Explanation Given Information: The provided function is: f ( x ) = 1 3 x 3 − x 2 − 3 x + 5 and the interval is [ − 2 , 0 ] . Formula used: Maximum-Minimum Principle: Suppose f is a function such that f ′ ( x ) exists for every x in an interval I and there is exactly one ( critical ) value c in I , for which f ′ ( c ) = 0 . Then f ( c ) is the absolute maximum value over I if f ″ ( c ) < 0 or f ( c ) is the absolute minimum value over I if f ″ ( c ) > 0 . Calculation: Consider the provided function: f ( x ) = 1 3 x 3 − x 2 − 3 x + 5 with interval [ − 2 , 0 ] Differentiate f ( x ) = 1 3 x 3 − x 2 − 3 x + 5 with respect to x : d d x f ( x ) = d d x ( 1 3 x 3 − x 2 − 3 x + 5 ) f ′ ( x ) = 1 3 d ( x 3 ) d x − d ( x 2 ) d x − 3 d x d x + d ( 5 ) d x = 1 3 ⋅ 3 ( x 3 − 1 ) − 2 x 2 − 1 − 3 ( 1 ) + 0 = x 2 − 2 x − 3 Substitute f ′ ( x ) = 0 in the above expression: x 2 − 2 x − 3 = 0 ( x − 3 ) ( x + 1 ) = 0 Apply zero product rule on above factors: ( x − 3 ) = 0 or ( x + 1 ) = 0 If ( x − 3 ) = 0 : ( x − 3 ) = 0 x − 3 = 0 x = 3 If ( x + 1 ) = 0 : ( x + 1 ) = 0 x + 1 = 0 x = − 1 For Maximum − Minimum Principal, differentiate f ′ ( x ) = x 2 − 2 x − 3 with respect to x : f ″ ( x ) = d ( x 2 ) d x − 2 d x d x − d ( 3 ) d x = 2 x 2 − 1 − 2 ( 1 ) − 0 = 2 x − 2 As, all points are 3, − 1 and interval [ − 2 , 0 ] Here, 3 would be excluded

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Chapter CR, Problem 37E

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To calculate: The absolute maximum and minimum value, if exist, of function f(x)=13x3−x2−3x+5 in interval [−2,0].

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Chapter CR, Problem 36E

Chapter CR, Problem 38E

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The absolute maximum and minimum value, if exist, of function f ( x ) = 1 3 x 3 − x 2 − 3 x + 5 in interval [ − 2 , 0 ] .

Solution Summary: The author calculates the absolute maximum and minimum value of the function f(x) = 13x