Explanation Given information: The provided function is: f ( x ) = x 2 − 4 x + 1 Formula used: Stationary Points: To locate stationary points put f ′ ( x ) = 0 and solve for x . Second derivative test: If the function f ( x ) has a stationary point at x = c and that f ″ ( c ) exists then evaluate the sign of f ″ ( c ) : A. If f ″ ( c ) < 0 , then f has a relative maximum at x = c . B. If f ″ ( c ) > 0 , then f has a relative minimum at x = c . Also, if f ″ ( c ) = 0 then the test is fails so use the first derivative test. Calculation: Consider the function, f ( x ) = x 2 − 4 x + 1 . Differentiate the above equation with respect to x , f ′ ( x ) = d d x ( x 2 − 4 x + 1 ) = 2 x − 4 Thus, f ′ ( x ) = 2 x − 4 Substitute 2 x − 4 for f ′ ( x ) in f ′ <

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ISBN: 9781337604963

Author: Stefan Waner, Steven Costenoble

Publisher: Cengage Learning

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Chapter 12.3, Problem 33E

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To calculate: The x-coordinates of all the critical points of the function f(x)=x2−4x+1 and check by first using the second derivative test and by some other method if the test fails, whether the each critical point is relative minimum, relative maximum or neither.

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Chapter 12.3, Problem 32E

Chapter 12.3, Problem 34E

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The x -coordinates of all the critical points of the function f ( x ) = x 2 − 4 x + 1 and check by first using the second derivative test and by some other method if the test fails, whether the each critical point is relative minimum, relative maximum or neither.

Solution Summary: The author calculates the x-coordinates of all the critical points of the function f(x)=x 2-4x+1.

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To calculate: The x-coordinates of all the critical points of the function f(x)=x2−4x+1 and check by first using the second derivative test and by some other method if the test fails, whether the each critical point is relative minimum, relative maximum or neither.

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