Mat 210 Updated on April 28, 2013

Transcription

1 Mt Brief Clculus Mt Updted on April 8, Alger: m n / / m n m n / mn n m n m n n ( ) ( )( ) n terms n n n n n n ( )( ) Common denomintor: ( ) ( )( ) ( )( ) ( )( ) ( )( ) Prctice prolems: Simplify using common denomintor: Function: Given f ( ) Given f( ) Given f ( ) C then f ( ) C, f() C, f ( h) C. h h then f ( h) f ( ) h ( h) ( h) ( h) then f ( h) f ( ) h h h h h h h h h h( h ) h Given f ( ) then f ( h) f ( ) h h h h h h h h h h h h h h h

2 Mt Brief Clculus Section. A Dsh of Limits (Pge# ) In this section we will formulte some importnt rules for limits. e Emple. (). Use conjecture to determine the limit lim. e Solution. We use clcultor to produce the tle of vlues. Consider F()= F() From the tle it ppers tht s gets closer to, the function F() gets closer to. e Therefore we conclude lim e Emple. (). Algericlly determine the limit lim. Solution. e lim = / / 6 lim lim( / / 6 higher order terms) Section. More on Limits (Pge# 8) Limits t infinity: we sy tht f() hs the limit l s tends to infinity if f() cn e mde ritrrily close to l y mking sufficiently lrge. We write lim f ( ) l. Limits tht do not eists: Emple. (). Consider the emple lim Solution. We use clcultor to produce the tle of vlues. Consider F()= F() Oserve tht ut. Thus limit does not eist Emple. (). Determine the limit of the rtionl function lim Solution. 7 9 lim Emple. (). Determine the limit of the rtionl function lim

3 Mt Brief Clculus 5 5 Solution. lim 7 9 lim Emple. (). Determine the limit of the rtionl function Solution. lim lim lim DNE, tht mens the limit does not eist. Emple. (4). Determine the limit of the function Solution. lim lim lim. Test the following limits: Eercise. (). lim ( ) Eercise. (). lim ( ) Eercise. (). lim Eercise. (4). lim 7 4 Eercise. (5). lim Eercise. (6). lim Eercise. (7). lim e Section. Continuity (Pge# 5) Definition: Geometriclly, function is continuous on n intervl if its grph is connected tht mens, it hs no reks (no holes, no jumps). One sided limit of function: Right hnd nd left hnd limit: The right hnd limit of f() t point is written s lim f ( ) nd the left hnd limit t is lim f ( ). h h

4 Mt Brief Clculus The limit of function t point eits if lim f ( ) = lim f ( ) nd we write lim f ( ) l, where l is finite numer. h h h Continuity of function in terms of limits: The function f() is continuous t point in n intervl if lim f ( ) f ( ), tht mens the function hs limit t the point h nd is equl to f(), the vlue of the function t. Otherwise the function is discontinuous. Emple. (). Let the function f() e defined for ll vlues of y, for f ( ), for Drw the grph of the function f() nd test the continuity from the grph. The function f() is discontinuous t = Becuse lim f ( ) nd lim f ( ) nd f ( ). In this cse lim f ( ) does not h h h eist. Emple. (). Determine the vlue of t which the function f ( ) is continuous. Solution. The domin of the function is (,). The function is not defined t =. Thus the function is continuous for ll vlues of in the intervl (,). See the grph of f().

5 Mt Brief Clculus Section.4 Intermedite Vlue Theorem. (Pge# ) Theorem: (Intermedite Vlue Theorem) Let f e function continuous in [, ] nd ssume tht f() nd f() hve different signs. Then there is t lest one c in (, ) such tht f(c) =, tht mens the grph of f crosses the is. This theorem is importnt in ssuring the eistence of solutions to some equtions tht cnnot e solved eplicitly. Emple.4 (). Prove tht the eqution 5e ( ) hs solution in the intervl (, 5). Solution. Consider tht f() = 5e ( ). Verify tht f() = -5, nd f(5) = 9.. By the Intermedite Vlue Theorem, since we hve f() nd f(5) hve different signs, it hs solution in the intervl (, 5). Note tht for the function f() which is strictly incresing/decresing with f() nd f() different signs hs ectly one solution in (, ). Emple.4 (). Prove tht the eqution 5e ( ) intervl (, 5), which is unique. hs solution in the Solution. In emple, we hve seen tht f() hs solution in (, 5). Oserve tht f '( ) 5e ( ) e 5e ( ) 5e ( ) for ll. The function is incresing. Section.5 The Derivtive (Pge # ) When we study the grph of function, we would like to hve precise mesure of the steepness of the grph t point. For the liner eqution y m, m is the slope. If m is lrge nd positive, then the line rises steeply from left to right; if m is lrge nd negtive, then the line flls steeply. For n ritrry function f, the steepness of the curve t prticulr point is the slope of the tngent line to the curve t tht point. f ( h) f ( ) The slope mpq of the secnt PQ is defined y m PQ is lso clled the h Newton quotient of the function f. Note tht when h =, the frction ecomes /, which is undefined. But choosing h = corresponds to letting Q = P. When Q moves towrd P long the grph of the function f, the coordinte of Q, which is +h, must tend to, nd so h tends to. Simultneously, the secnt line PQ tends to the tngent line to the grph t P. f ( h) f ( ) Thus the definition f '( ) lim h h is clled the slope of the tngent line t P to the grph of the function f. The numer f '( ) is lso clled the derivtive of the function f t the point.

