9Logarithmic. functions using calculus UNCORRECTED PAGE PROOFS

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1 9Logarithmic functions using calculus 9. Kick off with CAS 9. The derivative of f () = log e () 9. The antiderivative of f () = 9.4 Applications 9.5 Review

2 9. Kick off with CAS To come Please refer to the Resources tab in the Prelims section of our ebookplus for a comprehensive step-b-step guide on how to use our CAS technolog.

3 Units & 4 AOS Topic Concept 9. Rules for common derivatives Concept summar Practice questions The derivative of f() = log e () The proof for the derivative of = log e () relies heavil on its link to its inverse function = e. then, b the definition of a logarithm, If If = log e () e =. = e then appling the eponential derivative rule gives However, it is known that In summar, It is worth noting that so But so therefore, d d = e. d d = d d d d = e. e = d d =. d d (log e ()) =. d d (log e (k)) = also. This can be shown b appling the chain rule. If = log e (k) let u = k so that du d = k. Thus, = log e (u) and d du = u. d B the chain rule, d = du d d du d d = k when u = k. k d So d =. B using the chain rule it can also be shown that d d (log e (g())) = g () g(). 68 MATHS QUEST MATHEMATICAL METHODS VCE Units and 4

4 WOrKED EXAMPLE think In summar, d d (log e ()) = d d (log e (k)) = d d (log e (g())) = g () g() Note: The above rules are onl applicable for logarithmic functions of base e. Find the derivative of each of the following. a log e () b log e (!) c log e (sin()) d log e ( 7 + 6) e log e ( ) a Use the rule d d (log e (k)) = to differentiate the function. WritE a d d ( log e ()) = Simplif the answer. = f ( + ) log e ( ) b Rewrite the function using! = a. b log e (!) = log e b Simplif the function b appling log laws. = log e () Differentiate the function and simplif. c Use the rule d d (log e (g())) = g () g() to differentiate the function. State g() and g (). Substitute g() and g () into the derivative rule. = log e () d d a log e ()b = c If g() = sin(), g () = cos() = d d (log cos() e (sin())) = sin() or tan() Topic 9 LOgArITHMIC functions USIng CALCULUS 69

5 d Use the rule d d (log e (g())) = g () g() to differentiate the function. State g() and g (). Substitute g() and g () into the derivative rule. e Use the quotient rule to differentiate the function. Identif u and v. Note: Simplif the logarithmic function b appling log laws. Differentiate u and v. Substitute the appropriate functions into the quotient rule. d If g() = 7 + 6, g () = 7 d d ( 7 + 6) = e If = log e ( ), let u = log e ( ) = log e () and let v =. du d = dv d = v du d d = d udv d v ( ) (log e ( )) = ( ) ( ) (log e ( )) 4 Simplif. = ( ) f Use the product rule to differentiate the function. Identif u and v. Differentiate u and v. Substitute the appropriate functions into the product rule. 4 Simplif where possible. = log e ( ) ( ) f If = ( + ) log e ( ), let u = + and let v = log e ( ). du d = dv d = d d = udv d + vdu d = ( + ) + log e ( ) = ( + ) ( ) + log e ( ) Questions ma also involve the differentiation of logarithmic functions to find the gradient of a curve at a given point or to find the equations of the tangent at a given point. 7 MATHS QUEST MATHEMATICAL METHODS VCE Units and 4

6 WORKED EXAMPLE THINK The graph of the function f() =.5 log e ( ) is shown. a State the domain and range of f. b Find the value of the constant a given that (a, ) is the -ais intercept. c Find the gradient of the curve at (a, ). d Find the equation of the tangent at (a, ). a State the domain and range of the function. WRITE a Domain = (, ). Range = R. b To find the -intercept, f() =. b.5 log e ( ) = Solve.5 log e ( ) = for. log e ( ) = e = = = Answer the question. (a, ) (, ) a = c Determine the derivative of the function. c f () =.5 log e ( ) f () = = ( ) Substitute = into the derivative to find the gradient at this point. d State the general equation for a tangent. f () = ( ) = The gradient at = is. d The equation of the tangent is = m T ( ) State the known information. The gradient of the tangent at (, ) = (, ) is Substitute the values into the general equation. m T =. = ( ) 4 Simplif. = = =.5 log e ( ) (a, ) Questions involving the derivative of the logarithmic function ma involve maimum/ minimum applications. Topic 9 LOGARITHMIC FUNCTIONS USING CALCULUS 7

