Explanation Given: The function and the interval are: g ( x ) = x 3 + 6 x 2 − 15 x + 1 ; ( − 9 , 0 ) . Formula used: The derivative property as; d d x ( x n ) = n x n − 1 Calculation: g ( x ) = x 3 + 6 x 2 − 15 x + 1 Differentiate the provided function and get, g ′ ( x ) = 3 x 2 + 12 x − 15 Put g ′ ( x ) = 0 . That is 3 x 2 + 12 x − 15 = 0 ( x + 5 ) ( x − 1 ) = 0 x = − 5 , 1 So, − 5 and 1 are the critical points

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Chapter 12, Problem 38RE

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To calculate: The absolute extremum of the function g(x)=x3+6x2−15x+1 on the given interval.

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Chapter 12, Problem 37RE

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The absolute extremum of the function g ( x ) = x 3 + 6 x 2 − 15 x + 1 on the given interval.

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