Explanation Definition used: First derivative test: Suppose c to be a critical number for a function f . Assume that f is continuous on ( a , b ) and differentiable on ( a , b ) except at c . The interval ( a , b ) should have only one critical number of f . Relative maximum : A function is said to have a relative maximum f ( c ) at c if f ′ ( x ) > 0 in the interval ( a , c ) and f ′ ( x ) < 0 in the interval ( c , b ) . Relative minimum: A function is said to have a relative maximum f ( c ) at c if f ′ ( x ) < 0 in the interval ( a , c ) and f ′ ( x ) > 0 in the interval ( c , b ) . Calculation: The given function is f ( x ) = x 2 + 1 x . Obtain the derivative of the function as follows. f ′ ( x ) = ( x 2 + 1 x ) ′ = 2 x + ( − 1 ) x − 1 − 1 = 2 x − x − 2 = 2 x − 1 x 2 Set f ′ ( c ) = 0 and find the value of c . 2 c − 1 c 2 = 0 1 c 2 ( 2 c 3 − 1 ) = 0 1 c 2 = 0 or 2 c 3 − 1 = 0 Here, the derivative does not exist at a c that is not defined but the derivative is zero for c = 1 2 3 or 0

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Author: RITCHEY

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Chapter 13.2, Problem 26E

To determine

To find: The x-values of all points where the function f(x)=x2+1x has relative extrema and the value of relative extrema.

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Chapter 13.2, Problem 25E

Chapter 13.2, Problem 27E

Chapter 13 Solutions

EBK FINITE MATHEMATICS AND CALCULUS WIT

Ch. 13.1 - Find where the function is increasing and...Ch. 13.1 - Prob. 2YT Ch. 13.1 - Prob. 3YT Ch. 13.1 - Prob. 4YT Ch. 13.1 - Prob. 1WE Ch. 13.1 - Prob. 2WE Ch. 13.1 - Prob. 3WE Ch. 13.1 - Prob. 4WE Ch. 13.1 - Prob. 5WE Ch. 13.1 - Prob. 6WE

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The x -values of all points where the function f ( x ) = x 2 + 1 x has relative extrema and the value of relative extrema.

Solution Summary: The author explains that the relative minimum of 1.89 occurs at x=0.79.

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To find: The x-values of all points where the function f(x)=x2+1x has relative extrema and the value of relative extrema.

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Chapter 13 Solutions