ustion 3: How do you find th rlativ trma of a function? Th stratgy for tracking th sign of th drivativ is usful for mor than dtrmining whr a function is incrasing or dcrasing. It is also usful for locating th rlativ trma of a function. At a rlativ trma, a function changs from incrasing to dcrasing or dcrasing to incrasing. Th numbr lins in th prvious qustion allow us to s ths changs by obsrving changs in th sign of th drivativ of a function. Whn th drivativ of a function changs from positiv to ngativ, w know th function changs from incrasing to dcrasing. As long as th function is dfind at th critical valu whr th chang occurs, th critical point must b a rlativ maimum. If th drivativ of a function changs from ngativ to positiv, w know th function changs from dcrasing to incrasing. In this cas, th critical point is a rlativ minimum as long as th function is dfind thr. If th drivativ dos not chang sign at a critical valu, thr is no rlativ trma at th corrsponding critical point. Th First Drivativ Tst summarizs ths obsrvations and hlps us to locat rlativ trma on a function. 0
First Drivativ Tst Lt f b a non-constant function that is dfind at a critical valu c. If f changs from positiv to ngativ at c, thn a rlativ maimum occurs at th critical point c, f( c ). If f changs from ngativ to positiv at c, thn a rlativ minimum occurs at th critical point c, f( c ). If f dos not chang sign at trma at th corrsponding critical point. c, thn thr is no rlativ Eampl 6 Find th Rlativ Etrma of a Function Find th location of th rlativ trma of th function f 3 ( ) 4 8 5 Solution Th first drivativ tst rquirs us to construct a numbr lin for th drivativ so that w can idntify whr th graph is incrasing and dcrasing. Using th ruls for drivativs, th first drivativ of th function f ( ) is d d d d f d d d d 3 ( ) 4 8 5 43 8 0 So th drivativ is ( ) 4 8. f Us Sum / Diffrnc Rul and th Product with a Constant Rul. Us th Powr Rul for Drivativs and th fact that th drivativ of a constant is zro
W nd to us this drivativ to find th critical valus. St th drivativ, ( ) 4 8, qual to zro to find thos valus. f 4 8 0 6 7 3 0 6 3 0 0 30 3 St th drivativ qual to zro Factor th gratst common factor from ach trm Factor th trinomial To find whr th product is qual to zro, st ach factor qual to zro and solv for th variabl In gnral, critical valus may also com from valus whr th drivativ is undfind. Sinc f ( ) is a polynomial, it is dfind vrywhr so th drivativ is dfind vrywhr. Although this drivativ could b factord to find th critical valus, most quadratic drivativs ar not factorabl. In this cas, th quadratic quation yilding th critical valus can b solvd using th quadratic formula. This stratgy would yild th sam critical valus as factoring: 4 8 0 4 4 48 4 900 4 4 30 4 3, a b c Idntify a, b, and c for th quadratic b b 4ac formula and put a in th valus Simplify th numrator and dnominator To find th critical valus with mor complicatd drivativs, w may nd to solv th quation using a graph. Th solution to th quation
4 8 0 can also b found by locating th intrcpts on th drivativ f ( ). = 0 = 0 3 f( ) 6 3 Lik factoring or th quadratic formula, th critical valus ar locatd at and 3. All thr stratgis yild th sam critical valus. Kp in mind that a graph will giv approimat valus whil factoring or th quadratic formula yild act valus. If th drivativ is not factorabl, linar, or quadratic, anothr mthod will nd to b usd to dtrmin whr th drivativ is qual to zro. Evn though a graph of th drivativ only givs an stimat of th critical valus, it may b th only way to find th critical valus if th drivativ is complicatd. With th critical valus in hand, labl thm on a numbr lin so that w ar abl to apply th first drivativ tst. If w slct a tst point in ach intrval and dtrmin th sign of ach factor, w can complt th numbr lin and track th sign of th drivativ. 3