7 WOrKED EXAMPLE The graph of the function = e log e () is shown. = e log e think EErCisE 9. PraCtisE Work without Cas = (a, b) Use calculus to find the values of the constants a and b, where (a, b) is the local minimum turning point. Give our answer correct to decimal places. WritE Determine the derivative of the function. = e log e () d d = e The minimum turning point e occurs where d = d =. e = Use CAS to solve for. = Find the corresponding -value. When =.567, = e.567 log e (.567) =. 5 Write the answer. a =.567, b =. The derivative of f() = log e () WE Differentiate the following functions with respect to. a 7 log e a b b log e ( + ) c sin() log e ( ) d log e ( ) e log e (e e ) f " log e ( ) Differentiate the following functions with respect to and state an restrictions on. a = 5 log e () b = log e a b c = log e a + + b d = log e ( 6) 7 MATHS QUEST MATHEMATICAL METHODS VCE Units and 4

8 Consolidate Appl the most appropriate mathematical processes and tools WE The graph of the function f : (, ) R, f() = log e ( ) is shown. a State the domain and range of f. (a, ) b Find the value of the constant a, given that (a, ) is the -ais intercept. c Find the equations of the tangent at (a, ). d Find the equation of the line perpendicular to the curve at (a, ). = 4 Find the equation of the tangent to the curve = 4 log e ( ) at the point where the tangent is parallel to the line 6 + =. 5 WE The graph of the function f : R + R, f() = + log e () is shown. = + log e Use calculus to determine the coordinates of the minimum turning point. = 6 Use calculus to determine the local maimum or minimum value of the function defined b: a f() = log e (), > b f() = log e (), > c f() = log e a b, >. In each case, investigate the nature of the turning point to determine whether it is a maimum or a minimum. 7 Find the derivative of each of the following functions. a 4 log e a b b log e (! ) c log e ( + 7 ) d 6 log e (cos()) e " log e ( + ) f log e () e + 8 Find the derivative of each of the following functions. State an restrictions on. a ( + 7) log e ( ) b sin() log e ( ) c log e () ( ) d log e a 4 + b 9 Find the gradient of each of the following functions at the specified point. a = log 5 (); = 5 c = log 6 ( ); = b = log ( + ); = Find the equation of the tangent to each of the given curves at the specified point. a = log e ( ) at a, b b = log e () at (e, ) c = log e ( ) at (e, ) = log e ( ) Topic 9 Logarithmic functions using CALCULUS 7

9 The graph of the function defined b the rule = log e () is shown. a Find the derivative of with respect to. b Find the equation of the tangent = log e ( ) at a e, eb. ), ) The line = is a tangent to the curve = log e ( ) + b, where b is a constant. Find the possible value of b. The equation of a line perpendicular to the curve = log e (( )) has the equation = + k, where k is a constant. Find the value of k, correct to decimal place. 4 The graph of the function f : R R, f() = log e ( + ) is shown. a Determine the coordinates of the -intercepts, correct to decimal places. b Find the equations of the tangents at the -ais intercepts. c Find the coordinates of the minimum turning point. Give our answer correct to 4 decimal places. = = 5 a The function f is defined b f : [, ) R, f() = ( log e ()), and the function g is defined b g : [, ) R, f() = log e (). Find the coordinates of the points of intersection between f and g. b Find the gradient of each graph at the point (, ). c Sketch both graphs on the same set of aes. d For what -values is log e > ( log e ())? 6 The graph of the function f : R + R, f() = 8 log e () = = log e ( + ) is shown. a Find the coordinates of the -intercepts, giving our answers correct to 4 decimal places. b Find the gradient of the curve at the points found in part a, giving our answers correct to decimal places. = c Find the equation of the tangent at (, ) and the equation of the line perpendicular to the curve at (, ). d Show that the coordinates of the maimum turning point are a, log e () b. = 8 log e 74 MATHS QUEST MATHEMATICAL METHODS VCE Units and 4

