Explanation Given Information: The provided equation is f ( x ) = x 3 − x 2 − x − 1 . Formula used: First Derivative Test is, assume that for a critical point c of the continuous function f , 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 and negative to left and right of x = c respectively, then at the point x = c , f has a maximum. If f ' ( x ) is negative and positive to left and right of x = c respectively, then at the point x = c , f has a minimum. If f ' ( x ) has either both positive or negative at x = c , then f has neither a maximum nor a minimum at the point x = c . Stationary Points are the points in the interior of the domain where the derivative is zero. Singular Points are the points in the interior of the domain where the derivative is not defined. Closed intervals contain end points which are the end points of the functions and the function does not have end points if the interval is open as open intervals do not have any end points. Calculation: Consider the provided equation: f ( x ) = x 3 − x 2 − x − 1 Differentiate both sides of the equation with respect to x . f ' ( x ) = 3 x 2 − 2 x − 1 Now, locate stationary points. Recall that stationary Points are the points in the interior of the domain where the derivative is zero. f ' ( x ) = 0 3 x 2 − 2 x − 1 = 0 Factor 3 x 2 − 2 x − 1 . 3 x 2 − 2 x − 1 = ( 3 x + 1 ) ( x − 1 ) Thus, ( 3 x + 1 ) ( x − 1 ) = 0 . First consider ( 3 x + 1 ) = 0 . Solve for x . 3 x + 1 = 0 3 x = − 1 x = − 1 3 Next consider ( x − 1 ) = 0

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

7th Edition

ISBN: 9781337291293

Author: Waner, Stefan; Costenoble, Steven

Publisher: Brooks Cole

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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 5, Problem 2RE

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To calculate: The exact location of the relative and absolute extrema of the function f(x)=x3−x2−x−1 with domain (−∞,∞).

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Chapter 5, Problem 1RE

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

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The exact location of the relative and absolute extrema of the function f ( x ) = x 3 − x 2 − x − 1 with domain ( − ∞ , ∞ ) .

Solution Summary: The author explains how to calculate the exact location of relative and absolute extrema of the function f(x).

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To calculate: The exact location of the relative and absolute extrema of the function f(x)=x3−x2−x−1 with domain (−∞,∞).

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Chapter 5 Solutions