Explanation Given Information: The provided function is: g ( x ) = ( x − 1 ) 2 3 Formula used: First Derivative Test, suppose that c is a critical point of the continuous function f and its derivative is defined for x close to, and on both sides of x = c . Then, determine the sign of the derivative to the left and the right of x = c . If f ′ ( x ) is positive to the left of x = c and negative to the right, then f has a maximum at x = c . If f ′ ( x ) is negative to the left of x = c and positive to the right, then f has a minimum at x = c . If f ′ ( x ) has the same sign on both sides of x = c , then f has neither a maximum nor a minimum at x = c . Stationary Points are the points in the interior of the domain where the derivative is zero. To locate stationary points set f ′ ( x ) = 0 and solve for x . Singular Points are the points in the interior of the domain where the derivative is not defined. To locate singular points, find the interior points x where f ′ ( x ) is not defined but f ( x ) is defined. Closed intervals contain end points but open intervals do not have end points. Calculation: Consider the provided function, g ( x ) = ( x − 1 ) 2 3 Differentiating both sides of the function with respect to x as, g ′ ( x ) = 2 3 ( x − 1 ) − 1 3 Rewriting the equation without the negative exponent to get g ′ ( x ) = 2 3 ( x − 1 ) 1 3 The above equation can be zero only if the numerator is zero. Since the numerator is 2 which is not equal to zero, hence no stationary points. g ′ ( x ) = 2 3 ( x − 1 ) 1 3 The above equation is not defined for x = 1

Student Solutions Manual for Waner/Costenoble's Finite Math and Applied Calculus, 7th

7th Edition

ISBN: 9781337275972

Author: Waner, Stefan; Costenoble, Steven

Publisher: Cengage Learning

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In simple mathematics terms, a domain is defined as the space on which all the possible inputs for the function will lie. It can also be expressed as a set of departures for the required function.

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Chapter 12, Problem 5RE

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To calculate: The exact location of the relative and absolute extrema of the function g(x)=(x−1)23.

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Chapter 12, Problem 4RE

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Chapter 12 Solutions

Student Solutions Manual for Waner/Costenoble's Finite Math and Applied Calculus, 7th

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The exact location of the relative and absolute extrema of the function g ( x ) = ( x − 1 ) 2 3 .

Solution Summary: The author calculates the exact location of the relative and absolute extrema of a function g(x) = c and its derivative is defined for x close to, and on both sides

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To calculate: The exact location of the relative and absolute extrema of the function g(x)=(x−1)23.

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Chapter 12 Solutions