6 Mt Brief Clculus The eqution of the tngent line to the grph of y f () t the point (, f()) is y f ( ) f '( )( ) Def: f '( ) the slope of the tngent line to the curve y f () t the point (, f ( )) f ( ) f ( ) Rise Slope of line thru (, f ( )) nd (, f ( )) is Run Emple.5 (). Determine the slope of the lines given elow.. y 5. y 5 c. 4 y 8 Answer:. -,. -, c. -/8 Emple.5 (). The grph of f () is given elow. Find f '( ), f '(). 5 8 Answer: f '( ) 8. 75, f '() Emple.5 (). Compute f '( ) nd f '( ) of the given functions.. f ( ). f ( ) 6 c. f ( ) Emple.5 (4). Find the slope of the tngent line to the grph of f t the specified points nd lso find the eqution of the tngent line.. f ( ) t (, ). f ( ) 6 t (, ) c. f ( ) t (, ) Section.6 Rtes of Chnge nd Incresing nd decresing functions (Pge# ) The derivtive of function is defined s the slope of the tngent line to its grph t prticulr point. Suppose y is function of, which is defined s y f (). Let us consider two different vlues of s +h nd, the corresponding y vlues re f(+h) nd f(). In economics the chnge in y per unit chnge in hs prticulr nme,

7 Mt Brief Clculus The verge rte of chnge of f over the intervl from to +h is f ( h) f ( ) defined s h f ( h) f ( ) The instntneous rte of chnge of f t is defined s f '( ) lim h h f '( ) The reltive rte of chnge of f t is defined s f ( ) For the following functions C() = cost of producing units R() = p = revenue from selling units, where p is the selling price per unit P()= R() C() = profit from producing nd selling units Now we hve the following functions C'( ) R'( ) P'( ) C( h) C( ) lim = Mrginl cost t h h R( h) R( ) lim = Mrginl revenue h h P( h) P( ) lim = Mrginl profit h h Incresing nd decresing of functions: If f ( ) f ( ) whenever, then f is incresing If f ( ) f ( ) whenever, then f is strictly incresing If f ( ) f ( ) whenever, then f is decresing If f ( ) f ( ) whenever, then f is strictly decresing Emple.6 (). Emine where f ( ).5 is incresing/decresing. Solution. Using definition one cn esily find tht f '( ), which is lwys for ll nd for ll. Thus f() is incresing in [, ) nd decresing in (-, ]. Oserve the grph of the function.

8 Mt Brief Clculus Emple.6 (). Emine where f ( ) / is incresing/decresing. Solution. Oserve tht f '( ) 4 ( )( ). Using sign digrm one cn see tht f '( ) 4 ( )( ) in the intervl [, ] nd f '( ) 4 ( )( ) in the intervl (,] [, ). Oserve the grph elow. Emple.6 (). If the cost function of firm is C ( ), give economic interprettion of the constnts. Solution. The constnt, is interpreted s the vrile cost or mrginl cost in dollrs per unit of production nd the constnt is interpreted s the fied cost in dollrs. Emple.6 (4). The totl cost of producing units of commodity is C ( ) 9 75,. Compute the mrginl cost function. For which vlue of is the mrginl cost the lest? Solution.. The mrginl cost is C '( ) 8 75,. The mrginl cost function C' ( ) decreses in [,] nd increses in the intervl [, ). Thus the mrginl cost is lest for =. See the grph of C '( ) 8 75,.

