t = 12 – 2(1) = 12 – 2 = 10 parts per hour This rate of change discussion can be applied to many situations, including the position of an item as it’s thrown or moved from one place to another. So in general we say the instantaneous rate of change of a function f(x) lim f h x h0 h dy d The new name for this is the derivative of f which is also written f , y , , f among others. dx dx 3 When referring to the position of a moving item we call the rate of change the velocity of the object and instead f x s t of we use . Thus st hs t the average velocity of an object = h and The formula used to find the average velocity is also called the difference quotient. the instantaneous velocity of an object (velocity) = limh s t h0 h ****************************************************************************************** DEFINITION OF THE DERIVATIVE: For y f , we define the derivative of f at x, denoted by f′(x), tf bx lim fxh f x , if the h0 h f x limit exists. If exists for each x in the open interval (a, b), then f is said to be differentiable over (a, b). The process of finding the derivative is called differentiation. ****************************************************************************************** There are numerous applications where the derivative can be used. The three most prevalent are: a) the slope of the line tangent to the graph of a function at a particular point; b) the instantaneous rate of change of a function; c) the velocity of a moving object. NOTE: When referring to the slope of a function, , we are referring to the slope of the tangent line at a particular point on the graph of . EXAMPLE: Use the definition of the derivative to find the derivative 3x7 Always state the definition first. f h f x f lim h0 h First find the difference quotient, and then take the limit of the result as h approaches 0. fxh f x The difference quotient = h For our example that would be 3 h 3x7 3x3h73x7 3h h h h 3 The second step is to take the limit h 0 4 f lim 33 h0 It makes sense that the slope of a line tangent to a line will be the same as the slope of the given line. EXAMPLE: 2 Use the definition of the derivative to find the derivative of f x 3 4 f h x f lim h0 h 2 2 Step 1: fxh x 3 h 4 3x 4 h h 3 2xhh 2 4 3x 4 2 = h 3x 6xh3h 43x 4 2 6xh3h 2 = h h = h6x3h h 6x3h Step 2: h0 63h 6x Therefore f 6x Find the instantaneous rate of change of f (x) when x = –3. Find the slope of the line tangent to f (x) at x = –3 Find the point on the graph of f (x) when x = –3. 5 EXAMPLE: Use the definition of the derivative to find when f 45x2x 2 f lim fxh f x h0 h Step 1: ________________________________________ Step 2: lim (____________________) = h0 Therefore f (x) ___________________ Find the slope of the line tangent to (x)at x = 2. Find the instantaneous rate of change of f (xat x = 2. Find the point on the graph when x = 2. f (x) Write the equation of the line tangent to at x = 2. 6 EXAMPLE: Use the definition of the derivative to find when f 3 x7 fxh f x f lim h0 h Step 1: When the rate of change of f (x)is positive, the graph of (x)is rising, i.e. (xis increasing. When the rate of change of f (x)is negative, the graph of f (x is ______________, i.e. f (xis _________________ Look at the graphs you printed from Blackboard. EXAMPLE: P. 630 # 46 Are all functions differentiable everywhere? If not, how do we locate those places where it is NOT differentiable? Graphically, we look for points of discontinuity and places where the tangent is vertical or impossible to select. Algebraically, we’re looking for values of x that are not in the domain of the function or that are not in the domain off . Look again at the graphs. 