10 MastEr 7 a The function defined b the rule f : (, ) R, f() = log e ( ) has an inverse function, f. State the rule, domain and range for f. b Find the coordinates of the point(s) of intersection of f and f. Give our answers correct to 4 decimal places. c On the one set of aes, sketch the graphs of f and f, showing all relevant features. 8 The tangent to the curve = log e ( ) at = n intersects the -ais at =.5. Find the value of the integer constant n. 9. Units & 4 AOS Topic 4 Concept Rules for common antiderivatives Concept summar Practice questions WOrKED EXAMPLE 4 think The antiderivative of f() = We have found previousl that if = log e (), then d d =. Therefore, Also, a Remove as a factor. d = log e () + c, > a + b d = a log e (a + b) + c, > b a a Find d. b Evaluate 4 + d. WritE d = d Appl the integration rule. = log e () + c, > 4 b Remove 4 as a factor. + d = 4 + d Appl the integration rule. = c4 log e ( + ) d = log e ( + ) 4, >, > Substitute the end points and evaluate. = ( log e (() + )) ( log e (() + )) = log e (7) log e () = log e (7) Topic 9 LOgArITHMIC functions USIng CALCULUS 75

11 Integration b recognition Integration b recognition is used when we want to antidifferentiate more comple functions that we don t have an antiderivative rule for. This method involves finding the derivative of a related function and using this derivative to find the antiderivative. WOrKED EXAMPLE 5 Differentiate = 4 log e () and hence find log e ()d. think Use the product rule to differentiate the given function. Epress the answer in integral form. Note: There is no need to include +c as the question asked for an antiderivative. WritE = 4 log e () Let u = du, so 4 d =. Let v = log e (), so dv d =. d d = udv d + vdu d = 4 + log e () = 4 + log e () a 4 + log e ()bd = 4 log e (), > Separate the two parts of the integral. a 4 bd + a log e ()bd = 4 log e () 4 Subtract a bd from both sides to make 4 log e ()d the subject. (Remember we are determining log e ()d). 5 Antidifferentiate the function on the right side of the equation, 4. 6 Remove as a factor so that the function to be integrated matches the one in the question. log e ()d = 4 log e () a 4 bd log e ()d = 4 log e () 8 log e ()d = 4 log e () 8 7 Multipl the equation through b. log e ()d = a 4 log e () 8 b 8 State the answer. log e ()d = log e () 4, > 76 MATHS QUEST MATHEMATICAL METHODS VCE Units and 4

12 Area under and between curves Evaluating the area under a curve or the area between curves can also involve logarithmic functions as part of the method. WOrKED EXAMPLE 6 think The graph of =, > is shown. a Find the value of the constant a given that (a, ) is the -ais intercept. b Find the area between the curve and the -ais from = a to = 5. c Find the equation of the straight line that joins the points (, ) and (a, ). d Find the other point of intersection between the line and the curve. e Use calculus to find the area between the curve and the line to decimal places. a The -intercept is found b equating the function to zero. b State the integral needed to find the area under the curve from 7 to 5. Remember to account for the region being underneath the -ais. Antidifferentiate and evaluate. WritE a = = = ( ) = 6 7 = = 7 Therefore, a = 7. 5 b A = a bd 7 = log e ( ) (a, ) 5 = = = clog e (5 ) (5) alog e a 7 b a7 b b d = clog e () 5 log e a b + 7d Simplif. = log e () 8 log e ( ) 4 = log e () 8 + log e () 4 = log e () 84 = log e () + 8 units Topic 9 LOgArITHMIC functions USIng CALCULUS 77