9 Mt Brief Clculus Section.7 Simple Rules for Differentition (Pge# 7) f ( h) f ( ) The derivtive of function f is defined y the formul f '( ) lim, if h h this limit eists, we sy tht f is differentile t. The process of finding the derivtive is clled the differentition. Other symols used in differentition. The derivtive of function f is lso represented df f ( h) f ( ) y Df f '( ) lim d h h Formul. The power rule. For the function of the type then f '( ). Derivtive of ny constnt is. f ( ), where is constnt, Emple.7 (). Determine the derivtive of the given power functions.. f ( ). f ( ) c. f ( ) +5 Solution.. f '( ). f '( ) / c. f '( ) 4 / Section.8 Sums, Products nd Quotients (Pge# 4) Following re the differentition rules for the functions with different opertions. F( ) F( ) F( ) f ( ) g( ), then F'( ) f '( ) g'( ) f ( ) g( ), then F'( ) f '( ) g( ) f ( ) g'( ) f ( ) g( ) f '( ) f ( ) g'( ), then F'( ) g( ) [ g( )] Emple.8 (). Differentite the functions (ddition/sutrction).. f ( ). f ( ) 5 c. f ( ) Solution.. f '( ). f '( ) c. f ( ) / ' 4 Emple.8 (). Differentite the functions (multipliction/division). 5. f ( ) ( ). f ( ) ( )(5) c. f ( ) 9 9 Solution.. f '( ) ( ) ( ). f '( ) (6 )5 ( ) 5 (6) ( 5) c. f '( )

10 Mt Brief Clculus Section.9 Chin Rule (Pge# 47) The chin rule is dy d dy du du d dy Emple.9 (). Find y chin rule. d. y 5 u nd u. Solution... c. dy d dy d dy dy du d du d f ( ) ( ) c. f ( ) u ( 5 ) 5 ( 9 9 ( ) (6 ) 6 ( Section. Higher Order Derivtives (Pge# 5) dy The symols of higher order derivtives. The first derivtive of y is or y ' d d y The second order derivtive is d d y The third order derivtive is d d y Emple. (). Find. d. y 5. f ( ) ( ) Solution.. df d. dy d ( ) d y d d y d ) , 9, 7 (), d f d ()( ) 9 ( ), d f d Section. Eponentil nd Logrithmic Functions (Pge# 56) Formul. The eponentil rule. For the function of the type constnt, then f '( ) ln. Derivtive of f ) 9 ( ) e is or y " or y nd so on. ()()(9)( ) f f 8 ( ( ), where is '( ) e. ) dy Emple. (). Find. d. f ( ). f ( ) e e 5

11 Mt Brief Clculus Solution.. df 9 df 5 6 ln,. e 5e d d Formul. For the logrithmic function f ( ) ln, Derivtive of f ( ) log is dy Emple. (). Find. d f '( ). ln f '( ).. f ( ) ln. f ( ) log df df Solution..,. d d ln Section. Implicit Differentition (Pge# 67) 5 Functions like y 5y 9y, where cnnot e written s independent vrile with respect to nother vrile y. In this cse direct differentition is either difficult or impossile. To find derivtive we will use implicit differentition using chin rule. Following emples will illustrte the sitution. dy Emple. (). Find or y'. d. y 5y 9. y log 5y 9 Solution.. 6yy' 5y 5y' 9 y' 6y 5. yy' y' ln y ln Emple. (). Find the slope of the tngent line to the curve y 7 t (, ). Also compute the second derivtive. 4 y 4() () 5 Solution. y ' y 4 y' () For the second derivtive we consider gin y ' y 4 4 6y' 6y 4 4 Tking derivtive we find y " y' y' 4 y" Derivtive formuls: d n n ( ) n d ( e ) d e d ( ) ln (ln ) d d d d d ( c ) d Rules: Product Quotient

12 Mt Brief Clculus ( fg) fg f g f f g g f f g fg g g f f g g d f g f g g d Chin Rule: ( ( ( )) ( ( )) ( ) Tngent line: y f ( ) f ( )( ) Derivtive Prctice Prolems: Set A d 9. ( ) d. d 4. ( ) d d d ( ) d e e e e d where (, f ( )) is the point on y f ( ) 8. d d ( ) () ln. (7 ) 7 ln 7() d d d d ( ) 6. d d ( ) d d ( ) d d.. ln ln e d 4 d ( ) d 8 d ( ) 4 4 d 9. e e d d 4 d ( ) 4 4 d 4 d 6. 5 e ( ) 5 e ( ) d d 6 d e d 7 7 d 7 7(ln ) e 8. ln 9. ln d e d d d 7 7 d d ln( ) ln. ln. e 7 e d d d Set B d 9 d. ( ) d. ( ) ln. ( e ) d d d d d d / 4. ( ) 5. ( ) / d d d d d / d d 6. ( ) ( ) 7. ( / d p ) 8. p p d d d d d d (9 7 ) (8 7) d

