[ f x x 2 2 x, 4.1. Click here for answers. Click here for solutions. MAXIMUM AND MINIMUM VALUES

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1 SECTION. MAXIMUM AND MINIMUM VALUES. MAXIMUM AND MINIMUM VALUES A Click here for answers. S Click here for solutions. Sketch the graph of a function f that is continuous on [0, 3] and has the given properties.. Absolute maximum at 0, absolute minimum at 3, local minimum at, local maximum at. Absolute maximum at, absolute minimum at 3. is a critical number, but f has no local maximum or minimum. Absolute minimum at 0, absolute maximum at, local maxima at and, local minimum at Sketch the graph of f b hand and use our sketch to find the absolute and local maximum and minimum values of f. (Use the graphs and transformations of Section..) 5. f x x; x 6. f x x ; x 8 7. f x x ; 0 x 8. f x x ; 0 x 9. f x x ; 0 x 0. f x x ; 0 x. f x x ; x. f x x ; x 3. ; 0 x f x x. f cos ; 5. f sec ; f x x 5 f x x 8. f x e x, x Find the critical numbers of the function. 9. f x x 3x 0. f x 5 8x. f x x 3 3x. f t t 3 6t 3t 3. t x s 9 x. t x x 5. f x 5 6x x 3 6. f t t 3 3t 6t f x x 3 9x x 3 s t t 3 3t 6t s t t t 3 t 30. f r r r 3. f sin 3. t 33. V x xsx 3. T x x x 3 sin 35 5 Find the absolute maximum and absolute minimum values of f on the given interval. 35. f x x x, 0, f x x x,, 37. f x x 3 x, 3, f x x 3 5x x 7, 0, f x x 3 3x,, 0. f x 8x 5x x 3, 3,. f x x x, 3,. f x 3x 5 5x 3,, 3. f x x x, [, ]. f x s9 x,, 5. f x x, x, ; 6. Use a graph to estimate the critical numbers of f x x 3x x correct to one decimal place. ; 7. (a) Use a graph to estimate the absolute maximum and minimum values of f x x 3x 3 3x x on the interval 0,, correct to two decimal places. (b) Use calculus to find the exact maximum and minimum values. 8. Show that 0 is a critical number of the function f x x 5 but f has neither a local maximum nor a local minimum at Sketch the graph of a function on 0, that is discontinuous and has both an absolute maximum and an absolute Copright 03, Cengage Learning. All rights reserved.

2 SECTION. MAXIMUM AND MINIMUM VALUES. ANSWERS E Click here for exercises. S Click here for solutions.... Abs. max. f (0) = 3; abs.min.f () = 5 5. Abs. max. f () = 3 ;abs.min.f () = 6..3, 0.,. 7. (a), 0. (b), Abs. min. f ( ) = 6. Abs. max. f (8) = 3 7. None 8. Abs. min. f () = 0 9. Abs. max. f (0) = 0. Abs. max. f (0) = ; abs. min. f () = 0. Abs. and loc. max. f (0) = ; abs. min. f ( ) = 3. Abs. max. f ( ) = ; abs. and loc. min f (0) = 0 3. Abs. max. f () = 7; abs. and loc. min. f ( ) =0. Abs. and loc. max. f (0) = 5. Abs. and loc. min. f (0) = 6. None 7. Abs. and loc. max. f (0) = 8. Abs. min. f (0) = None. ±. ± ± 6. None 7., 8. ( ± 5 ) / 9. 0, ( 3 ± 5 ) / 30. ± 3. nπ/, n an integer 3. (n +)π, n an integer ,, Abs. max. f (3) = 5; abs. min. f () = 36. Abs. max. f ( ) = ; abs. min. f ( ) = Abs. max. f (5) = 66; abs. min. f () = Abs. max. f (3) = 6; abs. min. f () = Abs. max. f () = 9; abs.min. f ( ) = 0 Copright 03, Cengage Learning. All rights reserved. 0. Abs. max. f ( 3) = 89; abs.min. f ( ) = 9. Abs. max. f ( 3) = 7; abs.min.f ( ± ) =. Abs. max. f () = 55; abs. min. f ( ) = Abs. max. f () = 5; abs. min. f () = 3