13 c Find the equation of the line joining c (, ) = (, ) and (, ) = a (, ) and a 7, b, d Solve the two equations simultaneousl b equating the two equations. Identif the required -value and find the corresponding -value. m = 7 = 7 = m( ) = ( ) 7 = 7 + d + = 7 = 7 4 = 7 8 ( ) = 7 8( ) + 6 = = = ( 7)( 9) = 7, 9 = 9 = = 7 7, b State the answer. The point of intersection is a9, 7 b. e Write the rule to find the area between the curves from = 7 to = 9. 9 e A = a + a = a b d bb d Antidifferentiate. = c 4 log e ( ) + 4d Use CAS technolog to find the area. Note: If ou are asked to use calculus, it is important that ou write the rule and show the antidifferentiation, even when ou are using CAS technolog to find the result. The area between the curves is 7.4 units MATHS QUEST MATHEMATICAL METHODS VCE Units and 4

14 WOrKED EXAMPLE 7 Using calculus, find the area enclosed between the curve =, the -ais and the lines = and =. WritE Sketch a graph of the required area. think State the integral needed to find the area under the curve from = to =. The integral of = is = log e (). Negative values cannot be substituted, so smmetr must be used to find the area. = log e () units EErCisE 9. The antiderivative of f() = PraCtisE Work without Cas Questions 5, 7, 8 WE4 a Find 5 d. b Evaluate 4 d. a Find + d. b Find f() if f () = and f() = 4. A = A = a bd a b d 4 Antidifferentiate and evaluate. = Clog e ()D = log e () log e () = WE5 Differentiate = log e (cos()) and hence find an antiderivative of tan()d. 4 Differentiate = (log e ()) and hence evaluate e 4 log e d. Topic 9 LOgArITHMIC functions USIng CALCULUS 79

15 Consolidate Appl the most appropriate mathematical processes and tools 5 WE6 Consider f() =, >. + a Find the value of the constant a, where (a, ) is the -ais intercept. b Find the area between the curve and the -ais from = a to =. c A straight line given b = + intersects = f() in two places. What are 4 the coordinates of the points of intersection? d Use calculus to find the area between the curve and the line. 6 The graphs of =, > and + = + ( + ) = +, > are shown. ( + ) (, ) a Find the coordinates of the point of intersection of the two graphs. (, ) = + b Using calculus, find the area enclosed between the curves and the line =. Give our answer correct to 4 decimal places. 7 WE7 Using calculus, find the area enclosed between the curve =, the -ais and the lines = 4 and =. 8 Using calculus, find the area enclosed between the curve = +, the -ais and the lines = and =. 9 Antidifferentiate the following. a 4 c + + Evaluate: 4 b d + cos(4) a d b + 4 d 4 c ae + bd, correct to decimal places. a Given that d d = and = when =, find an epression for in terms of. b Given that d d = 5 and = when =, find an epression for in 5 terms of. = 8 MATHS QUEST MATHEMATICAL METHODS VCE Units and 4

16 If f() = log e (m), find f () and hence find log e (m)d where m is a constant. Differentiate log e () and hence find an antiderivative for log e (). 4 Differentiate log e ( 4) and hence evaluate 5 Differentiate log e (e + ) and hence find 5 6 The graph of = is shown. 5 + a Find the eact coordinates of the minimum and maimum turning points. b Find the derivative of log e (5 + ) and hence find an antiderivative for 5 +. c Find the area enclosed between the curve, the -ais, the line where equals the -coordinate of the maimum turning point, and the line = 6. 7 The graph of = 5 is shown. + The tangent to the curve at =.5 is also shown. a Find the equation of the tangent to the curve at =.5. b Find the derivative of log e ( + ) and 5 hence find an antiderivative for +. c Using calculus, find the area of the shaded region. Give our answer correct to 4 decimal places. 8 The graph of the function = + 4 is shown. The tangent to the curve at = is also shown. a Find the equation of the tangent to the curve at =. b Find the area of the shaded region. e 4 d. d, correct to 4 decimal places. e + = 5 + = 5 + = + 4 Topic 9 Logarithmic functions using CALCULUS 8