13 Mt Brief Clculus. p d p p p d ( ) d ln ln ln. d. d 6. d 5. ( ln ) 5 4 d ln d p p ln, p p d d d ln p 4. d ( ln ) ln d d d 4 4 ln d ( ). d d d d e e d ln ( ln ) d d. ln d d ln ln d. Section. Differentiting the Inverse (Pge# 74) If function y f ( ) is one-to-one (strictly incresing or strictly decresing) function over n intervl I, it hs n inverse function gover ( ) the rnge of y f ( ). A function, which is strictly incresing (or strictly decresing) in n intervl I, it is one-to-one. We hve the following importnt result: If the function y f ( ) is continuous nd strictly incresing (or strictly decresing) in n intervl I, then the function y f ( ) hs n inverse function y g( ), which is lso strictly incresing (or strictly decresing) in the intervl f() I, the rnge of y f ( ). If is n interior point of I nd y f ( ), then gis ( ) differentile t y f ( ) such tht g( y), where y f f ( ) ( ) y g( ) y ( y, ) y f ( ) (, y ) Emple. () Test the function f ( ) if it hs n inverse. Then find the inverse if ny. Consider point (,5) on the grph of the function f ( ). Find derivtive of the inverse function nd evlute t the point (5, ) on the inverse function.

14 Mt Brief Clculus Solution: The function f ( ) is strictly incresing; it is one-to-one (Verify the grph y horizontl line test). It hs inverse. y g( ) To find inverse follow the steps: Step. Write y y Step. Interchnge vriles: y y f ( ) Step. Solve for y: y Step 4. The inverse is f ( ) g( ) Now f ( ) g( ) We now verify y the formul g( y) g(5) f ( ) f () Emple. () The function f ( ) ln, (.68, ) hs n inverse function g(). Eplin why. And find g () e Solution: Grph the function using your clcultor nd find tht it hs minimum t.68 nd it is incresing on (.68, ). Tht is why the function hs n inverse. To find g () e, we first solve for from e ln e. Finlly we hve g() e f ( ) ln e Emple. () The function f ( ) ln hs n inverse function g(). Eplin why. And find g( e ). Find eqution of the tngent line to y = g() t (e+, e). Solution: The given function is incresing, it hs inverse. The eqution of the tngent line is e( e ) y e g( e )( e ) e ( e ) e f ( e) e 5 Emple. (4) The function f ( ) 6 hs n inverse function g(). Eplin why. Find g (7). Find eqution of the tngent line to y = g() t (7, ). Solution: Oserve tht 5 f ( ) 6 is one-to-one nd solving the eqution 5 7 6, we find. f() Thus (, 7) is point on the grph. g() Now g(7), f () (, 7) tngent line 4 where f ( ) (7,) The eqution of the tngent line to y = g() t (7, ) is

15 Mt Brief Clculus y ( 7) Note tht it is very esy to drw the grph for the inverse function, ut very difficult or impossile for us to find the function lgericlly. Look t the difficulty: Step. 5 y 6 5 Step. y y 6y Step. Solving for y is impossile lgericlly. But using the formul we re le to find derivtive of the inverse function t given point. Emple. (5) Given tht y f ( ) is one-to-one nd f(4) 5, f(4) /, find ( f ) (5) Solution: Oserve tht (4, 5) is point of y f ( ). Now ( f ) (5) f (4) Section. Liner Approimtions (Pge# 76) f ( h) f ( ) In chpter, we hve used the rule of derivtive f '( ) lim t point h h f ( ) f ( ) (, f()). We now use nother modified rule f '( ) lim s generl rule of h derivtive t point (, f()). From this generl form we omit the limit nd use the result f ( ) f ( ) s n pproimtion like f '( ). After simplifiction we find f ( ) f ( ) f '( )( ), which is clled the liner pproimtion (lso clled tngent line pproimtion) of the function f out =, ( is close to ). Emple. (). Prove tht for close to, nd illustrte this pproimtion y drwing the grphs of y nd y on the sme screen. We consider f ( ) f '( ) f '(), s Now using liner pproimtion formul we find f ( ) f ( ) f '( )( ) f ( ) f () f '()( )

16 Mt Brief Clculus Emple. (). Prove tht to find pproimtion of ( ) m m for close to nd use this pproimtion / ( 6.95) Emple. (). Use the liner nd qudrtic pproimtion of f ( ) to / ( 6.95). pproimte the vlue of / Solution. From f ( ) we hve f '( ) nd f "( ) 9 The Liner Approimtion t =, is f ( ) f ( ) f '( )( ) Consider = 7, then (7) / f 7, f '(7) (7) 7 / Then (6.95) (6.95 7) Rules for Differentils. The nottion d( f ( )) is clled the differentil of the function f. We hve the following rules. d( f ( ) g( )) df ( ) dg( ) or simply we write. d( f. d( f g) df. d( fg) gdf / g) fdg dg gdf fdg g Emple. (4). Find the differentils (p, q, nd r re constnts):. d( ). d( ln ) c. d( p q) r Solution:.. c. d( ) 6d d d( ln ) 6ln d d ( ) r r d p q r( p q) pd 5 /