7 Assignment: Pp. 577 ff 1, 3, 5, 7, 9, 11, 14, 15, 18, 30, 31–34 all; 35, 36, 37, 42, 43, 45, 48, 50 CLASS NOTES 9.4 DERIVATIVE FORMULAS d I. Constant Rule – 0 dx Proof: EXAMPLES: d 5 = __________________ D [–14] = ________________ y 2, y ___________ dx x II. Power Rule If y un where u is a differentiable function of x and n is a real number, then dy nu n1 u For the functions used in the next few sections the functions used will be dx limited to those in which x and sou 1 . d Proof thatdx 1 : Let f = x. Using the definition of the derif lim fxh f x , we have h0 h (xh) x h f lim lim lim 11. h0 h h0 h h0 EXAMPLES: 3 3 y x y x y x y 1 x III. CONSTANT MULTIPLE RULE d n d n n1 dx cu c dx u c nu EXAMPLES: 5 4 y 2x y 3x y 1 x 6 y 3 2 x y 2x 2 PAY ATTENTION TO PARENTHESES!!!! EXAMPLE: 6 6 y 2 y 2 5x 2 IV. SUM AND DIFFERENCE RULE d d d dx f g dx f dx f g EXAMPLES: D x 2x x 3x1 = _________________________ 1 D x 4t 1 = __________________________ y 4x x3 3 2 f 6x 3x 2x1 x f 3 5x x 2 2 EXAMPLE: Find the equation of the tangent line to the graph of f x 52 at x = 4 3 EXAMPLE: 1 Find the equation of the tangent line to when x = –1 3x2 EXAMPLE: 1 2 Determine the points, if any, at which the tangen in2 to 5x is horizontal. EXAMPLE: P. 642 #50 4 Assignment: Pp. 588 590 1–10 all; 12 – 33 mult of 3; 45 – 53 odd 5 CLASS NOTES 9.5 PRODUCT AND QUOTIENT RULES When the function with which we’re working is actually the product of two other functions, we might have to use another method for finding the derivative of it. PRODUCT RULE d If u and v are differentiable functions of x, en v vu dx Let’s test it! Keep in mind that this is not a proof of the Product Rule but only an example. Find y when y 3x 5x 9 Using the Product Rule Using the Power Rule y 3x 5 2 9 3 y 6x 17x 45 6x 106x 27 12x 17 y 12x 17 We got the same derivative using both methods. You may use whichever method is appropriate for the function in question, unless instructed otherwise. EXAMPLE: Use the product rule to find the derivativ f x 1 2x5 EXAMPLE: Use the product rule to find the derivativ f 2x 1 x13 1 QUOTIENT RULE We also need a special rule for finding the derivatives of rational functions. This rule is called the Quotient Rule. d vu u v QUOTIENT RULE – If u and v are differentiable functions of x, then 2 , where v 0. dx v EXAMPLE: Find the derivative of 1 1 x 2 In this case u = 1 and v 1 x2 , andu 0 and v 2x . 1 x2 02x 2x That means that y 2 . This can then be simplified to 2 2 2 2 1 x 2 1 x 1 x EXAMPLE: 2x 5 Find the derivative of 3x 2 u ___________, u ______________, v __________, v ________ 3x 2 2x5 3 The Quotient Rule gives us 2 =____________________ = ____________________ 3x 2 EXAMPLE: 2t 3 Find D x 2 u _____________, u _____________ v ____________, v ________ t 5t 6 Therefore the derivative is __________________ = ________________________ = ___________________ 2 We can use a combination of all that we’ve learned thus far in calculus as well as in algebra to find the derivatives of quotients. EXAMPLE: Find dy when y 4 We’ve found this derivative using the power rule before, but we can also find dx 5x 2 the derivative using the Quotient Rule where 4, u 0, v _____________v ___________ Then dy _____________________ = ___________________ = _________________ dx EXAMPLE: We can also use the rules we’ve learned from algebra to simplify a function before finding its derivative. t 2 f 2 t 5t 6 This can be simplified by factoring the denominator and then looking for common factors. t 2 f ________________________ Then use quotient rule to find the derivative. t 2 3 Now the Quotient and Product Rules are tools that we can use. EXAMPLE: 1 Find the equation of the line tangent to 2 at when x 1 5 x 3 EXAMPLE: x2 Find the values of x where there is a horizontal ngent to f x x 1 EXAMPLE: P. 650 #40 EXAMPLE: P. 651 # 46 4 Pp. 596 7 3 – 24 mult of 3; 36 – 51 mult of 3 5
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