3.1 Extreme Values of a Function

Transcription

1 .1 Etreme Values of a Function Section.1 Notes Page 1 One application of te derivative is finding minimum and maimum values off a grap. In precalculus we were only able to do tis wit quadratics by find te verte. Derivatives will allow us to find tese values for functions wit iger powers. Te word etrema refers to min and ma values. Etrema of a Function 1.) f (c) is te minimum of f on interval I if f ( c) f ( ) for all in I..) f (c) is te maimum of f on interval I if f ( c) f ( ) for all in I. A function is guaranteed to ave a minimum and maimum value if te function is continuous on a closed Te following are eamples of wen a function may or may not ave a min or ma: From te pictures above we see tat in picture (a) we ave a closed interval and tere is a defined maimum and minimum value. Te igest point in and interval is call te absolute maimum and te lowest point in te interval is called a absolute minimum. Picture (b) as an absolute minimum, but not a maimum since tat point is not included in our Picture (c) as a absolute maimum but not a minimum on [-1, ] because tis grap is not continuous on tis Local Etreme Values Local etreme values means tere is a ill or valley in te grap. Tis is not necessarily te absolute maimum or minimum of te grap. In te picture to te rigt we see a ill and valley but tey are not te igest and lowest point of te grap. Te igest or lowest possible points on te grap are called absolute etrema, and tese are labeled on te grap.

2 Critical Number Section.1 Notes Page Let f be defined at c. Te point = c is a critical number if eiter of te following occurs wic is sown in te pictures below: 1.) f (c) is undefined OR.) f ( c) = 0 AND f (c) is defined. Te derivative would not eist for any graps tat ave a corner. Te absolute value grap would apply ere because tis as a sarp corner. Here s wy: EXAMPLE: Find te derivative of y = at te point (0, 0) by using te it process. 0 f ( + ) f ( ) Start wit te it formula. 0 f (0 + ) f (0) In tis case = We put in for and we know tat f ( 0) = 0. 0 As approaces 0 from te left, tis epression will equal -1. As approaces 0 from te rigt, tis epress will equal 1. Since tis approaces two different values, te it does not eist and terefore te derivative does not eist at (0, 0). How to find etrema on a closed interval [a, b] 1.) Take te first derivative and set it equal to zero to find te critical point. You may also need to see if te derivative is undefined anywere in te interval, because tis will also give you a critical point..) Evaluate f at eac critical point you find in te.) Evaluate f at te endpoints of your (In oter words, find f (a) and f (b) ). 4.) Te least of tese values is te absolute minimum. Te greatest is te absolute maimum.

3 EXAMPLE: Let Section.1 Notes Page 4 f ( ) = on [-4, ]. Find all critical numbers and absolute etrema on tis Te first ting to do is to find te derivative. We will get f ( ) = We need to set tis to zero to find te critical point: = 0. We need to factor tis and set te factors equal to zero. After factoring you sould get 1 ( + ) = 1 ( 1)( + ). Setting tis equal to zero we will get te critical points 0, 1, and -. Tere are no places were te derivative will be undefined, so tese are te only critical points. Now we need to set up a table wit our critical points and also te endpoints. We will use te values 0, 1, and - since tey are te critical points. We also need to ceck = -4 and = since tese are te endpoints of our So we will put all tese -values into te ORIGINAL function to get te y-values. DO NOT use te derivative. For eample, to get 0 on te table below I put -4 into te original equation for : 4 f ( 4) = ( 4) + 4( 4) 1( 4) = f() After you complete te table we need to find te absolute etrema. Te largest y-value will be te absolute maimum and te smallest y-value will be te absolute minimum. You would write te following as te answer: Te absolute maimum is 0 wen = -4. Te absolute minimum is - wen = -. EXAMPLE: Let f ( ) = 1 8 on [-1, ]. Find all critical numbers and absolute etrema on tis We can rewrite tis as; f ( ) = 1 8. Wen we find te derivative we can just use te power rule. You 1 8 will get: f ( ) = 8 8, wic can be rewritten as f ( ) = 8. By observation we can see tat a zero would cause te derivative to be undefined, so we know tat = 0 is a critical point since tis is defined in te 8 original function. We need to set tis equal to zero to find te oter critical point: 0 = 8. We can add 8 te 8 to bot sides to get 8 =. Cross multiplying we will get 8 = 8. Ten divide bot sides by 8 to get = 1. After cubing bot sides we get = 1 wic is our second critical point. Wen we make our table we will be using te = -1, 0, 1, and. To get te y-values, we will put tese in for in f ( ) = f() We see tat te absolute maimum is 0 wen = -1 and te absolute minimum is 0 wen = 0.

