Explanation Given Information: The provided function is, h ( x ) = 1 x + 1 x 2 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 equation, h ( x ) = 1 x + 1 x 2 Differentiate the function h ( x ) = 1 x + 1 x 2 on both sides with respect to x as, h ′ ( x ) = − 1 x 2 − 2 x 3 Locate the singular points by f ′ ( x ) = 0 to get, − 1 x 2 − 2 x 3 = 0 Add, 1 x 2 to both sides of the equation to get, − 1 x 2 − 2 x 3 + 1 x 2 = 1 x 2 − 2 x 3 = 1 x 2 Cross multiplying to get, − 2 x 2 = x 3 Subtracting x 3 from both sides of the equation to get − 2 x 2 − x 3 = x 3 − x 3 − 2 x 2 − x 3 = 0 Multiply both sides by − 1 . − 2 x 2 − x 3 = 0 2 x 2 + x 3 = 0 Taking x 2 common from both sides of the equation to get x 2 ( 2 + x ) = 0 Thus, x 2 = 0 x = 0 Or 2 + x = 0 x = − 2 The equation h ′ ( x ) = − 1 x 2 − 2 x 3 is not defined at x = 0 and x = 0 is not an interior point of the domain. Thus, there are no singular points. Now, locate End points. Since, the domain is all real numbers except x = 0 . Thus, there are no end points. The equation has a stationary point at x = − 2

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ISBN: 9781337291248

Author: Waner, Stefan.

Publisher: Cengage Learning,

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Chapter 5, Problem 7RE

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To calculate: The exact location of the relative and absolute extrema of the function h(x)=1x+1x2.

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Chapter 5, Problem 6RE

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The exact location of the relative and absolute extrema of the function h ( x ) = 1 x + 1 x 2 .

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To calculate: The exact location of the relative and absolute extrema of the function h(x)=1x+1x2.

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