17 Mt Brief Clculus Section.4 Why We Use Elsticity (Pge # 8) The elsticity of function y f ( ) with respect to is defined s E El f ( ) f ( ) f( ) Interprettion: The mening of E % is tht n increse of % in the price would led to decrese of % of demnd, if ll other fctors tht influence the demnd remined constnt. A Emple.4 () Find the elsticity of y f ( ), A is constnt, nd interpret your nswer. A / A 5/ Solution: y f ( ) A f ( ) A 5/ Now E El f ( ) f ( ).5 f( ) A An increse of % in the price the demnd would decrese y.5%, when other fctors re unchnged. Emple.4 () The reltion etween trffic volume nd ependiture of uilding rods.6 re given s T( K).4K, K is ependiture. Find the elsticity of T with respect to K. Find lso the consequences if ependiture increses y %. K K.6 Solution: E El f ( ) T( ).4.6K.6.6 T( K).4K The increse in % of ependiture would led to.6% increse in trffic volume. The % increse in price would led to.6% increse in trffic volume. Emple.4 () Prove tht p Solution: p E El f ( ) pel f ( ), where p is constnt. p El f ( ) p( f ( )) f ( ) pelf ( ) p ( f( )) Emple.4 (4) Find E El 5 Solution: ( ) E El ( ) ( )( ) Section. Single Vrile Optimiztion (Pge# 95) Introduction: Mimum nd Minimum Vlues of function.

18 Mt Brief Clculus The points in the domin of function where it reches its lrgest nd its smllest vlues re usully clled mimum nd minimum vlues. These re sometimes clled etreme vlues. Suppose the function f() domin D, then c D is mimum point for f iff f ( ) f ( c) for ll in D d D is minimum point for f iff f ( ) f ( d) for ll in D f(c) is clled mimum vlue of f t = c, nd f(d) is clled minimum vlue of f t = d. 8 Emple. (). Determine the points where the function f ( ) hs mimum 4 or minimum vlues. 8 Solution. The function f ( ) hs mimum vlue when the denomintor is lest. 4 Note tht the denomintor 4 is positive for ll vlues of in the domin. The denomintor is lest only if =. Therefore f() hs mimum vlue t = nd f()=8 is the mimum vlue. Section. Simple Tests for Etreme Points (Pge# 98) In this section we will study the sign of first derivtive of f() to determine mimum nd minimum vlues of the function. Also we will confirm our results y grphing the function. First derivtive test for mimum/minimum vlues of f(). If If f '( ) for c nd f '( ) for c, then c is mimum point for f. f '( ) for c nd f '( ) for c, then c is minimum point for f. Emple. (). Find possile etreme points for the function nd verify your result. f ( ) ln for (, ). Grph Solution. We hve f ( ) ln f '( ) ln / e.7 Now using rel line test for the sign of f (), we get f '( ) for.7, nd f '( ) for.7. By the first derivtive test we see tht c e /. 7 is the minimum point for f().

19 Mt Brief Clculus Second derivtive test (Concve nd conve functions) for mimum/minimum vlues of f(). If f() is concve in n intervl I nd c is sttionry point (tht mens f '( c) ) for f, then c is mimum point of f. If f() is conve in n intervl I nd c is sttionry point (tht mens f '( c) ) for f, then c is minimum point of f. Rememer tht If f "( ), f is concve nd if f "( ), f is conve. When f "( ) then f ' decreses nd f is concve ( like concve down) When f "( ) then f ' increses nd f is conve ( like concve up) Emple. (). Find possile etreme points for f ( ) e e using second derivtive test. Grph the functions nd verify your result. Solution. We f ( ) e e f '( ) e e f "( ) e 4e for ll. The function f() is conve. It hve minimum vlue when ln f '( ) e e. c, which is clled the sttionry point. Thus we hve the minimum vlue of f ( ) e e =.89 for c =.. Section. Locl Etreme Points (Pge# ) Definition: The function f hs locl mimum (or minimum) t point c, if there eists n intervl (, ) out c such tht f ( ) f ( c), ( or f ( ) f ( c)) for ll in (, ) which re in the domin of f. Function vlues corresponding to locl mimum (minimum) points re clled locl mimum (minimum) vlues.

