Explanation Given: The given function is g ( x ) = x 3 − 3 x 2 + 5 . Definition used: If the derivative of function f ′ ( x ) changes sign from positive to negative as the function’s graph move across a critical number c , then function f has relative maximum at x = c . If the derivative of function f ′ ( x ) changes sign from negative to positive as the function’s graph move across a critical number c , then function f has relative minimum at x = c . Calculation: To get the relative minimum and maximum values first find the derivative of function. d d x g ( x ) = d d x ( x 3 − 3 x 2 + 5 ) = d d x ( x 3 ) − d d x ( 3 x 2 ) + d d x ( 5 ) = 3 x 2 − 6 x = 3 x ( x − 2 ) The derivative of function is g ′ ( x ) = 3 x ( x − 2 ) . To get the zero, set the derivative of function is equal to zero. g ′ ( x ) = 0 3 x ( x − 2 ) = 0 The zeroes of the derivative of function is x = 0 and x = 2 . The critical points of the function are x = 0 and x = 2

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Chapter 4.1, Problem 60E

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The relative maximum and relative minimum for the function g(x)=x3−3x2+5 .

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Chapter 4.1, Problem 59E

Chapter 4.1, Problem 61E

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The relative maximum and relative minimum for the function g ( x ) = x 3 − 3 x 2 + 5 .

Solution Summary: The author calculates the relative minimum and relative maximum values for the function g(x).

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The relative maximum and relative minimum for the function g(x)=x3−3x2+5 .

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