3 SECTION. MAXIMUM AND MINIMUM VALUES 3. SOLUTIONS E Click here for exercises f (x) = x, 0 x<. Absolute maximum f (0) = ; no local maximum. No absolute or local 0. f (x) = x, 0 x f (x)=+x, x. Absolute minimum f ( ) = ; nolocal No local or absolute maximum. Absolute maximum f (0) = ; no local maximum. Absolute minimum f () = 0;nolocal. f (x) = x, x. Absolute and local maximum f (0) =. Absolute minimum f ( ) = 3; nolocal. (_, ) 0 x f (x) = x, x. Absolute maximum f ( ) = ; no local maximum. Absolute and local minimum f (0) = f (x) =x, x 8. Absolute maximum f (8) = 3; no local maximum. No local or absolute 3. 7 f (x) = x, 0 x. Absolute maximum f () = 7; no local maximum. Absolute and local minimum f ( ) =0since x 0 for all x. 7. f (x) = x, 0 <x<. No maximum or. 0 x f (θ) =cos(θ/), π <θ<π.localand _¹ ¹ absolute maximum f (0) =. No local or absolute Copright 03, Cengage Learning. All rights reserved. 8. f (x) = x, 0 <x. Absolute minimum f () = 0; no local No absolute or local maximum. 5. =_ ¹ _ ¹ ( _3, ) (0, ) 0 f (θ) =secθ, π <θ π. 3 Local and absolute minimum f (0) =. No absolute or local maximum.

4 SECTION. MAXIMUM AND MINIMUM VALUES Copright 03, Cengage Learning. All rights reserved. 6. f (x) =x 5.Nomaximumor 7. f (x) = x. Local and absolute maximum f (0) =. No local or absolute 8. f (x) = e x, x 0. Absolute minimum f (0) = 0; no local No absolute or local maximum. 9. f (x) =x 3x f (x) = 6x =0 x = 3. So the critical number is f (x)=5+8x f (x) =8 0. No critical number.. f (x) =x 3 3x + f (x) =3x 3=3 ( x ) =3(x +)(x ). So the critical numbers are ±.. f (t) =t 3 +6t +3t f (t) =3t +t +3=3 ( t +t + ). B the quadratic formula, solutions are t = ± = ± 3. Critical numbers are t = ± g (x) = 9 x = x /9 g (x) = 9 x 8/9 = 9 9 x 0, 8 but g (0) doesnotexist,sox =0is a critical number.. g (x) = x + g (x) =if x>, g (x) = if x<,butg ( ) does not exist, so x = is a critical number. 5. f (x) =5+6x x 3 f (x) =6 6x =6(+x)( x). f (x) =0 x = ±,so± are the critical numbers. 6. f (t) =t 3 +3t +6t + f (t) =6t +6t +6. But t + t +=0has no real solution since b ac = ()() = 3 < 0. No critical number. 7. f (x) =x 3 9x x +3 f (x) =x 8x =6 ( x 3x ) =6(x +)(x ) f (x) =0 x =, ; so the critical numbers are x =,. 8. s (t) =t 3 +3t 6t + s (t) =6t +6t 6=6 ( t + t ). B the quadratic formula, the critical numbers are t = ( ± 5 ) /. 9. s (t) =t +t 3 +t s (t) =t 3 +t +t =t ( t +3t + ) =0when t =0or t +3t +=0. B the quadratic formula, the critical numbers are t =0, 3 ± 5. r 30. f (r) = r + ( r f + ) r (r) (r) = = r + (r +) (r +) =0 r = r = ±, so these are the critical numbers. Note that f (r) alwasexistssincer f (θ) =sin (θ) f (θ) =sin(θ)cos(θ)() =(sinθcos θ) =[sin( θ)] =sinθ =0 f (θ) =0 sin θ =0 θ = nπ, n an integer. So θ = nπ/ are the critical numbers. 3. g (θ) =θ +sinθ g (θ) =+cosθ =0 cos θ =. The critical numbers are θ = π +nπ =(n +)π, n an integer. 33. V (x) =x x V (x) = x x + x V ( ) does not exist. For x> [the domain of V (x)], V (x) > 0,sois the onl critical number. 3. T (x) =x (x ) /3 T (x) =x (x ) /3 + x ( ) 3 (x ) /3 (). So T ( ) does not exist. T (x) =x (x ) ( /3 x + x) 3 =x (x ) /3 ( 8 3 x ) =0 x =0or x = 3 8. So the critical numbers are x =0, 3 8,and. 35. f (x) =x x +, [0, 3]. f (x) =x =0 x =. f (0) =, f () =, f (3) = 5. So f (3) = 5 is the absolute maximum and f () = is the absolute 36. f (x) = x x, [, ]. f (x) = x =0 x =. f ( ) = 7, f ( ) =, f () =. So f ( ) = 7 is the absolute minimum, f ( ) = is the absolute maximum. 37. f (x) =x 3 x +, [ 3, 5]. f (x) =3x = 3 ( x ) =3(x +)(x ) = 0 x = ±. f ( 3) = 0, f ( ) = 7, f () = 5, f (5) = 66. So f () = 5 is the absolute minimum and f (5) = 66 is the absolute maximum.