17 Master 9.4 Interactivit Area under a curve int The graph of the function f : (, ) R, f() = log e ( ) is shown. a State the domain and range of f. b Find the value of the constant a, given that (a, ) is the -ais intercept. c Find the area between the curve and the -ais from = a to = 5, correct to 4 decimal places. The graph of the function f : (, ) R, f() = 5 log e ( ) is shown. a Find the area of the shaded region, correct to decimal places. b Find the rule for the inverse function, = f (). c Verif our answer to part a b finding the area enclosed between the curve = f (), the -ais and the line = 5 log e (). Applications (4, ) 6 Application problems can involve real-life applications of logarithms. We cannot antidifferentiate a logarithmic function without technolog, so if we want to find the area under a logarithmic curve, we require another method. One option is to use integration b recognition. Another is to link the areas bound b the curve of the inverse of the required function and the aes. To find the inverse of a function, all components relating to of the original function will relate to of the inverse. Similarl, all components relating to of the original function will relate to of the inverse. This is also true for areas bound b the curve and the aes. If f() = log e (), f () = e. The area bound b the curve of f() and the -ais from = to = is shown. This area is equivalent to the area bound b the curve of f () and the -ais from = to =. = f () = f() = 5 log e () = log e ( ) (a, ) = = f() = 8 MATHS QUEST MATHEMATICAL METHODS VCE Units and 4

18 WOrKED EXAMPLE 8 The graph of the function f : c, b R, f() = log e () is shown. a Find f (). b Calculate f ()d. c Hence, find the eact area of the shaded region. think a To find the inverse, swap and, and solve for. WritE a Let = f(). Swap and : = log e () e = = e f () = e b Set up the appropriate integral and antidifferentiate. b f ()d = = c e d e d Evaluate. = e e c The required shaded area is f()d, the blue area. This is equivalent to the area bound b the curve of f () and the -ais from = to =. e c, ( ) = e = f () = f() e ( e ), ( e ) = f(), (, ) e To find the area bound to the -ais, the green shaded area, we need to find the area of the A rectangle = = rectangle with coordinates (, ), (, ), a, e b, and (, ). Topic 9 LOgArITHMIC functions USIng CALCULUS 8

19 Subtract the area underneath f (), from = to = (worked out in part b). This answer is the required green shaded area. A = A rectangle f ()d = a e b = 4 e 4 State the answer. f()d = 4 e units Eercise 9.4 PRactise Work without CAS Question Applications WE8 The graph of the function f : c, b R, 4 f() = log e (4) is shown. a Find f (). log e 8 b Calculate f ()d. c Hence, find the eact area of the shaded region. Part of the graph of the function h : ( 5, ) R, h() = log e ( + 5) + is shown. a Find the coordinates of the aial intercepts. b Find the rule and domain for h, the inverse of h. c On the one set of aes, sketch the graphs of = h() and = h (). Clearl label the aial intercepts with eact values and an asmptotes. d Find the values of, correct to 4 decimal places, for which h() = h (). e Find the area of the region enclosed b the graphs of h and h. Give our answer correct to 4 decimal ploaces. e = log e ( + 5) + ( 4, ) = f() (, log e (8)) = 5 (, ) 84 MATHS QUEST MATHEMATICAL METHODS VCE Units and 4