20 Mt Brief Clculus The First Derivtive Test: Suppose c (, ) is sttionry point for y f ().. If f '( ) on left of c nd f '( ) on right of c, (tht mens f is incresing on left of c nd decresing on right of c.) then = c is locl mimum point of f.. If f '( ) on left of c nd f '( ) on right of c, (tht mens f is decresing on left of c nd incresing on right of c.) then = c is locl minimum point of f. The Second Derivtive Test: Suppose f e twice differentile function nd c (, ) is sttionry point for y f ().. If f '( c) nd f "( c) then = c is strict locl mimum point.. If f '( ), nd f "( c) then = c is strict locl minimum point.. If f '( ), nd f "( c) nd there is chnge of concvity out the point c, then = c is n inflection point. Emple. (). Clssify the sttionry points of f ( ). 9 6 Solution. We hve f ( ) f '( ), nd lso 9 6 f "( ) with f "( ), nd f "(). We conclude tht the sttionry point c = - is locl mimum point nd c = is locl minimum point. Emple. (). Determine the possile etreme points nd etreme vlues of the 5 function f ( ) 5. Solution. Hints. The sttionry points re c =,, c = is point of inflection (verify), c is locl mimum point nd c is locl minimum point. Emple. (). A function f is given y the formul f ( ) 6 ) Find the domin of f nd the open intervl where f is positive ) Find the possile etreme points c) Emine f s, nd d) Determine the limit of f( ) s. Hs f mimum or minimum in the domin? Solution:

21 Mt Brief Clculus ) The domin of f is D [ 6,) (, ). One my look t the grph of f or check the sign of f using test points to find intervl where f is positive. We produce oth elow. f() is positive on ( 6, ) (, ) From grph Using test point: From clcultor y = ( / ) ( 6) grph nd trce t the test points: Sign of f: ) f ( ) 6 6 In clcultor input y = ( 6) / ( / ) ( 6) grph nd find zeros nd lso vlues for which the derivtive is undefined to get sttionry points. The sttionry points re 6, -4 nd 6 (Note tht is not sttionry point). Now use line test for intervl of increse or decrese. The point = -4 is point of mimum, the point = -6 is minimum nd the point = 6 is lso minimum. c) f s, nd f s nd f s. When, f M/min hs discussed in prt ). Section.4 The Etreme Vlue Theorem (Pge# 9) If f is continuous function over ounded closed intervl [, ], then there is point d [, ] where f hs minimum, nd point c [, ] where f hs minimum, such tht f ( d) f ( ) f ( c) for ll [, ]

22 Mt Brief Clculus The Men Vlue Theorem If f is continuous function over ounded closed intervl [, ], nd differentile in the open intervl (, ) then there eists t lest one interior point c (, ), such tht f ( ) f ( ) f '( c) Emple.4 (). Find the mimum nd minimum nd drw the grph of f ( ) 4 4 8, [, 8]. Solution. Since the function is polynomil, it is continuous. Using etreme vlue theorem we see tht the given function hs oth m nd min within the given domin. We pply second derivtive test to emine m/min of the function. f ( ) f '( ) 8 4 f "( ) 8. Note tht f "( ) 8, so the function hs minimum t c = 5, where 5 [, 8]. Emple.4 (). Find ll the numers c [, 8] for the function f ( ) f ( ) f '( c). Solution. We hve f ( ) f ( ) 64 f '( c) c 8 c 4 Section.5 Business nd Economic Models (Pge# 6) f ( ) such tht In this section we consider rel life prolems from usiness nd economics. Emple.5 (). The revenue nd cost function of product re given elow: R ( Q) Q Q /, nd C( Q) 5 Q, Q [, ]. Find the vlue of Q tht mimizes profit. Find lso the profit. Solution. The profit function is P( Q) R( Q) C( Q) Q Q / 5 Q P'( Q) 8 Q /5 P"( Q) /5 By the second derivtive test we see tht P(Q) hs mimum vlue t Q =4. The mimum profit is P(Q) =. Section.6 Inflection points (Pge# 5) Definition: The point c is clled n inflection point for the function f if there eists n intervl (, ) out c such tht:. f "( ) in (, c) nd f "( ) in (c, ) or

23 Mt Brief Clculus. f "( ) in (, c) nd f "( ) in (c, ) Emple.6 (). Find locl etreme points nd inflection points for the function ln f ( ). ln ln ln Solution: Hints. We hve f ( ) f '( ). Find 4 f "( ) nd verify tht the sttionry point is c = e. 65 is locl mimum point. It does not hve point of inflection. Emple.6 (). Find locl etreme points nd inflection points for the function f ( ) ( ) e. Solution. We hve f ( ) ( ) e f '( ) e ( ) e. The only sttionry point is c = -. The point c = - is locl mimum point. The locl mimum vlue of the function is f ( ) ( ) e = e =.7. See the grph. Prctice Prolems for Chpter. Find ) f '( ) ) f "( ) c) All sttionry points d) Clssify sttionry points e) All locl mimum nd minimum points f) All locl mimum nd minimum vlues of the function g) All inflection points for the following functions:. f ( ) 5. f ( ). f ( ) f ( ) 5 5. f ( ) e 6. f ( ) ln f ( ) f ( ) 9. f ( ) 4 4. Verify etreme vlue theorem for the function f ( ) 4, [-, ].. Verify etreme vlue theorem for the function f ( ) 6, [, 8]. Chpter 4. Integrtion (Pge# 4) Differentition hs een the min topic in the preceding three chpters. Some prolems in mthemtics require finding function from informtion out its derivtive. This cn e regrded s inverse of derivtives. The process of reconstructing function from its derivtives is clled the integrtion.

