MATH 1910 If high school precalculus and ACT math of at least 26 contact 694
MATH 1910 If high school precalculus and ACT math of at least 26 contact 694 MATH 1910
pellissippi state community college
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MATH 1910 Section 23 Calculating Limits Using the Limit Laws Look at the limit laws beginning on page 111 of the textbook An important note is that the limit laws can only be applied to functions whose limits EXIST Ex Given that lim fx 2 lim gx 4 lim hx 1 xgta xgta xgta evaluate the following limits if they exist a 1im3fx hx b limgx2 c lim 83 fx xgta xgta xgta 39 39 3 d 111 m 6 11121111496 3gx Ex Evaluate the following limits if they exist without the aid of a calculator a limZ b lim Vx32x7 xgt71x 4x 3 xgt71 There are times when you have to do some algebraic maneuvering such as factoring both sides of a rational expression rationalizing the NUMERATOR of a rational expression before you can evaluate a limit Ex Evaluate the following limits if they exist without the aid of a calculator xZ x IZ x2 1 Jsh J a 11m b 11m c 11m xgt73 x3 xgt1x3 1 mo h mwvmm x 4x3 xlx l lim lim lim xgt73 x3 xgtlx 1x2x1 mo h 5h 1 cancel out the x 3 lim 9 rationalize the numerator xel x x 1 2 5 h 5 lim x 4 77 lim H7 3 1H0 h 5 h J3 lim haO h 5 h 43 Ii 1 l m haO 5h 2J3 1 1 Ex Evaluate the one sided limit lim using your knowledge of the absolute value function xgt0 x x MATH 1910 Section 33 Rates of Change in the Natural and Social Sciences Ex The position of a particle moving on the x axis is modeled by the equation st 36t 712tZ I t 2 0 where t is measured in seconds and rt is measured in inches a What is the velocity ofthe particle at time I What is the velocity after 2 seconds When is the particle at rest When is the particle moving forward to the right Find the total distance traveled in the rst 10 seconds Draw a onedimensional position diagram showing the path of the particle as it moves on the x axis Find the acceleration of the particle at time t and a er 2 seconds Graph all three lnctions position st velocity s t and acceleration s t When is the particle speeding up When is it slowing down 3 3 E SOLUTION Here s what we ll need for this problem Position Jnction 9 50 36t 712tZ I t 2 0 the inequality is just to let you know to only use positive time value velocity Jnction 9 vt s39t 36 7 24f 3tZ rst derivative acceleration Jnction 9 at v39t squott 724 6t second derivative a Velocity at time t is s39t 36 7 24f 3tZ b Velocity a er 2 seconds is v2 s392 367 242 32Z 0 inches second c The particle is said to be at rest when the velocity is zero so set velocity 0 and solve for time t t 0 9 solutions are at t 2 sec and t 6 sec This particle is at rest at t 2 sec and t 6 sec d Right and up hnnld L 39 I 39 39 emu clULil aiue Le and down are negative vector velocity values This particle is moving right when s39t gt 0 look at the graph of s39t 36 7 24t 3tZ The velocity is positive on these intervals 0 2 U 6 00 l 39 39 39 Iigntun L 39 va 0 2 u 6 00 seconds MATH 1910 Section 33 Rates of Change in the Natural and Social Sciences e Becarefulwiththisoneinthe rst10 A 39 39 39 39 n a 39 ato se onds You need to nd the total distance in intervals 50 0 inches 6 the particle starts at 0 inches 52 32 inches 6 the particle is 32 inches to the right after 2 seconds 56 0 inches 6 the particle is back at 0 inches where it started s10 160 inches 6 at 10 ecnnd thenarticlei 39 So the total distance traveled is 32 32 160 224 inches t The derivative graph helps to show the motion ofthe particle moving right between t 0 and 2 seconds moving left between t 2 and t 5 seconds moving right attero seconds 39 39 39 ofthis particle 7 L x x n n l 7 6 r r r r r i t 2 t 0 H 0 32 inches inches g The acceleration a er 2 seconds is a2 s 2 724 62 712 inchessec2 h um f 367 24r3I st 36I712IZ 3 21 v tt s itil46 n g The particle is speeding up when its acceleration and velocity have the same sign Speeding up on the time interval between an 4 seconds and again a er t 5 seconds to in nity Slowing down between t 0 and t 2 seconds and between t 4 andt 6 seconds MATH 1910 Section 33 Rates of Change in the Natural and Social Sciences 411quot3 Ex A spherical balloon is being in ated Find the rate of change of the volume V with respect to the radius r when r is a 10 cm b 11 cm c 12 cm What conclusions can you make SOLUTION 4W3 defines the volume of the balloon as a function of its radius The function V The derivative of this function gives the rate of change of volume with respect to the radius 411quot3 47239 V 9 V39 3r2 3 3 V39 47239quot2 a V3910 47239102 4007239 cm3 per cm meaning the volume is growing at a rate of 4007239 cm3 per cm of radius length when the radius is 10 cm long b V3911 47239112 4847239 cm3 per cm c V3912 47239122 5767239 cm3 per cm Since the rate of change is increasing as the radius increases the volume is obviously growing