(+)(-)(-) = + = 0 (+)(+)(-) = - = 0 (+)(+)(+) = + incrasing dcrasing 3 incrasing f( ) 6 3 f ( ) Whn a continuous function changs from incrasing to dcrasing, w hav a rlativ maimum at th critical valu. Whn a continuous function changs from dcrasing to incrasing, w hav a rlativ minimum at th critical valu. In this cas, th rlativ maimum is locatd at and th rlativ minimum is locatd at 3. To find th ordrd pairs for th rlativ trma, w nd to substitut th critical valus into th original function f( ) to find th corrsponding y valus: 3 Rlativ Maimum: f 4 8 5 37 4 3 Rlativ Minimum: f 3 4 3 3 8 3 5 Th rlativ maimum is locatd at locatd at 3,. 37 4, and th rlativ minimum is 4
Eampl 7 Find th Rlativ Maimum Find th location of th rlativ maimum of th function g ( ) Solution Th drivativ of gis ( ) found with th uotint Rul for Drivativs with u v u v Put ths prssions into th uotint Rul: g( ) Th uotint Rul is d u vu uv d v v To mak it asir to find th critical valus, simplify th drivativ. Factor from th numrator Rduc th common factor in th numrator and dnominator Now w can find th critical valus by dtrmining whr this fraction is qual to zro or undfind. Any fraction is qual to zro whr th numrator is qual to zro. In this cas, this is whr 0 or. Fractions ar undfind whr th dnominator is qual to zro. Th dnominator for this fraction is and is always positiv so thr ar no valus whr th fraction is undfind. 5
To apply th first drivativ tst, labl a numbr lin with this critical valu and tst th first drivativ on ithr sid of th critical valu: incrasing = 0 dcrasing f( ) Sinc this function is continuous and th drivativ changs from incrasing to dcrasing at, th critical point is a rlativ maimum. Th y valu for th rlativ maimum coms from th function g ( ) and is g(). Th ordrd pair for th rlativ maimum is,. Eampl 8 Find th Minimum Avrag Cost Th total daily cost to produc units of a product is givn by th function C 0.005 0 000 dollars Th avrag cost function C is found by dividing th total daily cost function C by th quantity. a. Find th avrag cost function C. C. Substitut th total daily cost function into th numrator of this fraction to yild Solution Th avrag cost function is dfind by C 6
C 0.005 0 000 b. Find th quantity that yilds th minimum avrag cost. Solution Th minimum avrag cost is found by locating th rlativ minimum of th avrag cost function. To us th first drivativ tst to find this rlativ minimum, w nd to tak th drivativ of C using th uotint Rul for Drivativs. Th numrator, dnominator and thir drivativs ar u 0.005 0000 v u0.000 v Th drivativ of th avrag cost function is C vu uv 0.000 0.005 0000 v Sinc w nd to us th drivativ to find th critical valus, w simplify th drivativ as much as possibl: C 0.00 00.005 0000 0.005 000 Simplify th numrator by carrying out th multiplication and subtraction Combin lik trms Any fraction is qual to zro whn th numrator is qual to zro and undfind whr th dnominator is qual to zro. 7
St th numrator qual to 0 0.005 000 0 0.005 000 00000 00000 447 St th dnominator qual to 0 0 0 Sinc quantitis producd must b positiv, only 447 is a rasonabl critical valu for this function. Th numbr lin for this function only includs positiv quantitis. Tsting on ithr sid of th critical numbrs yilds th bhavior of C. 0 dcrasing = 0 447 incrasing C 0.005 000 Sinc th function dcrass on th lft sid of th critical valu and incrass on th right sid of th critical valu, th quantity at approimatly 447 units is a rlativ minimum. Th minimum avrag cost is obtaind from C 0.005 0000 and is calculatd as 8
C 00000 0.005 00000 0 00000 000 00000 4.47 Sinc th total daily cost is in dollars and w ar dividing by th numbr of units to gt th avrag cost, th units on th avrag cost ar dollars pr unit. This mans that a production lvl of about 447 units givs th lowst avrag cost of 4.47 dollars pr unit. Figur 6 - Th rlativ minimum for th avrag cost function. 9