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LESSON 14 LOCAL MAXIMUM AND MINIMUM VALUES Definition Let a function f be defined on an interval ( a , b ) and let x 1 and x 2 be any numbers in the interval. Then (1) f is increasing on the interval if f ( x 1 ) f ( x 2 ) whenever x 1 x 2 , (2) f is decreasing on the interval if f ( x 1 ) f ( x 2 ) whenever x 1 x 2 , (3) f is constant on the interval if f ( x 1 ) f ( x 2 ) for every x 1 and x 2 in the interval. Part (1) of the definition is saying that the function f is increasing on the interval ( a , b ) if the functional values, which are given by the y-coordinates of the points on the graph of y f ( x ) are increasing as you move left to right across the interval on the x-axis. Part (2) of the definition is saying that the function f is decreasing on the interval ( a , b ) if the functional values, which are given by the y-coordinates of the points on the graph of y f ( x ) are decreasing as you move left to right across the interval on the x-axis. Example Determine the interval(s) on which the function f is increasing and decreasing. f is increasing on ( a , b ) ( c , d ) f is decreasing on ( , a ) ( b , c ) ( d , ) Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1850 Example Determine the interval(s) on which the function f ( x ) 2 x 2 12 x 19 is increasing and decreasing. 2 Then y 2 x 12 x 19 . We can find the vertex of the 2 parabola by completing the square on y 2 x 12 x 19 . Thus, Set y f ( x ) . y 2 x 2 12 x 19 y 19 __ 2 ( x 2 6 x __ ) y 19 18 2 ( x 2 6 x 9 ) y 1 2 ( x 3 ) 2 Thus, the vertex of the parabola is V ( 3 , 1 ) and the parabola opens down since a 2 0. NOTE: You could have also found the x-coordinate of the vertex of the parabola b 12 ( 3 ) 3 . The y-coordinate of the vertex of the by x 2a 4 parabola is given by y ( 3 ) 18 36 19 1 . 2 The graph of y 2 x 12 x 19 : Answer: Increasing: ( , 3 ) Decreasing: ( 3 , ) Theorem If a function f is continuous on the closed interval [ a , b ] and differentiable on the open interval ( a , b ) , then Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1850 1. if f ( x ) 0 for all x in the interval ( a , b ) , then the function f is increasing on the interval ( a , b ) , 2. if f ( x ) 0 for all x in the interval ( a , b ) , then the function f is decreasing on the interval ( a , b ) . Proof Will be provided at a later date. Examples Find the interval(s) on which the following functions are increasing and decreasing. 1. f ( x ) 2 x 2 12 x 19 f ( x ) 4 x 12 4 ( 3 x ) NOTE: The domain of the function f is the set of all real numbers. Sign of f ( x ) : + 3 Answer: Increasing: ( , 3 ) Decreasing: ( 3 , ) 2. y 3x 9 x2 NOTE: The domain of the function y is the set of all real numbers such that x 3 and x 3 . We differentiated this function in Lesson 9 obtaining that y 3( 9 x2 ) 3( 9 x2 ) 3( 9 x 2 ) = = [ ( 3 x ) ( 3 x ) ]2 ( 9 x2 )2 ( 3 x )2 ( 3 x )2 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1850 Sign of y : + + + 3 3 Answer: Increasing: ( , 3 ) ( 3 , 3 ) ( 3 , ) Decreasing: Nowhere 3. y 6 4t 3t 22 2 NOTE: The domain of the function y is the set of all real numbers such that 11 t and t 2 . 4 We differentiated this function in Lesson 9 obtaining that 6 ( 8t 3 ) 6 ( 8t 3 ) 6 ( 8t 3 ) dy 2 = 2 2 = dt ( 4t 3t 22 ) [ ( t 2 ) ( 4t 11) ] ( t 2 ) 2 ( 4t 11) 2 Sign of dy : dt + + Answer: 11 4 Increasing: , 3 8 3 8 4. 