Explanation Differentiate the function with respect to the x . d d x ( f ( x ) ) = d d x ( x 3 + 6 x 2 ) f ′ ( x ) = 3 ( x 2 ) + 6 ( 2 x ) = 3 x 2 + 12 x Second derivative of the function y = x 3 + 6 x 2 . f ′ ( x ) = 3 x 2 + 12 x f ″ ( x ) = 3 ( 2 x ) + 12 = 6 x + 12 For the point of inflection, substitute f ″ ( x ) = 0 . 6 x + 12 = 0 6 x = − 12 x = − 12 6 x = − 2 Hence, at x = − 2 the curve changes its concavity. Substitute, x = − 2 in the function y = x 3 + 6 x 2 . y = ( − 2 ) 3 + 6 ( − 2 ) 2 = − 8 + 24 = 16 Hence, point of inflection is ( − 2 , 16 ) or ( − 4 , 32 ) . For, local maxima and local minima substitute, f ′ ( x ) = 0 in 3 x 2 + 12 x = 0 . 3 x 2 + 12 x = 0 3 x ( x + 4 ) = 0 3 x ( x + 4 ) = 0 x ( x + 4 ) = 0 Hence, Local maxima or local minima occurs when x = 0 , − 4 . Substitute, x = 0 in f ″ ( x ) = 6 x + 12 . f ″ ( x ) = 6 x + 12 f ″ ( 0 ) = 6 × 0 + 12 = 12 Substitute, x = 0 in y = x 3 + 6 x 2 . y = ( 0 ) 3 + 6 ( 0 ) 2 = 0 Hence, at ( 0 , 0 ) _ local minima occur. Substitute, x = − 4 in f ″ ( x ) = 6 x + 12

Pearson eText for Basic Technical Mathematics with Calculus -- Instant Access (Pearson+)

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ISBN: 9780137554843

Author: Allyn Washington, Richard Evans

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Chapter 24 Solutions

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Chapter 24 Solutions