24 Mt Brief Clculus Emple. Assume tht the mrginl cost function is C'( ) 5 nd it is lso known tht the fied costs re $. Find the cost function. To solve this prolem we need to know the inverse of derivtive to construct the cost function. All the sections in this chpter will e discussing different rules nd methods of deriving such functions. Section 4. Indefinite Integrls (Pge# ) In this section we will use only the derivtive of functions to derive the most generl form of the integrl. Emple 4. (). Determine the generl integrl form of f '( ) y inspection. Solution. Rememer tht the derivtive of of of f ( ) C, where C is constnt is f '( ) is f ) C (. f ( ) is f '( ). Also the derivtive f '( ). Therefore the generl integrl Definition nd Symol: Suppose f() nd F() re two functions of hving the property tht f ( ) F'( ). Here F' ( ) is clled the derivtive of F(), so F() could e clled the ntiderivtive of F '( ). We write F '( ) d f ( ) d F( ). The most generl form is F '( ) d f ( ) d F( ) + C, where C is constnt, sometimes clled integrl constnt. Rules of Integrls. d, where d ln C e d e C d C, where, ln e e d C Emple 4. (). Use derivtive rule to prove tht ln d ln C, for >. We consider f ( ) ln C, then f '( ) ln ln. From definition we know tht the integrl of f '( ) ln is f ( ) ln C. In symol we write

25 Mt Brief Clculus ln d ln C. Emple 4. (). Find the generl integrl of the following functions. ) d ) d c) d Solution. ) d 5 / d ) d c) d C 7 / 7 / C d / d d) C / d) d C Emple 4. (4). Find the generl integrl of the following functions. ) e 4 ( ) d ) d c) d d) d 4 4 e Solution. ) e d C 4 ) d C ln ( ) c) d d 4 d 4 4ln C d) d d d C ln( ) Section 4. Riemnn Sums. (Pge# 8) Section 4. Are nd Definite Integrls. (Pge# 4) This section will show how the concept of integrl cn e used to clculte the re of mny plne regions. The symol f ( ) d is clled the definite integrl which represents the re under the grph of f() nd the ordintes =, nd =. Emple 4. (). Find the re under the prol The required re is More detils will e given in the clss. f ( ) over the intervl [, ]. d F() F(), where F( ) the integrl of f(). Section 4.4 Properties of Definite Integrls (Pge# 49)

26 Mt Brief Clculus From the definition of definite integrl numer of properties hve een derived in this section. If f is continuous function in n intervl tht contins the constnts,, c, then f ( ) d f ( ) d f ( ) d kf ( ) d k f ( ) d, where k is constnt ( ) d f ( ) d c f f ( ) d f '( ) e f ( ) d e c f ( ) f ( ) Differentition with respect to the Limits of Integrtion From the definition of integrl we hve tht, if F' ( t) f ( t) then f ( ) d F( t) F( ) t d Now f ( ) d F'( t) dt f ( t) In other words, the derivtive of the definite integrl w. r. to the upper limit of integrtion is equl to the integrnd evluted tht point. This is lso known s the Fundmentl Theorem of Clculus Prt I. More generl form of Fundmentl Theorem of Clculus Prt I is like u( t) d f ( ) d F'( u( t)) u'( t) dt Emple 4.4 (). Evlute the following definite integrls. ) d ) e d Solution. ) Then d 5 / d F() F() d 7 7 c) d 7 / C, where d) t F( ) 7 d 7 /

27 Mt Brief Clculus ) ( ) e d e e, using formul f '( ) e c) d ln( ) C, where F ( ) ln( ) Then d F() F() ln 9 ln ln 9 / d) d C Then d F() F( ), where f ( ) d e F( ) Emple 4.4 (). Find the following derivtive of the limit of integrls. d ) d dt t Solution. ) d t dt t d t ) dy dt t y d t d d d ) dy dy dy dt y dt y dt t t y d dt t dy y d dt t dy y t t / t f ( ) f ( ) Section 4.5 Integrtion y Sustitution (Pge# 5) In this section we shll see how the chin rule for differentition leds to n importnt method for evluting mny complicted integrls. We illustrte this with emples. Some of these types of prolems cn lso e solved y the method of integrtion y prts. For Integrtion y sustitution we do not use ny formul. One only need to choose the right form of sustitution. Emple 4.5 (). Evlute e d Solution. We consider u du d u u Now e d e du e C e C Emple 4.5 (). Show tht 8 ( ) Solution. We consider u du 9 d 8 89 ( ) C