2 11 3 11 , 4 8 4 Decreasing: , 2 ( 2 , ) g( w) 3w 22 4w 2 3w 22 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1850 NOTE: The domain of the function g is the set of all real numbers such that 11 w and w 2 . 4 We differentiated this function in Lesson 9 obtaining that g ( w ) 4w ( 44 3w ) 4w ( 44 3w ) = ( 4w 2 3w 22 ) 2 [ ( w 2 ) ( 4w 11) ]2 Sign of g ( w ) : + Answer: 11 4 Increasing: ( 0 , 2 ) 2 , Decreasing: , 5. 0 + 2 44 3 44 3 11 11 44 , 0 , 4 4 3 7 x 3 13x h( x ) 4 x 8 x 2 16 NOTE: The domain of the function h is the set of all real numbers such that x 2 and x 2 . We differentiated this function in Lesson 9 obtaining that 7 x 6 17 x 4 232 x 2 208 h( x ) ( x 4 8 x 2 16 ) 2 We used the Factor Theorem and synthetic division to factor the numerator of this fraction in Lesson 9 obtaining that Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1850 (1 x ) (1 x ) ( 7 x 2 52 ) h( x ) ( x 2 4 )3 Sign of h ( x ) : + Answer: 1 1 2 + 2 Increasing: ( 2 , 1) (1, 2 ) Decreasing: ( , 2 ) ( 1, 1) ( 2 , ) 6. y 5 x 17 9 x2 NOTE: The domain of the function y is the set of all real numbers such that x 3 and x 3 . We differentiated this function in Lesson 9 obtaining that y ( x 5 ) ( 5x 9 ) ( 9 x2 )2 Sign of y ( x 5 ) ( 5x 9 ) : [ ( 3 x ) ( 3 x) ] 2 + Answer: + + 3 9 5 3 5 Increasing: ( , 3 ) 3 , 9 ( 5, ) 5 9 , 3 ( 3, 5 ) 5 Decreasing: Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1850 7. h( z ) 10 3 ( z 2 3z 18 ) 2 NOTE: The domain of the function h is the set of all real numbers. We differentiated this function in Lesson 10 obtaining that h( z ) 20 ( 2 z 3 ) 20 ( 2 z 3 ) = 3 ( z 2 3z 18 )1 / 3 3[ ( z 6 ) ( z 3 ) ]1 / 3 Sign of h( z ) : 6 Increasing: 6 , Answer: + 3 3 2 3 ( 3, ) 2 Decreasing: ( , 6 ) 8. + 3 , 3 2 2 x 2 11 f ( x) 4 x 2 16 NOTE: The domain of the function f is given by ( , 4 ) ( 4 , ) 2 since we need that x 16 0 . 2 x 2 11 f ( x) ( x 2 16 )1 / 4 1 4 x ( x 2 16 )1 / 4 ( 2 x 2 11) ( x 2 16 ) 3 / 4 2 x 4 f ( x ) = 2 1/ 2 ( x 16 ) Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1850 1 x ( x 2 16 ) 3 / 4 [ 8 ( x 2 16 ) ( 2 x 2 11) ] 2 = ( x 2 16 )1 / 2 x ( 8 x 2 128 2 x 2 11) 3x ( 2 x 2 39 ) x ( 6 x 2 117 ) = = 2 ( x 2 16 ) 5 / 4 2 ( x 2 16 ) 5 / 4 2 ( x 2 16 ) 5 / 4 39 However, 0 is not in the domain of the 2 39 39 4 and 4 , then the numbers 2 2 f ( x ) 0 x 0 , x function f . Since 39 are in the domain of the function f. 2 f ( x ) is undefined when x 4 Sign of f ( x ) : + 39 2 Answer: Increasing: 4 4 39 2 39 , 4 2 Decreasing: , 9. + 39 2 39 , 2 4, 39 2 s( t ) 2 t 3 13t 2 20 t 18 NOTE: The domain of the function s is the set of all real numbers. s( t ) 6 t 2 26 t 20 = 2 ( 3 t 2 13 t 10 ) = 2 ( t 5 ) ( 3 t 2 ) Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1850 Sign of s ( t ) : + Answer: 2 3 5 10. Increasing: , Decreasing: + 2 ( 5, ) 3 2 , 5 3 y x 4 8x 3 12 NOTE: The domain of the function y is the set of all real numbers. dy 4 x 3 24 x 2 = 4 x 2 ( x 6 ) dx Sign of dy : dx + 6 Answer: + 0 Increasing: ( 6 , ) Decreasing: ( , 6 ) 11. g ( x ) 5x 4 x2 NOTE: The domain of the function g is the set of all real numbers such that x 0. g ( x ) 5x 4 x 2 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1850 g ( x ) 5 8 x 3 = x 3 5x 3 8 (5 x 8 ) = x3 3 3 3 NOTE: g ( x ) 0 5 x 8 0 x Sign of g ( x ) : + 3 Answer: + 0 2 8 2 x 3 5 5 5 2 ( 0, ) Increasing: , 3 5 2 , 0 Decreasing: 3 5 12. f ( x ) 4 x (15 x) NOTE: The domain of the