20 Consolidate Appl the most appropriate mathematical processes and tools Part of the graph of the function f : R + R, f() = log e () is shown. a Find the coordinates of the point where the graph intersects the -ais when >. b Find the derivative of log e (). c Use our answer to part b to find the area of the shaded region, correct to decimal places. 4 Part of the graph of the function g : c 5, b R, g() = log e (5) is shown. a Find the coordinates of the point where the graph intersects the -ais. b If = log e (5) +, find d d. = log e (5) (, ) c Use our result from part b to find the area of the shaded region. 5 Let h be the graph of the function h : D R, h() = log e ( 4), where D is the largest possible domain over which h is defined. a Find the eact coordinates of the aial intercepts of the graph = h(). b Find D as the largest possible domain over which h is defined. c Use calculus to show that the rate of change of h with respect to is alwas negative. d i Find the rule for h. ii State the domain and range of h. e On the one set of aes, sketch the graphs of = h() and = h (), clearl labelling intercepts with the - and -aes with eact values. Label an asmptotes with their equations. 6 a If = log e (), find d d. Hence find the eact value of log e ()d. b If = (log e ()) m where m is a positive integer, find d d. c Let I m = e (log e ()) m d for m >. Show that I m + mi m = m e. d Hence, find the value of e (log e ()) d. = log e () e (, ) Topic 9 Logarithmic functions using CALCULUS 85

21 7 The graph of = 4 log e ( + ) is shown. a Find the equation of the tangent to the curve at =.5. b Differentiate = ( ) log e ( + ) with respect to. c Hence, find the area enclosed between the curve, the -ais and the lines =.5 and =.5. 8 The graph of the function m : R R, m() = log e ( + ) is shown. a Find the gradient of the curve at =. b Determine the area enclosed between the curve, the -ais and the lines = and =, correct to 4 decimal places. c Use our result to part b to find the area enclosed between the curve, the -ais and the lines = and =, correct to 4 decimal places. 9 Bupramorphine patches are used to assist people with their management of pain. The patches are applied to the skin and are left on for the period of a week. When a patient applies a patch for the first time, the amount of morphine in their blood sstem can be modelled b the equation C = 5 log e ( +.5t), where C mg is the amount of morphine in the subject s blood sstem and t is the time in das since the patch was applied. a Find the concentration of morphine in the patient s blood sstem, C 7, seven das after the patch was applied to the skin, correct to decimal place. b Sketch the graph of C versus t. c Find the rate at which the morphine is released into the blood sstem after three das. d Use the inverse function of C to determine the total amount of morphine released into the patient s blood sstem over the seven das. That is, find 7 5 log e ( +.5t)dt, correct to decimal place. The shaded area in the diagram is the plan of a mine site. All distances are in kilometres. Two of the boundaries of the mine site are in the shape of graphs defined b the functions with equations f : R R, f() = e and g : R + R, g() = log e a b, where g() is the inverse function of f(). = 4 log e ( + ).5.5 (, ) = log e ( + ) 4 = e = = log e ( ) (, ) = 4 86 MATHS QUEST MATHEMATICAL METHODS VCE Units and 4

22 Calculate the area of the region bounded b the graphs of f and g, the -ais and the lines = and = 4. Give our answer correct to decimal place. A patient has just had a medical procedure that required a general anaesthetic. Five minutes after the end of the procedure was completed, the patient starts to show signs of awakening. The alertness, A, of the patient t minutes after the completion of the procedure can be modelled b the rule A = 4.6 log e (t 4). The graph of the function is shown. A a Find the value of the constant a, given that (a, ) is the -ais intercept. A = 4.6 log e (t 4) b When the patient has an alertness of 5, the are allowed to have water to sip, and 5 minutes later the can be given a warm drink and something to eat. How long does it take for the patient to reach an alertness of 5? Give our answer correct to the nearest minute. c Find the rate at which the alertness of the patient is changing minutes after the completion of the medical procedure. (a, ) t d Use the inverse function of A to determine the total change of alertness for minutes after the completion of the medical procedure. That is, calculate the area between the curve and the t ais from t = 5 to t =. The graphs of the functions g : R + R, g() = log e () and = log e () h : (, ) R, h() = log e ( + ) are shown. a Find the coordinates of the point of = log e ( + ) intersection of the two graphs to the right of the origin. Give our answer (, ) correct to 4 decimal places. b Find the area enclosed between the curves between the points of intersection found in part a, correct to 4 decimal places. Topic 9 Logarithmic functions using CALCULUS 87