5 SECTION. MAXIMUM AND MINIMUM VALUES f (x) =x 3 5x +x +7, [0, 3]. f (x) =x 30x +=6(x ) (x ) = 0 x =,. f (0) = 7, f ( ) = 39, f () = 3, f (3) = 6. So f (3) = 6 is the absolute maximum and f () = 3 the absolute 39. f (x) =x 3 +3x +, [, ]. f (x) =6x +6x =6x (x +)=0 x =, 0. f ( ) = 0, f ( ) = 5, f (0) =, f () = 9. So f () = 9 is the absolute maximum and f ( ) = 0 is the absolute 0. f (x) =8x +5x x 3, [ 3, ]. f (x) =8+30x x =6(3 x)(+x) =0 x =3,. f ( 3) = 89, f ( ) = 9, f (3) = 8, f () = 56. So f ( 3) = 89 is the absolute maximum and f ( ) = 9 is the absolute. f (x) =x x +, [ 3, ]. f (x) =x 3 8x =x ( x ) =0 x =0, ±. f ( 3) = 7, f ( ) =, f (0) =, f ( ) =, f () =,sof ( ± ) = is the absolute minimum and f ( 3) = 7 is the absolute maximum.. f (x) =3x 5 5x 3, [, ]. f (x) =5x 5x =5x (x +)(x ) = 0 x =, 0,. f ( ) = 57, f ( ) =, f (0) =, f () = 3, f () = 55. Sof ( ) = 57 is the absolute minimum and f () = 55 is the absolute maximum. 3. f (x) =x + x, [, ]. f (x) =x x =x3 x =0 x 3 =0 (x ) ( x + x + ) =0, but x + x + 0,sox =. The denominator is 0 at x =0, but not in the desired interval. f ( ) = 7, f () = 3, f () = 5. Sof () = 3 is the absolute minimum and f () = 5 istheabsolutemaximum.. f (x) = 9 x, [, ]. f (x) = x / 9 x =0 x =0. f ( ) =, f (0) = 3, f () = 5.So f () = 5 is the absolute minimum and f (0) = 3 is the absolute maximum. 5. f (x) = x, [, ]. x + f (x +) x (x) = = 0 no critical (x +) (x +) number. f () = and f () = 3,sof () = is the absolute minimum and f () = is the absolute maximum We see that f (x) =0at about x =.3, 0.,and.. Since f exists everwhere, these are the onl critical numbers. 7. (a) From the graph, it appears that the absolute maximum value is f () =, and that the absolute minimum value is about f (0.5) = 0.. (b) f (x) =x 3x 3 +3x x f (x) =x 3 9x +6x =(x ) (x ). So f (x) =0 x = or x =. Now f () = =0 (neither maximum nor minimum) and ) 3 ( ) 3 +3 ( ) = 7 56 f ( ) ( = (minimum). At the right endpoint we have f () = =(maximum). 8. f (x) =x 5. f (x) =5x f (0) = 0 so 0 is a critical number. But f (0) = 0 and f takes both positive and negative values in an open interval containing 0,so f has neither a local maximum nor a local minimum at Copright 03, Cengage Learning. All rights reserved.

3.3. Click here for solutions. Click here for answers. DERIVATIVES AND THE SHAPES OF GRAPHS

3.3. Click here for solutions. Click here for answers. DERIVATIVES AND THE SHAPES OF GRAPHS SECTION DERIVATIVES AND THE SHAPES OF GRAPHS DERIVATIVES AND THE SHAPES OF GRAPHS A Click here for answers S Click here for solutions 6 a Find the intervals on which f is increasing or decreasing b Find