function f is the interval [ 0 , ) . f ( x ) x1 / 4 (15 x) = 15 x1 / 4 x 5 / 4 f ( x ) 5 3/ 4 5( 3 x ) 15 3 / 4 5 1 / 4 x x x ( 3 x ) = = 4 4 4 4 x 3/ 4 f ( x ) 0 x 3 and f ( x ) is undefined when x 0 . Sign of f ( x ) : 0 + 3 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1850 Answer: Increasing: ( 3 , ) Decreasing: ( 0 , 3 ) Definition Let c be a number in the domain of the function f. Then 1. f ( c ) is a local maximum of f if there exists an open interval ( a , b ) containing c such that f ( x ) f ( c ) for all x in the interval ( a , b ) . 2 f ( c ) is a local minimum of f if there exists an open interval ( a , b ) containing c such that f ( x ) f ( c ) for all x in the interval ( a , b ) . On the graph of y f ( x ) , f ( c ) is the y-coordinate of the point with an xcoordinate of c. Part (1) of the definition is saying that f ( c ) is a local maximum of the function f if it is the largest y-coordinate of all the y-coordinates of points on the graph of y f ( x ) in a neighborhood of x c . On the graph of y f ( x ) , f ( c ) is the y-coordinate of the point with an xcoordinate of c. Part (1) of the definition is saying that f ( c ) is a local minimum of the function f if it is the smallest y-coordinate of all the y-coordinates of points on the graph of y f ( x ) in a neighborhood of x c . Theorem If a function f has a local maximum or a local minimum at a number c, then either f ( c ) = 0 or f ( c ) is undefined. The converse of this theorem is not true. The converse of this theorem would read: If f ( c ) = 0 or f ( c ) is undefined, then the function f has a local maximum or a local minimum at a number c. Unfortunately, most calculus students use this statement. The theorem above tells you where to look for local maximums and local minimums. It doesn’t guarantee that you will find one. Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1850 The theorem works like the following. If I tell you at 4:00 today, I will either be in my office, in the math conference room, UH 1000, or in the hallway between two of these three places, then you would not go to the library looking for me. And if you go to the math conference room, you will not find me there if I am in my office. Definition A number c in the domain of a function f is called a critical number of f if either f ( c ) = 0 or f ( c ) is undefined. NOTE: Critical numbers are possible places where local maximums or local minimums occur. NOTE: The critical numbers of a function f are a subset of the number(s) on the number line used in the three-step method for finding the sign of f . Theorem (The First Derivative Test for Local Maximums and Minimums) Suppose that c is a critical number of a function f and ( a , b ) is an open interval containing c. If f is continuous on the closed interval [ a , b ] and differentiable on the open interval ( a , b ) , except possibly at c. Then 1. if f ( x ) 0 for all x in the interval ( a , c ) and f ( x ) 0 for all x in the interval ( c , b ) , then f ( c ) is a local maximum of the function f, 2. if f ( x ) 0 for all x in the interval ( a , c ) and f ( x ) 0 for all x in the interval ( c , b ) , then f ( c ) is a local minimum of the function f, 3. if f ( x ) 0 or f ( x ) 0 for all x in the interval ( a , b ) , except possibly at c, then f ( c ) is neither a local maximum nor a local minimum. Proof Will be provided at a later date. In