MATH section 3.1 Maximum and Minimum Values Page 1 of 7

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1 MATH section. Maimum and Minimum Values Page of 7 Definition : Let c be a number in the domain D of a function f. Then c ) is the Absolute maimum value of f on D if ) c f() for all in D. Absolute minimum value of f on D if ) c f() for all in D. Definition : The number c ) is a local maimum value of f if ) c f() for all is near c.. local minimum value of f if ) c f() for all is near c.. Theorem : The Etreme Value Theorem If f is continuous on a closed interval [ ab, ], then f attains an absolute maimum value c ) and an absolute minimum value d ) at some numbers c and d in [ ab],. Theorem : Fermat s Theorem If f has a local maimum or minimum at c, and if f ( c) eists, then f () c =. Definition 6: A critical number of a function f is a number c in the domain of f such that either f () c = or f ( c) does not eist. Note: If f has a local maimum or minimum at c, then c is a critical number of f. The Closed Interval Method To find the absolute maimum and minimum values of a continuous function f on a closed interval [ ab], : ) Find the values of f at the critical numbers of f in ( ab),. ) Find the values of f at the endpoints of the interval. ) The largest of the values from Steps and is the absolute maimum value; the smallest of those values is the absolute minimum value. Additional eamples: -a) -b) The Etreme Value Theorem The Closed Interval Method 6) Absolute maimum value: none because the point does not eist Absolute minimum value: g () = local maimum values: g () = and g (6) = local minimum values: g () = and g () = 6) f( ) = (See figure to the right) 6 8 Absolute maimum: ) = ( ) = + = Absolute minimum: none Local maimum: none Local minimum: none

2 MATH section. Maimum and Minimum Values Page of 7 8) ) ) ) 6) f() t = cost t Absolute maimum: ) = Absolute minimum: f( ) = ) = Local maimum: ) = Local minimum: f( ) = f( ) = f( ) = < < Absolute maimum: none Absolute minimum: none Local maimum: none Local minimum: none if < f( ) = if Absolute maimum: none because ) and Absolute minimum: ) = Local maimum: none because ) is at the endpoint Local minimum: ) = ) = domain: (, ) [ ] 6[ ] 5[] 5 = + = + = ( + 5)( ) = = = = = ( + 5) = 5 = The critical numbers are = 5 and =. ) < lim f( ) ) = + + domain: (, ) [ ] [ ] [] 6 = + + = + + = = = ( + +) The trinomial above is not factorable, so we must use the quadratic formula to see if we have a real number solution. () ± () ()() ± ± = ( + + ) = = = () 6 6 Since this epression, does not have real number solution, there are no critical numbers. +

3 MATH section. Maimum and Minimum Values Page of 7 8) gt () = t domain: (, ) (t ) if t t if t gt () = t = gt () = (t ) if (t ) < t + if t < dg [] = if t > = dt [] = if t < Since dg is a constant for all dt and dg dt does not eist for t, the critical number is t =. t =. p ) hp ( ) = p + V.A.? p + = no solution ; so, domain: (, ) dh [] ( p + ) ( p )[ p] ( p + ) ( p p) p + p + dp = = = p p + + p + ( ) ( p + ) ( ) ( ) dh p + p + = = dp = p + p+ ( ) ± ( ) ()( ) ± + () p = = () = ( p p ) ± (+ ) ± 5 = = = ± 5 The critical numbers are p = 5 and p = + 5. ) g ( ) = = ( ) Critical number: due to restriction of ( ), the domain of g is (,) (, ) ( ) + ( ) ( ) ( ) dg = = + = + = ( ) ( ) ( ) dg + = = = + = ( ) 5 dg The value of =. = does not eist but is not in the domain of g. Therefore, the critical number is 5