but it is growing at an increasing rate Ex Boyle s Law states that when a sample of gas is compressed at a constant temperature the product of the pressure in C atmospheres atm and the volume remains constant P a Find the rate of change of the pressure with respect to the volume b What does the sign of the result from part a signify SOLUTION C a P V C 9 rewrite so that pressure is a function of volume 9 V C The rate of change of pressure with respect to volume is 9 P39 C V72 9 P39 2 V b The negative sign indicates the fact that pressure as a function of volume is a decreasing function The more volume increases the more pressure decreases which is also suggested by the inverse proportional relation P V APPLICATIONS OF DIFFERENTIATION 42 MAXIMUM AND MINIMUM VALUES 2 lt Rememberthat the maximum or minimum of a function is the yvalue A Plurals maxima or minima B Extrema all maximum and minimum values he absolute global maximum of a function is the one largest value of the function on a given 39nterval A May occur at more than one point in the interval y sin x has an absolute maximum of 1 it occurs atx 2k k e Z B May occur at an endpoint y x2 on 74 3 has an absolute maximum of 16 it occurs at x 4 This is not a local maximum see IV below C Some functions do not have an absolute maximum y eX The absolute global minimum of a function is the smallest value of the function on a given interval A May occur at more than one point in the interval y x4 716x2 has an absolute minimum of 64 at x HE B May occur at an endpoint y J has absolute minimum of 0 it occurs at x O C Some functions do not have an absolute minimum y In x The local relative maximum of a function is the largest value of the function in a small neighborhood an open interval containing the maximum point A May occur at more than one point in the interval y cos x has a local maximum of 1 it occurs atx x 2k k e Z B Cannot occur at an endpoint because the function does not exist in an open interval containing the point y x2 on 74 3 has no local maxima even though it does have an absolute maximum C Some functions do not have a local maximum y x3 The local relative minimum of a function is the smallest value of the function in a small neighbor hood an open interval containing the minimum point y x has a local minimum of O at x O A May occur at more than one point in the interval y 25x4 x3 7 5x2 has a local minimum of 9375 at x 5 and a local minimum ofx 8 at x 2 B Cannot occur at an endpoint because the function does not exist in an open interval containing the point y J has no local minima even though it does have an absolute maximum C Some functions do not have a local minimum y x3 has no extrema of any kind Extreme Value Theorem lff is continuous on a closed interval a b then fattains an absolute maximum value fc and an absolute minimum value fd at some numbers c and d in a A An absolute extremum is not a local extremum if it occurs at an endpoint y J 1 VII Fermat s Theorem Iff has a local maximum or minimum at c and if f c exists then f c O The converse of Fermat s Theorem is not necessarily true A If fx x3 then f390 0 butfhas no minimum or maximum B If fx x then f0 0 but f 0 does not exist VIII A critical number of a function f is a number c in the domain of f such that either f c O or f c does not exist This is why I have been insisting on factored form for derivativesl A Another way to state Fermat s Theorem Iff has a local maximum or minimum at c then c is a critical number of f NOT REVERSIBLEII B Find all critical numbers of y This was 5 on the Differentiation Practice worksheet 4x3 1 I We found 3 iii f1 in factored form 3 0 2 72x2x3 71 0 Soeitherx0 or 2x3710 2x37102x32x z794 Check to make certain the denominator is not equal to O for either of these values 3 is undefined if 4x3 1 0 2 x3 7 2 x E e 7630 Hmmmmma critical number must be in the domain of the function so that our work in the last step was unnecessary The critical numbers are x O and x f z 794 C Find all critical numbers of y x Inx This was 23 on the DP worksheet CI yInx102Inx712e 1x2xlz368 dx e cos2 x D FInd all crItIcal numbers of y m ThIs was 18 on the DP worksheet 3 cosx 0 2 x g k7r k e Z however some of these values will make sin x 1 and the original function will be undefined sinx 1 2 x 2k k e Z The critical numbers 37r are x 72k7z k e Z E Find all critical numbers of y tanxcotx This was 11 on the DP worksheet Since 3 0 all numbers which satisfy the original function are critical numbers The function is undefined whenever cos x O or sin x 0 so we must rule out all multiples of g The critical numbers are all real numbers except x k e Z F Find all critical numbers of y eyx This was 29 on the DP worksheet We found e dy ex 1 a 7 X2 7 e X but we need to put the derIvatIve In fractIonal form In order to fInd the 4 X X 2 critical numbers d y 7e 7 l u 0 2 7exeyx 7 x2 0 2 x2 7exelt dx X2 ex X2ex 2