Part (1) of the theorem, the interval ( a , c ) is on the left-hand side of the critical number c and the interval ( c , b ) is on the right-hand side of the critical number c. Thus, the statement f ( x ) 0 for all x in the interval ( a , c ) is saying that the function f is increasing on the left-hand side of the critical number c, and the Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1850 statement f ( x ) 0 for all x in the interval ( c , b ) is saying that the function f is decreasing on the right-hand side of the critical number c. If you would visualize this, you would see the local maximum being formed. In Part (2) of the theorem, the statement f ( x ) 0 for all x in the interval ( a , c ) is saying that the function f is decreasing on the left-hand side of the critical number c, and the statement f ( x ) 0 for all x in the interval ( c , b ) is saying that the function f is increasing on the right-hand side of the critical number c. If you would visualize this, you would see the local minimum being formed. Examples Find the local maximum(s) and local minimum(s) of the functions in the previous examples using the First Derivative Test. 1. f ( x ) 2 x 2 12 x 19 f ( x ) 4 x 12 4 ( 3 x ) NOTE: The domain of the function f is the set of all real numbers. Inc + Sign of f ( x ) : Dec 3 Critical Number(s): 3 NOTE: A local maximum occurs when x 3 and the local maximum is f ( 3) . f ( 3 ) 18 36 19 1 Answer: Local Maximum(s): 1 Local Minimum(s): None Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1850 2. y 3x 9 x2 NOTE: The domain of the function y is the set of all real numbers such that x 3 and x 3 . We differentiated this function in Lesson 9 obtaining that 3( 9 x2 ) 3( 9 x2 ) 3( 9 x 2 ) y = = [ ( 3 x ) ( 3 x ) ]2 ( 9 x2 )2 ( 3 x )2 ( 3 x )2 Sign of y : + + 3 + 3 NOTE: 3 and 3 are not in the domain of the function y Critical Number(s): None Answer: Local Maximum(s): None Local Minimum(s): None 3. y 6 4t 2 3t 22 NOTE: The domain of the function y is the set of all real numbers such that 11 t and t 2 . 4 We differentiated this function in Lesson 9 obtaining that 6 ( 8t 3 ) 6 ( 8t 3 ) 6 ( 8t 3 ) dy dt ( 4t 2 3t 22 ) 2 = [ ( t 2 ) ( 4t 11) ] 2 = ( t 2 ) 2 ( 4t 11) 2 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1850 dy Sign of : dt Inc Inc Dec Dec + + NOTE: 11 4 2 3 8 11 and 2 are not in the domain of the function y 4 Critical Number(s): 3 8 NOTE: A local maximum occurs when t 3 and the local maximum is 8 3 y . 8 6 6 3 y = = , then ( t 2 ) ( 4t 11) 4t 2 3t 22 8 6 6 6 96 . 361 = 3 3 = 19 19 = 361 2 11 16 8 2 8 2 Since y Answer: Local Maximum(s): 96 361 Local Minimum(s): None 4. g( w) 3w 22 4w 3w 22 2 NOTE: The domain of the function g is the set of all real numbers such that 11 w and w 2 . 4 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1850 We differentiated this function in Lesson 9 obtaining that g ( w ) 4w ( 44 3w ) 4w ( 44 3w ) = ( 4w 2 3w 22 ) 2 [ ( w 2 ) ( 4w 11) ]2 Sign of g ( w ) : Dec NOTE: Inc + 11 4 0 Inc + 2 Dec 44 3 11 and 2 are not in the domain of the function g 4 Critical Number(s): 0 and 44 3 NOTE: A local minimum occurs when w 0 and the local minimum is 44 g ( 0 ) . A local maximum occurs when w and the local maximum is 3 44 g . 