REVIEW LAB ANSWER KEY
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1 REVIEW LAB ANSWER KEY. Witout using SN, find te derivative of eac of te following (you do not need to simplify your answers): a. f x 3x 3 5x x 6 f x 3 3x 5 x 0 b. g x 4 x x x notice te trick ere! x x g x 4 x x x c. x 3 x 0 6x 5/3 6 use te fact tat 3 x 0 x 0 3 x 0 3 and 6 6x 4 to get: x 0 0 x x x 4 d. f x x 7x 5 3 x 5x we ll do te last 3 of tese in class... e. f x 7x 7x 3 0 we ll do te last 3 of tese in class... f. f x x 7 3x 3x4 3x 6 7 5x we ll do te last 3 of tese in class.... Let f x x. x a. Find te average rate of cange of f x between x and x 5. To find average rate of cange, we need te y coords of points on te grap of f wit x coordinates and 5. So first I defined te function in SN and ten I ad it calculate: f 5 5 f 4 Av r.o.c f 5 f 5 b. Fill in te blanks: Your answer to (a)_ is te slope of te _secant line between te points,_4_ and _5,_ 5. c. Find te instantaneous rate of cange of f x at x. Since I aven t asked you to do it yourself, you can just ask SN to evaluate f x.it returns, f x x. x x f d. Fill in te blanks: Your answer to (c)_- is te slope of te tangent line at te point,_4_ 3. You are given te following table of values for f x.
2 f x a. Evaluate f 5 0 b. Evaluate lim f x x 5 36 c. Evaluate lim f x x 5 49 d. Evaluate limf x x 5 Does not exist, since left and rigt limits are different. e. Is f x continuous at x 5? Explain your answer by referring to te formal (limit) definition of continuity. No, since te limit doesn t exist at 5 and terefore te limit and te function can t be equal tere. 4. Given is te grap of a function f x. y Te grap of f x. a. Find te following limits and values (if tey exist). No work needed. x f 3 f f f 0 f f DNE 0 4 3
3 lim f x DNE ( ) lim x 3 f x 0 x lim f x x lim f x x lim f x DNE ( ) lim x 3 f x 0 x lim f x x lim f x 3 x lim x 3 f x DNE lim x f x 0 lim f x x lim f x DNE x b. Assume tat te domain of tis function includes all x values to te left and rigt of te viewing window. Wit tat information, determine: lim x f x 3 lim x f x 0 notice tat on te left and side of te picture, te grap is swooping down and approacing te x-axis. Tat is wy I determined te limit at - is 0, since te x-axis is te line y 0. On te rigt and side te function as become orizontal wit eigt 3 so te limit at is 3. c. Is te function f continuous at x? Justify your answer by explaining weter or not f satisfies eac of te conditions in te formal definition of continuity. Te function as value 4 at - but te limit at - was. Since te value of te function is not equal to te limit, te tird condition of te definition of continuity is not met. (Visually we ave a ole in te grap and a point elsewere) so te function f is not continuous at x -. d. Use te definition of te derivative to determine weter or not f is differentiable at x 0. Notice visually tat te function as a "sarp corner" at 0. But to use te definition of te derivative I need to turn "sarp corner" into a matematical argument. It sould go someting like tis: From te left, te slope of te tangent lines at 0 sould be positive (eading up). f x f x Tis suggests te left and limit of sould be positive. But on te rigt of 0, te function is eading sarply down, wic means tat te rigt and limit of f x f x would be negative. Since tese limits aren t te same, f x f x lim 0 DNE so te derivative does not exist tere. Tat means f is not differentiable at x (Tan 3. #73)Te body mass index (BMI) measures body weigt in relation to eig. A BMI of 5 to 9.9 is considered overweigt, a BMI of 30 or more is considered obese, and a BMI of 40 or more is morbidly obese. Te percent of te U.S. population tat is obese is approximated by te function P t t t 0.8t 0 t 3 were t is measured in years, wit t 0 corresponding to te beginning of 99. a. Wat percentage of te U.S. population was deemed obese at te beginning of 99? P percent 3