4 MATH section. Maimum and Minimum Values Page of 7 ) g( θ ) = θ tanθ Critical number: due to restriction of tanθ, θ = θ = + k tan undefined θ = θ = + k where k is any integer θ = θ = + k The domain of g is { θ real number and θ + k where k is any integer} dg = [] sec θ () = sec θ dθ dg ± = = sec θ = = cos θ = cosθ = dθ cos θ cos θ θ = θ = + k cosθ = where k is any integer 5 5 θ = θ = + k θ = θ = + k cosθ = where k is any integer θ = θ = + k The values of where dg is undefined are the same values where domain does not eist; so they are not dθ critical numbers. We have infinite amount of critical numbers and they are θ = + k, θ = + k, θ = + k, and θ 5 = + where k is any integer. k 6) f( ) = 5+ 5 [,] = + 5[] = 5 6 = 6( + )( ) = = = = = 6( 9) = = both values are in the interval [,] ) = 5 + 5( ) ( ) = 5 5() + 5 = 5 5( ) = 5 5() = 5 8 = f = + = + = + = + = + = ) = 5+ 5() () = 5 () 5 5() () 5 5() 5 5 5( ) 5 5() 5 8 ( ) f = + = + = + = + = + = () 5 5() () 5 () 7 () 5 ()(7 6) 5 8() Absolute minimum: ) = 5 Absolute maimum: ) =

5 MATH section. Maimum and Minimum Values Page 5 of 7 8) ) f( ) = [,5] = 6[ ] + = = = = = both values are in the interval [,5] = = = ( ) ) = () 6() + 5 = 5 ) = () 6() + 5 = () ( 6) + 5 = 6( ) + 5 = + 5 = 7 ) = ( ) 6( ) + 5 = () ()() + 5 = () ( ) + 5 = () ( ) = 76 f = + = + = + = + = Absolute minimum: ) = 76 Absolute maimum: ) = 5 (5) (5) 6(5) 5 (5) (5 6) 5 5( ) f( ) = ( ) [,] = ( ) ( ) = 6 ( ) = = 6 ( ) = ( + ) (6 )( ) ( + ) = ( ) = = 6 (( + )( )) 6 = + = = = 6 ( + ) ( ) = = = ) = (( ) ) = ( ) = () = ) = (() ) = ( ) = ( ) = all values are in the interval [,] ) = (() ) = ( ) = () = Now we test the endpoints of the interval (we only need to test the right endpoint because the left endpoint is one of the critical numbers above): ) = (() ) = ( ) = () = 7 Absolute minimum: ) = Absolute maimum: ) = 7 ) f( ) = [,] + Since f is a rational function, we need to check if we have vertical asymptotes because if we do, then our domain will not be (, ). ( ) ± ( ) ()() ± ± + = = = = no real number solution and our () domain of function is (, ) and it is continuous and differentiable in the interval [,]. Now we can proceed.

6 MATH section. Maimum and Minimum Values Page 6 of 7 []( + ) ( )[ ] ( + ) ( ) = = = ( + ) ( + ) ( + ) = = = ( + )( ) ( + ) + = = = = = ( ) = discard () ) = = = = () () + + () ) = = = () () + () ) = = () () = 7 Absolute minimum: ) = Absolute maimum: ) = only one value is in the interval [,] 7 ) f() t = t+ cot t [, ] Since a part of f is a trigonometric function, we need to check if the trigonometric part is one piece or 7 not in the interval [, ]. We know that cotθ is a single piece in the interval (, ), so t = t = cot t = undefined (, ) where cot t is one piece t = t = 7 Our interval of [, ] is inside the interval (, ). So our function f is continuous and differentiable 7 in the interval [, ]. Now we can proceed. = [] + csc t csc t dt = csc t csc t = = = sin t = sin t = dt = ± csc t = sin t sin t sin t = = We have multiple solutions of t for each case solved above. sin t = 6

7 MATH section. Maimum and Minimum Values Page 7 of sin t = 7 7 We need to discard both solutions of sin t = 7 interval [, ]. ) = + cot cot.578 = + = + ) = + cot cot.7 = + = ) = + cot cot.996 = ) = + cot cot.86 = + 8 Absolute minimum: ) = + cot = Absolute maimum: ) = + cot =.7 discard discard because the values of t are not part of our initial It is not easy to visualize our function, ) t = t+ cot t in the 7 interval [, ], so a picture of the function is provided. Note that in this interval the absolute maimum and minimum occurs at the critical numbers.