3 g( 0 ) 22 1 22 22 44 g 2 = 2 22 3 44 44 4 4 3 22 3 3 3 2 = 44 22 1 1 22 = = 352 9 = 4 22 = 88 4 22 2 4 1 22 4 22 1 9 9 9 9 9 22 4 22 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1850 1 9 361 = 361 or 9 Since g ( w ) 3w 22 3w 22 44 = = , then g 4w 3w 22 ( w 2 ) ( 4w 11) 3 2 44 22 22 11 9 22 9 = = = = 38 209 44 6 176 33 19 ( 209 ) 38 ( 209 ) 3 3 3 3 3 3 1 9 9 . 19 (19 ) 361 Answer: Local Maximum(s): 9 361 Local Minimum(s): 1 5. 7 x 3 13x h( x ) 4 x 8 x 2 16 NOTE: The domain of the function h is the set of all real numbers such that x 2 and x 2 . We differentiated this function in Lesson 9 obtaining that 7 x 6 17 x 4 232 x 2 208 h( x ) ( x 4 8 x 2 16 ) 2 We used the Factor Theorem and synthetic division to factor the numerator of this fraction in Lesson 9 obtaining that Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1850 (1 x ) (1 x ) ( 7 x 2 52 ) h( x ) ( x 2 4 )3 Sign of h ( x ) : Inc + Dec Inc + 1 1 2 2 NOTE: 2 and 2 are not in the domain of the function h Critical Number(s): 1 , 1 NOTE: A local maximum occurs when x 1 and the local maximum is h( 1 ) . A local minimum occurs when x 1 and the local minimum is h(1 ) . h( 1) Answer: 7 13 6 2 1 8 16 9 3 Local Maximum(s): h(1 ) 7 13 6 2 1 8 16 9 3 2 3 Local Minimum(s): 2 3 NOTE: If you didn’t find the domain of the function h to know that the number 2 is not in the domain of the function, then by the First Derivative Test, you would have concluded that a local minimum is occurring at x 2 . You would have recognized the domain problem when you would have tried to find the local minimum by calculating h( 2 ) . You would have division by zero. In the same way, if you didn’t find the domain of the function h to know that the number 2 is not in the domain of the function, then by the First Derivative Test, you would have concluded that a local maximum is occurring at x 2 . You would have recognized the domain problem when you would have tried to find the local maximum by calculating h( 2 ) . You would also have division by zero. Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1850 6. y 5 x 17 9 x2 NOTE: The domain of the function y is the set of all real numbers such that x 3 and x 3 . We y differentiated ( x 5 ) ( 5x 9 ) ( 9 x2 )2 Sign of y this function in Lesson 9 obtaining that ( x 5 ) ( 5x 9 ) : [ ( 3 x ) ( 3 x) ] 2 Inc + + Dec Dec Inc + 3 9 5 3 5 NOTE: 3 and 3 are not in the domain of the function y 9 ,5 5 Critical Number(s): NOTE: A local maximum occurs when x 9 and the local maximum is 5 9 y . A local minimum occurs when x 5 and the local minimum is 5 y( 5 ) . 8 25 9 9 17 8 ( 25 ) 8 ( 25 ) 8 ( 25 ) y = = = = = 81 25 5 9 81 9 ( 25 ) 81 9 ( 25 9) 9 (16) 9 25 25 1 ( 25 ) 25 = 9 ( 2) = 18 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1850 y( 5 ) 25 17 8 1 9 25 16 2 Answer: Local Maximum(s): 25 18 Local Minimum(s): 1 2 NOTE: A graph of this function was given in Lesson 9. 7. h( z ) 10 3 ( z 2 3z 18 ) 2 NOTE: The domain of the function h is the set of all real numbers. We differentiated this function in Lesson 10 obtaining that h( z ) 20 ( 2 z 3 ) 20 ( 2 z 3 ) = 3 ( z 2 3z 18 )1 / 3 3[ ( z 6 ) ( z 3 ) ]1 / 3 Sign of h( z ) : Dec Inc + 6 Critical Number(s): 6 , Dec Inc + 3 2 3 3 ,3 2 NOTE: A local minimum occurs when z 6 and the local minimum is 3 h( 6 ) . A local maximum occurs when z and the local minimum is 2 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1850 3 h 2 h( 3 ) . . A local minimum occurs when x 3 and the local minimum is h( 6 ) 10 3 ( 36 18 18 ) 2 = 10 3 0 = 0 3 9 9 h 10 3 18 2 4 2 10 3 81 4 2 10 3 4 38 3 = 10 10 3 36 = 43 3 81 4 3 2 = 10 3 2 = 10 = 2 = 4 ( 34 ) 2 812 3 = 10 = 43 42 5 32 4 32 4 3 9 18 72 4 4 4 3 49 2 45 3 36 = 2 h( 3 ) 10 3 ( 9 9 18 ) 2 = 10 3 0 = 0 45 3 36 Local Maximum(s): 2 Answer: Local Minimum(s): 0 8. 