4 b. Wat percentage of te U.S population was deemed obese at te beginning of 004? t 3 P percent c. Wat was te average rate of cange of te percent of te population deemed obese from 99 to 004? P 3 P percent per year d. How fast was te percent of te U.S. population tat is deemed obese canging at te beginning of 99? IusedSNtocalculateformetatP t 0.00t 0.007t 0.8. P so te percent of obese people was canging by 0.8percent/year at te beginning of 99. e. How fast was te percent of te U.S. population tat is tat is deemed obese canging at te beginning of 004? 004 corresponds to t 3. P So te percent of obese people was canging by.0694 percent/year at te beginning of 004. f. Tere is an error in tis question. If you do wat you sould, te question is asking you to solve P t 0. But tere are no solutions to tis (SN will tell you so). 6. Let f x x 4. UseSNtogiveacarefulsketcoff x for te following two windows. Notice tat te one on te left is zoomed in more y -3-4 x y x -0 a. Use te limit definition of te derivative to find f x. You may use SN to simplify te difference quotient but you sould evaluate te limit "by and," using continuity. f x f x First I find te difference quotient for f x. I write out by and wat it sould be, f x f x x 4 x 4 x 4 Ten I use SN to simplify te difference quotient. x 4 x 4 x 4 x x 4 x 3 8 x x 8x 3 x By and (versus using SN), I take te limit as 0. lim 0 f x f x lim x x 8x 3 x x 0 x 0 8x 3 x 8x 3 x So, f x 8x 3 x. 4
5 Alternatively, I could use te 4-step metod to do tis problem: f x x 4 x 4.0 f x f x x 4 x x f x f x x x 4 x 3 8 x x 8x 3 x Now I take te same limit as before... f x f x lim 0 lim x x 8x 3 x x 0 x 0 8x 3 x 8x 3 x So, f x 8x 3 x. b. For wat x value(s) is te tangent line to f x orizontal? Draw tose line(s) in te grap on te left. 8x 3 x 0, Solution is:,0, So draw te orizontal lines tangent to te grap troug te points wit x coords.5, 0, -.5 See tem on te picture? c. Find te equation of te tangent line to f x at x. Draw your tangent line in te grap on te rigt. f 60 slope of tangent line. f 4.0 equation: y 60 x 4 d. By using your answer to part (b) and looking at te grap of f x, over wat interval(s) is f x 0 and over wat intervals is f x 0? Te derivative is positive wen te function is increasing, so between - and 0 and to te rigt of. Te derivative is negative wen f is decreasing, left of, 0to. In interval notation te correct answers are:, 0, f 0.,0,,f 0. REVIEW LAB, PAGE 5 7. Evaluate eac limit algebraically and sow your steps. If you use SN to simplify an expression (and you are encouraged to do so) please indicate at wic step. You may use SN to ceck your answer, but solely giving te answer is not enoug work in general. a. lim x 4x 70 x 3x 30x First try plugging in on top and bottom I evaluate in SN and get and since tese are numbers and te bottom is nonzero, teir 64 quotient is te limit, b. lim x 4x 70 x 5 3x 30x 75 DNE First try plugging in on top and bottom tis is a 0/0 situation so I need to ave SN factor te top and bottom. SN factors te top: x 4x 70 x 7 x 5 SN factors te bottom: 3x 30x 75 3 x 5 Now I look at my new quotient: x 4x 70 3x 30x 75 x 7 x 5 3 x 5 top and bottom again. x 7 3 x 5 plug 5 into 5
6 we ave 4/0 so tat means te expression is undefined and te limit DNE, c. lim x 4x 70 x 7 3x 30x 75 0 First try plugging in on top and bottom since tese are numbers and te bottom is nonzero, teir quotient is te limit, d. lim 0x 3 35x 50x 4 DNE x 3 6x 3 5x 4x 36 First try plugging in on top and bottom tis is a 0/0 situation so I need to ave SN factor te top and bottom. 0x 3 35x 50x 4 x 4 x 3 x x 6x 3 5x 4x 36 x x x 3 So now we divide and cancel like terms, x 4 x 3 x x x x x 3 x x 4 x x 3 Note now if you plug in -3 to te top, you get (-)( ) - and in te bottom you get (-5)(0) 0. Since tis is in te form (nonzero number)/0, te limit will not be defined and tus you write DNE (SN will evaluate it as "undefined"). 0x 3 35x 50x 4 x 6x 3 5x 4x 36 e. lim Notice tis is te same function as in part d, but we re taking te limit at a different point. So to start wit I m just going to use te simplified, factored form of x4 0x 3 35x 50x 4 x x 4 tat we got above:. Notice wen you plug 6x 3 5x 4x 36 x x 3 in - everywere, you get so tis is te limit. f. lim x 4 3 x 3 x x 3 0 Tis is a limit at infinity. So to do it remember ow I taugt you in class: divide out te top and bottom by te largest power of x present anywere. In tis case, we divide te top and bottom by since tat s te largest present. Note tat SN won t be able to elp you wit tis sort of dividing; you ll ave to use your own knowledge of exponents and division to do it. x 3 x x 3 x,, x x 3 x x3 x x 3 x Now we know tat:, x allgoto0wenx wile is still at infinity. so we ve got: 0 is te limit at infinity. 6
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