MATH 12002 - CALCULUS I § 3.1: Maximum and Minimum Values - Examples Professor Donald L. White Department of Mathematical Sciences Kent State University D.L. White (Kent State University) 1 / 7
Definitions & Theorems Extreme Value Theorem If y = f ( x ) is a continuous function on a closed interval [ a , b ] , then f has both an absolute maximum and an absolute minimum on [ a , b ] . Theorem If y = f ( x ) is continuous on a closed interval [ a , b ] , then its absolute maximum and minimum on [ a , b ] must occur at one of the endpoints or at a point c, a < c < b, where f has a local maximum or minimum. Theorem If y = f ( x ) has a local maximum or minimum at x = c, then either f ′ ( c ) = 0 or f ′ ( c ) is undefined. A number c in the domain of f where f ′ ( c ) = 0 or f ′ ( c ) is undefined is called a critical number for f . D.L. White (Kent State University) 2 / 7
Definitions & Theorems Putting these theorems together, we have the following procedure for finding absolute maximum and minimum values of a continuous function on a closed interval: Let y = f ( x ) be continuous on the closed interval [ a , b ]. 1 Find all critical numbers c of f in ( a , b ); that is, all c with a < c < b such that f ′ ( c ) = 0 or f ′ ( c ) is undefined. 2 Compute the y values f ( a ) and f ( b ) of f at the endpoints a and b and the y value f ( c ) for every critical number c found in Step 1. 3 Compare the y values found in Step 2. The largest is the absolute maximum value of f on [ a , b ] and the smallest is the absolute minimum value of f on [ a , b ]. D.L. White (Kent State University) 3 / 7
Examples Example 1 Find the absolute maximum and absolute minimum values of f ( x ) = x 4 − 2 x 2 + 3 on the interval [ − 2 , 3] . Solution We first need to find the critical numbers. We have f ′ ( x ) = 4 x 3 − 4 x = 4 x ( x 2 − 1) = 4 x ( x − 1)( x + 1) . Hence f ′ ( x ) is defined everywhere and is 0 when one of the factors of f ′ ( x ) is 0 , so at x = 0 , x = 1 , and x = − 1 . Therefore, the critical numbers for f are 0 , 1 , and − 1 . [Continued → ] D.L. White (Kent State University) 4 / 7
Examples Example 1 Solution [continued] The maximum and minimum must occur at one of the critical numbers 0 , 1 , − 1 , or at an endpoint − 2 or 3 . For f ( x ) = x 4 − 2 x 2 + 3 , we compute 0 4 − 2(0 2 ) + 3 = 3 f (0) = 1 4 − 2(1 2 ) + 3 = 1 − 2 + 3 = 2 f (1) = ( − 1) 4 − 2(( − 1) 2 ) + 3 = 1 − 2 + 3 = 2 f ( − 1) = ( − 2) 4 − 2(( − 2) 2 ) + 3 = 16 − 8 + 3 = 11 f ( − 2) = 3 4 − 2(3 2 ) + 3 = 81 − 18 + 3 = 66 . f (3) = Therefore, the absolute maximum value of f ( x ) is y = 66 at x = 3 , and the absolute minimum value of f ( x ) is y = 2 at x = 1 and x = − 1 . D.L. White (Kent State University) 5 / 7
Examples Example 2 Find the absolute maximum and absolute minimum values of f ( x ) = x (1 − x ) 2 / 5 on the interval � 1 � 2 , 2 . Solution We first need to find the critical numbers. We have 1 · (1 − x ) 2 / 5 + x · 2 5 (1 − x ) − 3 / 5 ( − 1) f ′ ( x ) = 2 x (1 − x ) 2 / 5 − = 5(1 − x ) 3 / 5 5(1 − x ) 3 / 5 (1 − x ) 2 / 5 − 2 x = 5(1 − x ) 3 / 5 5(1 − x ) − 2 x 5 − 7 x = = 5(1 − x ) 3 / 5 . 5(1 − x ) 3 / 5 [Continued → ] D.L. White (Kent State University) 6 / 7
Examples Example 2 Solution [continued] 5 − 7 x We have f ′ ( x ) = 5(1 − x ) 3 / 5 , and so f ′ ( x ) = 0 when 5 − 7 x = 0 , so at x = 5 / 7 , and f ′ ( x ) is undefined when 5(1 − x ) 3 / 5 = 0 , so at x = 1 . Thus the critical numbers for f are 5 / 7 and 1 . The maximum and minimum must occur at one of the critical numbers 5 / 7 or 1 , or at an endpoint 1 / 2 or 2 . For f ( x ) = x (1 − x ) 2 / 5 , we compute 7 ) 2 / 5 = 5 7 ) 2 / 5 ≈ 0 . 43 f ( 5 5 7 (1 − 5 7 ( 2 7 ) = 1 · (1 − 1) 2 / 5 = 1 · 0 2 / 5 = 0 f (1) = 2 ) 2 / 5 = 1 2 ) 2 / 5 = ( 1 2 ) 7 / 5 ≈ 0 . 38 f ( 1 1 2 (1 − 1 2 ( 1 2 ) = 2 · (1 − 2) 2 / 5 = 2 · ( − 1) 2 / 5 = 2 · 1 = 2 . f (2) = Therefore, the absolute maximum value of f ( x ) is y = 2 at x = 2 , and the absolute minimum value of f ( x ) is y = 0 at x = 1 . D.L. White (Kent State University) 7 / 7
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