2 x 2 11 f ( x) 4 x 2 16 NOTE: The domain of the function f is given by ( , 4 ) ( 4 , ) 2 since we need that x 16 0 . We differentiate this function in Example 8 above obtaining that Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1850 3x ( 2 x 2 39 ) f ( x ) 2 ( x 2 16 ) 5 / 4 Sign of f ( x ) : Dec Inc + Dec 39 2 Inc + 4 4 39 2 NOTE: The domain of the function f is given by ( , 4 ) ( 4 , ) 39 , 2 Critical Number(s): 39 2 39 and the local minimum 2 NOTE: A local minimum occurs when x f is . A local minimum occurs when x 39 . minimum is f 2 39 2 f 39 2 28 4 7 4 2 4 4 = 39 2 11 39 2 = 2 39 4 16 2 f 28 4 4 7 2 = 28 4 2 4 7 4 73 4 73 = 39 11 4 39 32 2 2 28 4 2 7 3 4 39 and the local 2 74 686 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1850 = 28 = 4 7 2 28 4 2 343 7 = = Answer: Local Maximum(s): None Local Minimum(s): 4 4 686 9. s( t ) 2 t 3 13t 2 20 t 18 NOTE: The domain of the function s is the set of all real numbers. s( t ) 6 t 2 26 t 20 = 2 ( 3 t 2 13 t 10 ) = 2 ( t 5 ) ( 3 t 2 ) Inc + Sign of s ( t ) : Dec Critical Number(s): Inc + 2 3 5 2 ,5 3 NOTE: A local maximum occurs when t 2 and the local maximum is 3 2 s . A local minimum occurs when t 5 and the local minimum is 3 s( 5 ) . 16 52 40 16 156 360 486 674 2 s 18 = = 27 9 3 27 27 27 27 27 3 s( 5 ) 250 325 100 18 157 Answer: Local Maximum(s): 674 27 Local Minimum(s): 157 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1850 10. y x 4 8x 3 12 NOTE: The domain of the function y is the set of all real numbers. dy 4 x 3 24 x 2 = 4 x 2 ( x 6 ) dx Sign of dy : dx Dec Inc Inc + + 6 0 Critical Number(s): 6 , 0 NOTE: A local minimum occurs when x 6 and the local minimum is y ( 6 ) . Neither a local maximum nor a local minimum occurs at x 0 since the function y is increasing on the left-hand side of x 0 and it is still increasing on the right-hand side of x 0 . Using a calculator: y ( 6 ) 1296 1728 12 444 or y( 6 ) 6 4 8 63 12 = 6 ( 6 3 8 6 2 2 ) = 6[ 6 2 ( 6 8 ) 2 ] = 6 [ 36 ( 2 ) 2 ] = 6 ( 72 2 ) = 6 ( 74 ) = 444 Answer: Local Maximum(s): None Local Minimum(s): 444 11. g ( x ) 5x 4 x2 NOTE: The domain of the function g is the set of all real numbers such that x 0. Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1850 g ( x ) 5 8 x 3 = x 3 5x 3 8 (5 x 8 ) = x3 3 Inc + Sign of g ( x ) : Dec 0 2 3 + 5 NOTE: 0 is not in the domain of the function g Critical Number(s): 2 3 5 NOTE: A local maximum occurs when x 2 3 5 and the local maximum 2 g is 3 . 5 8 5 4 5x 4 4 2 5 g g ( x ) 5 x Since , then 3 5 = 4 x2 x2 3 25 3 8 4 = 4 3 25 Answer: 12 4 3 = 12 3 25 4 = 3 3 25 25 Local Maximum(s): 3 3 25 Local Minimum(s): None 12. f ( x ) 4 x (15 x) Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1850 NOTE: The domain of the function f is the interval [ 0 , ) . f ( x ) x1 / 4 (15 x) = 15 x1 / 4 x 5 / 4 f ( x ) 5 3/ 4 5( 3 x ) 15 3 / 4 5 1 / 4 x x x (3 x) = = 4 4 4 4 x 3/ 4 Dec Inc + Sign of f ( x ) : 0 3 Critical Number(s): 3 NOTE: A local minimum occurs when x 3 and the local minimum is f ( 3 ) . Since f ( x ) 4 x (15 x) , then f ( 3 ) 12 4 3 Answer: Local Maximum(s): None Local Minimum(s): 12 4 3 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1850