23 Master At the Roal Botanical Gardens, a new area of garden is being prepared for native Australian plants. The area of garden has two curved walking paths as borders. One of the paths can be modelled b the rule f() = e ( ) +. a The other curved walking path is defined b the rule for the inverse of f, f. State the rule for f. = f () = e ( ) + The other borders are given b = 5 and = 5 as shown. The remaining border is formed b the -ais, as shown. All measurements are in metres. b Determine the respective ais intercepts of the graphs of f and f. c Find the area of the garden above the -ais, as shown in the diagram below, b calculating 5 5 ( ) ae + bd ( log e ( ) + )d. 5 5 e.5 + = f () Give our answer correct to decimal places. f () = e ( ) + 5 f () 5 88 MATHS QUEST MATHEMATICAL METHODS VCE Units and 4

24 d Find the area of the garden below the -ais, as shown in the diagram below, correct to decimal places. 5 e Hence, determine the total area of the garden correct to decimal place. 4 A oung couple are participating on a realit television show in which the are renovating an apartment. The have commissioned an up-andcoming artist to create a modern art piece to be featured in their living/ dining room. The artist has decided to use eponential and logarithmic curves as well as some straight lines to create the art piece. The curves and lines are shown along with the colour sketch for the finished piece. 5 5 = f () = f() 5 = f () Topic 9 Logarithmic functions using CALCULUS 89

25 (.94,.9) = = e + = 4 II = e + = log e ( + ) a Find the coordinates of the points of intersection between each of the following pairs of graphs. Give our answers correct to decimal place. i = e + and = e + ii = e + and = log e ( + ) iii = and = e + iv = and = log e ( + ) v =.94 and = log e ( + ) vi = and = e + b Calculate, correct to 4 decimal places, the area of: i region I ii region II iii region III. III I 9 MATHS QUEST MATHEMATICAL METHODS VCE Units and 4

26 ONLINE ONLY 9.5 Review the Maths Quest review is available in a customisable format for ou to demonstrate our knowledge of this topic. the review contains: short-answer questions providing ou with the opportunit to demonstrate the skills ou have developed to efficientl answer questions without the use of CAS technolog Multiple-choice questions providing ou with the opportunit to practise answering questions using CAS technolog ONLINE ONLY Activities to access ebookplus activities, log on to Interactivities A comprehensive set of relevant interactivities to bring difficult mathematical concepts to life can be found in the Resources section of our ebookplus. Etended-response questions providing ou with the opportunit to practise eam-stle questions. a summar of the ke points covered in this topic is also available as a digital document. REVIEW QUESTIONS Download the Review questions document from the links found in the Resources section of our ebookplus. studon is an interactive and highl visual online tool that helps ou to clearl identif strengths and weaknesses prior to our eams. You can then confidentl target areas of greatest need, enabling ou to achieve our best results. Units & 4 Logarithmic functions using calculus Sit topic test Topic 9 LOgArITHMIC functions USIng CALCULUS 9

27 9 Answers Eercise 9. a c e a b c d 7 sin() + cos() log e ( ) d (e + ) e d d = 5, (, ) d d =, (, ) b f ( + 4) + ( ) log e ( ) ( ) ( )" log e ( ) d d =, (, ) (, ) ( + )( + ) d d =, (, ) (, ) 6 a Domain = (, ), range = R b f(a) = log e (a ) c = a + log e(a ) d log e ( ) 4 Tangent: = log e () 6 5 Minimum turning point at (., log e ()) 6 a Local minimum at a e, e b 7 a b Local maimum at a e, e b c Local maimum at a e, e b c e f ( + )" log e ( + ) e + 4e log e () (e + ) b 4( ) d 6 tan() 8 a ( ) log e ( ) + ( + 7), a, b cos() log e ( ) + sin() b, (, ) log e () + log e () c, (, )/{} ( ) d 6, (, 4) (4 )( + ) 9 a b c d d = d ; at = 5, log e (5) d = 5 log e (5) d d = d ; at =, log e ()( + ) d = 9 log e () d d = d ; at =, log e (6)( ) d = log e (6) a = b = e c = e d a d = b = k = a =.84,.795 b = 4 e + e b At (.84, ); = At (.795, ); = c Minimum turning point = (.4,.484) 5 a (, ) and (e.5, ) b For f : at (, ), d d =. c For g : at (, ), d d =. = = ( log e ()) (, ) d { : < < e.5 } 6 a =.47,.864 = log e () b At (,47, ), d d = 7.9; at (,864, ), d d = 6.5. c Tangent: = Perpendicular line: = 7 or 6 = d The turning point occurs where d d =. 9 MATHS QUEST MATHEMATICAL METHODS VCE Units and 4

