Explanation Definition used: First derivative test: Suppose c to be a critical number for a function f . Assume that f is continuous on ( a , b ) and differentiable on ( a , b ) except at c . The interval ( a , b ) should have only one critical number of f . Relative maximum : A function is said to have a relative maximum f ( c ) at c if f ′ ( x ) > 0 in the interval ( a , c ) and f ′ ( x ) < 0 in the interval ( c , b ) . Relative minimum: A function is said to have a relative maximum f ( c ) at c if f ′ ( x ) < 0 in the interval ( a , c ) and f ′ ( x ) > 0 in the interval ( c , b ) . Calculation: The given function is f ( x ) = x 3 + 3 x 2 − 24 x + 2 . Obtain the derivative of the function as follows. f ′ ( x ) = ( x 3 + 3 x 2 − 24 x + 2 ) ′ = ( 3 x 2 ) + 3 ( 2 x ) − 24 ( 1 ) + 0 = 3 x 2 + 6 x − 24 Set f ′ ( c ) = 0 and find the value of c . 3 c 2 + 6 c − 24 = 0 3 ( c 2 + 2 c − 8 ) = 0 c 2 + 4 c − 2 c − 8 = 0 c ( c + 4 ) − 2 ( c + 4 ) = 0 Simplify further as follows. c ( c + 4 ) − 2 ( c + 4 ) = 0 ( c + 4 ) ( c − 2 ) = 0 c + 4 = 0 or c − 2 = 0 c = − 4 or c = 2 Therefore, the critical numbers are c = − 4 and c = 2 . These points determine the intervals ( − ∞ , − 4 ) , ( − 4 , 2 ) and ( 2 , ∞ ) . Consider the points from each of these intervals and find the sign of the derivative function at these intervals. Consider a point x = − 5 in the interval ( − ∞ , − 4 ) and find the sign of the derivative at this point. f ′ ( − 5 ) = 3 ( − 5 ) 2 + 6 ( − 5 ) − 24 = 3 × 25 − 30 − 24 = 75 − 54 = 21 Consider a point x = 0 in the interval ( − 4 , 2 ) and find the sign of the derivative at this point

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Chapter 13.2, Problem 16E

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To find: The x-values of all points where the function f(x)=x3+3x2−24x+2 has relative extrema and to find the value of relative extrema.

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Chapter 13.2, Problem 15E

Chapter 13.2, Problem 17E

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EBK FINITE MATHEMATICS AND CALCULUS WIT

Ch. 13.1 - Find where the function is increasing and...Ch. 13.1 - Prob. 2YT Ch. 13.1 - Prob. 3YT Ch. 13.1 - Prob. 4YT Ch. 13.1 - Prob. 1WE Ch. 13.1 - Prob. 2WE Ch. 13.1 - Prob. 3WE Ch. 13.1 - Prob. 4WE Ch. 13.1 - Prob. 5WE Ch. 13.1 - Prob. 6WE

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The x -values of all points where the function f ( x ) = x 3 + 3 x 2 − 24 x + 2 has relative extrema and to find the value of relative extrema.

Solution Summary: The author explains how to find the x-values of all points where a function f has relative extrema.

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To find: The x-values of all points where the function f(x)=x3+3x2−24x+2 has relative extrema and to find the value of relative extrema.

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