4 Section.1 Notes Page 4 EXAMPLE: Let f ( θ ) = secθ on, 6. Find all critical numbers and absolute etrema on tis First te derivative is f ( θ ) = secθ tanθ. By looking at tis we find tat θ = would make te derivative undefined. If tis were in our interval ten tis would be a critical point. Since it is not in our interval we will ignore tis and just set te derivative equal to zero: 0 = secθ tanθ. Wen we do tis we ave sec θ = 0 and 1 tan θ = 0. If we try to set te first equation equal to zero we can write sec θ =. so our equation becomes cosθ 1 0 =. Cross multiplying we will get 0 = 1, wic is not true so tis first equation won t give us any critical cosθ sinθ points. Te second equation, tan = 0 wic we can solve to get θ = 0 θ can be rewritten as = 0. Cross multiplying we will get sin θ = 0 cosθ. Tis will be te only critical point. So we will make our table wit our endpoints and also θ = 0. Wen we put in θ = we will get = It is okay to use te decimal in 6 te table so we can easily compare te results. 0 6 f() We see tat te absolute maimum is wen =. Te absolute minimum is 1 wen = 0. EXAMPLE: Let f ( ) = e on [-, 1]. Find all critical numbers. Ten find te absolute etrema on tis First te derivative is f ( ) = e. So f ( ) = e. Now we set te derivative equal to zero: 0 = e. Wen we do tis we ave = 0 and e = 0. So one critical number is = 0. For te second equation, te only way to solve tis would be to take te natural log of bot sides to clear out te e. However wen we do tis we cannot take te natural log of zero. Terefore tis equation as no solution. Terefore te only critical number is = 0. Now we need to set up a table wit our critical points and also te endpoints. We will use te values -, 0, and 1. So we will put all tese -values into te ORIGINAL function to get te y- values. 0 1 f() 4 e 1 1 e So we see te absolute maimum will be 0 and tat occurs at = 1. Ten te absolute minimum is occurs at =. 4 e and tat

5 Section.1 Notes Page 5 EXAMPLE: Let 5 f ( ) = + on [-1, 5]. Find all critical numbers. Ten find te absolute etrema on tis We can rewrite tis as will get: f ( ) = 5 f ( ) = + by dividing bot tings on top by. Now take te derivative and you. Tere are no places were te derivative will be undefined and we can t set te derivative equal to zero because tere is no variable. Because of tis we know tat tere are no critical points. If tere are no critical points ten te only numbers we will use in our table will be te endpoints, = -1 and = f() 1 5 EXAMPLE: Find te etreme values of f() and were tey occur if f ( ) =. 1 1 We can rewrite te above as f ( ) = ( 1). Now we want to take te derivative. You will use te cain 4 rule. You will get f ( ) = ( 1) (). Tis can be rewritten as f ( ) =. We will set tis equal ( 1) 4 to zero to get a critical point. 0 =. After cross multiplying you will get 0 = 4, so = 0. If tere is ( 1) a point tat causes te derivative function to be undefined, ten tis is anoter critical point. In tis problem = 1 and = -1 would cause te bottom to be zero. We cannot ave any etrema at tese points since tey are not defined on te original function. Terefore te only critical number is = 0. We need to decide if tis is a local minimum or maimum. Let s make a table of values close to 0: f() We see tat as we approac zero from te left and te rigt te y values are getting bigger. Tis means tere is a local maimum of at = 0. Does tis ave an absolute maimum or absolute minimum? Te answer is no. Since we ave vertical asymptotes at = 1 and = 1 tis means as we very close to 1 or 1 te y values will approac positive or negative infinity. Tis is not an eact number terefore tere is no absolute maimum or minimum. So te only etrema is a local maimum of at = 0. More on net page

6 EXAMPLE: Find te etreme values of f() and were tey occur if ( ) Section.1 Notes Page 6 1 f ( ) = cos e. u We will take te derivative. We will use te formula f ( ) = were u = e. You will get: 1 u e e f ( ) =. Tis can be written as: f ( ) =. Now we need to find te critical points. Te 1 ( e ) 1 e e first ting we will do is to set te derivative equal to zero: 0 =. We cross multiply to get 0 = e. 1 e To solve we need to take te natural log of bot sides, owever we cannot take te natural log of zero. Terefore we did not find any critical numbers tis way. Te oter way to find critical numbers is to see wat number makes te derivative undefined but still defined in te original function. Te derivative is a rational epression wic means te way it can be undefined is if te denominator is zero. So we will set te denominator equal to zero: 0 = 1 e. After squaring bot sides: 0 = 1 e. Ten we can isolate te e: e = 1. Take te natural log of bot sides: ln e = ln1. Tis simplifies to = 0, so = 0. Tis is defined in te original function so tis is a critical number. We now need to decide if tis is a local minimum or maimum. Let s make a table of values close to 0: f() We see ere tat since te numbers are getting constant te farter we go to te left, tis means tere is a orizontal asymptote. Specifically, tis is y =. Tis means tere will not be an absolute ma because te grap will get very close to but never touc y =. So we only ave an absolute minimum. Terefore te absolute minimum is 0 at = 0.