28 8 = 8 = 4 = ( )( + ) = =, but > =, = a b 8 log e a b = 8 log e ( ) = log e () The maimum turning point is at a, log e () b. 7 a f : R R, f () = e (+) ; domain = R, range = (, ) b ( , ), and (.7467,.7467) c 8 n = = f ʹ() (.86, ) Eercise 9. a (,.95) (.95, ) = f() (,.86) log 5 e () + c, > b log 4 e a a + log e () + + c, > b f() = 4 4 log e (), > tan()d = log e (cos()) 4 d d = log 4 e () and log e ()d = e 5 a a = b log e (4) c Q, R, a, 4 b d 6 6 log e() 6 a a 5 +!, 6 b units 6! + b or a 5 +! 6 b,! b 7 log e () units 8 log e a b + 9 a 4 log e () + c, > b 4 log e (4 + 7) + c, > 7 4 c + + log e () + + c, > d log e ( ) + sin(4) + c, < 4 a log e a 7 b b log e () c a = 5 log e (( + )) + b = 5 log e ( 5) + f () = log e (m) + and log e (m)d = log e (m) + c f () = + log e () and log e ()d = log e () 4 d d = 9 4 and 4 d = 9 log e a77 b 5 d d = e (e + ) and e d =.695 e a Maimum turning point = (!5,!5), minimum turning point = (!5,!5) b 5 log e (5 + ) c 5 log e a 4 b units 7 a = 4 or 5 = b 5 log e ( + ) c.695 units 8 a = 4 b 4 + log e () units 9 a Domain = (, ), range = R b a = c 5.94 units a units b f () = e +, R c units Eercise 9.4 a f () = 4 e b 7 units c 4 log e (8) 7 units a (, log e 5) and (e.5 5, ) b h : R R, h () = e ( ) 5 Topic 9 Logarithmic functions using CALCULUS 9

29 c See figure at foot of this page.* d = , e 7.76 units a a, b b d d = log e () + c.67 units 4 a a 5, b b d d = log e (5) c log e () 9 5 units 5 a (, log e ()) and a 4, b b D = a, b c d d = where > alwas, since <, so d d < alwas. d i h () = 4 ( e ) e 6 a b ii Dom = R, range = a, b = h() (, 4 (log e (), ) ( = ( (, log e ()), 4 ( = = h () d d = log e () + and log e ()d = e + d d = (log e ()) m + m(log e ()) m * c = 5 = log e ( + 5) + e ( log e (5) +, ) = () h () (, e 5) c Check with our teacher. e d (log e ()) d = e a = ( 4 log e () 4) b d d = 4 log e ( + ) + ( ) = 4 log e ( + ) + c.5 log e () units 8 a.8 b.458 units c 6.87 units 9 a C 7 = 7.6 mg b c C C = 5 log e ( +.5t) 7 t dc dt = 5 mg/da d 6.4 mg (7, 7.6) 5. km a a = 5 b minutes c units/minute d units a (.57,.97) b.796 units a f () = log e ( ) + b (, e + ) and (e +, ) c 6.58 m d 8.9 m e 45.5 m 4 a i (.4,.) ii (.9,.) iii (., 5.7) iv (.,.7) v (.4,.7) vi (.,.) b i A I =.55 units ii A II = units iii A III =.556 units ( e 5, ) (, log e (5) + ) ( ) = e 5 = 5 94 MATHS QUEST MATHEMATICAL METHODS VCE Units and 4

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