To determine To find: The absolute extrema of the function f ( x ) = 3 x 2 3 − 3 x 5 3 on the interval [ 0 , 8 ] . Explanation Obtain the critical number of the function f ( x ) = 3 x 2 3 − 3 x 5 3 . Differentiate the function f ( x ) = 3 x 2 3 − 3 x 5 3 as follows. f ′ ( x ) = d d x ( 3 x 2 3 − 3 x 5 3 ) = 3 ( 2 3 ) x 2 3 − 1 − 3 ( 5 3 ) x 5 3 − 1 = 2 x − 1 3 − 5 x 2 3 = 2 x 1 3 − 5 x x 1 3 That is, f ′ ( x ) = 1 x 1 3 ( 2 − 5 x ) . The function 1 x 1 3 ( 2 − 5 x ) equal to zero only when 2 − 5 x = 0 . That is, the critical point is x = 2 5 . Now, evaluate the value of the function at critical point x = 2 5 and the endpoint 0 and 8. Substitute x = 0 in the function f ( x ) = 3 x 2 3 − 3 x 5 3 . f ( 0 ) = 3 ( 0 ) 2 3 − 3 ( 0 ) 5 3 = 0 − 0 = 0 That is, f ( 0 ) = 0

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Chapter 6.1, Problem 1YT

To determine

To find: The absolute extrema of the function f(x)=3x23−3x53 on the interval [0,8] .

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Chapter 5, Problem 77RE

Chapter 6.1, Problem 2YT

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Chapter 6 Solutions

EBK CALCULUS WITH APPLICATIONS, BRIEF V

Ch. 6.1 - YOUR TURN 1 Find the absolute extrema of the...Ch. 6.1 - Prob. 2YT Ch. 6.1 - Prob. 1WE Ch. 6.1 - Prob. 2WE Ch. 6.1 - Prob. 1E Ch. 6.1 - Find the locations of any absolute extrema for the...Ch. 6.1 - Prob. 3E Ch. 6.1 - Prob. 4E Ch. 6.1 - Prob. 5E Ch. 6.1 - Prob. 6E

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