28 Mt Brief Clculus Now 8 ( ) d 8 u 9 du Emple 4.5 (). Evlute 4 d Solution. We consider u Now 4 u C 8 89 ( ) C udu d d udu ( 4 4 d 4 d (4u u ) du ( 4 ) 9 5 ) 5 C Section 4.6 Integrtion y Prts (Pge# 57) There re some functions, which cnnot e integrted y the rules we hve discussed in erlier sections. Integrtion of such function pper s multipliction or division of two or more functions. In this section we will discuss the integrtion of such functions nd its ppliction in usiness nd economics. Formul of integrtion y prts. We consider the formul nd illustrte y emples. f ( ) g( ) d udv uv vdu where u f ( ) du f '( ) d nd dv g( ) d v g( ) d Emple 4.6 (). Evlute e d Solution. We consideru du d, nd Now e d udv uv vdu e dv e d v e e d e e C m m m Emple 4.6 (). Show tht ln d ln C, where m m ( m ) m m Solution. We consider u ln du d, nd dv d v m Now m m m m m ln d udv uv vdu ln d ln m m m ( m ) Emple 4.6 (). Evlute ln( ) d Solution. We consider u ln( ) du Now let us find d, nd dv d v C

29 Mt Brief Clculus F( ) ln( ) d udv uv vdu ln( ) d ln( ) d ln( ) ln( ) 4 Using definite integrl rule we hve ln( ) d F() F( ) Section 4.7 Infinite Intervls of Integrtion (Improper Integrtion) (Pge# 6) Suppose f is function tht is continuous for ll. Then f ( ) d is defined for ech. If the limit of this integrl s eists nd is finite, then we sy tht f is integrle over [, ), nd we define f ( ) d lim f ( ) d The improper integrl f ( ) d lim f ( ) d l, numer l. Otherwise the improper integrl is sid to diverge. is sid to converge to finite We define lso f ( ) d lim f ( ) d nd lim c c c f ( ) d lim f ( ) d lim f ( ) d Emple 4.7 (). Determine e d, if converges. Solution. e d lim e d lim ( e e ), the result converges to. Emple 4.7 (). Determine Solution. e e d, if converges d lim lim e d e e

30 Mt Brief Clculus Section 4.8 Business nd Economic Applictions (Pge# 7) This section will show the importnce of integrls in usiness nd economics. We will consider rel prolems from usiness nd economics. Consumer nd Producer Surplus Economists re interested in studying how consumer s nd producer s welfre re influenced y chnges in economic prmeters. Consumer s surplus nd Producer s surplus: The supply nd demnd curves re plotted considering price p long y is nd quntity q long is. The symol p indictes the equilirium price nd q represents the equilirium quntity, where demnd nd supply curve intersects. CS is for consumer s surplus nd PS is for producer s surplus. Using the concept of integrtion, one cn esily find the CS is equl to the re under the demnd curve with ordinte q minus the rectngle with sides equl to equilirium price nd equilirium quntity. On the other hnd the PS is equl to the re of the rectngle with sides equl to equilirium price nd equilirium quntity minus the re under the supply curve, with ordinte q. At equilirium point E, demnd is equl to supply. The corresponding equilirium price p is the one which induces the consumers to purchse (demnd) precisely the sme mount tht the producers re willing to offer (supply) t tht price. According to the demnd curve in the figure, there re consumers who re willing to py thn p per unit. In fct, even if the price is lmost s high s p, some consumers still wish to uy some units t tht price. The totl mount sved y ll the consumers uying the commodity t price lower thn wht they mimlly re willing to py, is clled the CONSUMER S SURPLUS. Most producers lso derive positive enefit from selling t the equilirium price ecuse they re willing to supply the commodity t price lower thn the equilirium price. The mount of this enefit is clled the surplus ccruing to the producer. In our figure, even if the price is lmost s low s p, some producers re still willing to sell it for, we cll the PRODUCER S SURPLUS. p p p CS PS E Supply Curve Demnd Curve q

31 Mt Brief Clculus We hve the following formuls: Suppose S(q) nd D(q) re supply nd demnd curves respectively, then PS = q p q S( q) dq nd CS = D( q) dq pq q 4 Emple 4.8 (). The demnd nd supply curves re D ( q) q nd S ( q). q 6 Find consumer s surplus nd producer s surplus. 4 Solution. For equilirium D( q) S( q) q q q nd q 6 p D q ) S( q ) ( Now PS = And CS = q pq S( q) dq ( q ) dq 6 4 q D( q) dq p q Income Strem Emple 4 8 dq 4ln q 6 6 Totl vlue Future vlue 4 5dt 5*4 5, 4.4(4 t) 5e dt

32 Mt Brief Clculus The rte is 5% (where r =.5).