Explanation Theorem used: “If ( c , f ( c ) ) is the inflection point of f , then c is a partition number for f ”. Calculation: The inflection point is a point on the graph where the concavity changes. That is, the point where f ″ ( x ) change its sign from positive to negative or vice versa. Since x 2 − 4 x + 5 > 0 for any values of x , the domain of the function ln ( x 2 − 4 x + 5 ) is ( − ∞ , ∞ ) . Obtain the second derivative of the function in order to get the inflection point. f ″ ( x ) = d 2 d x 2 ( ln ( x 2 − 4 x + 5 ) ) = d d x ( d d x ( ln ( x 2 − 4 x + 5 ) ) ) = d d x ( 1 x 2 − 4 x + 5 ( 2 x − 4 ) ) = d d x ( 2 x − 4 x 2 − 4 x + 5 ) On further simplification, f ″ ( x ) = ( x 2 − 4 x + 5 ) d d x ( 2 x − 4 ) − ( 2 x − 4 ) d d x ( x 2 − 4 x + 5 ) ( x 2 − 4 x + 5 ) 2 = ( x 2 − 4 x + 5 ) ( 2 ) − ( 2 x − 4 ) ( 2 x − 4 ) ( x 2 − 4 x + 5 ) 2 = 2 x 2 − 8 x + 10 − ( 2 x − 4 ) 2 ( x 2 − 4 x + 5 ) 2 = 2 x 2 − 8 x + 10 − ( 4 x 2 − 16 x + 16 ) ( x 2 − 4 x + 5 ) 2 Further simplified as, f ″ ( x ) = 2 x 2 − 8 x + 10 − 4 x 2 + 16 x − 16 ( x 2 − 4 x + 5 ) 2 = − 2 x 2 + 8 x − 6 ( x 2 − 4 x + 5 ) 2 Thus, f ″ ( x ) = − 2 x 2 + 8 x − 6 ( x 2 − 4 x + 5 ) 2 . Equate f ″ ( x ) to 0 and obtain the partition number. − 2 x 2 + 8 x − 6 ( x 2 − 4 x + 5 ) 2 = 0 − 2 x 2 + 8 x − 6 = 0 − 2 ( x 2 − 4 x + 3 ) = 0 x 2 − 4 x + 3 = 0 Further simplified as, ( x − 3 ) ( x − 1 ) = 0 ( x − 3 ) =0 and ( x − 1 ) = 0 x = 3 and x = 1 Thus, the partition numbers are x = 3 and x = 1

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Chapter 11.2, Problem 38E

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To find: The intervals where the graph of the function f(x)=ln(x2−4x+5) is concave upward and concave downward and the x coordinate of the inflection points.

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Chapter 11.2, Problem 37E

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Ch. 11.1 - Matched Problem 1 Repeat Example 1 for f(x) = x2 –...Ch. 11.1 - Matches Problem 2 Find the critical numbers of f,...Ch. 11.1 - Matched Problem 3 Find the critical numbers of f,...Ch. 11.1 - Prob. 4MP Ch. 11.1 - Prob. 5MP Ch. 11.1 - Prob. 6MP Ch. 11.1 - Prob. 7MP Ch. 11.1 - Prob. 8MP Ch. 11.1 - Prob. 9MP Ch. 11.1 - Skills Warm-up Exercises In Problems 1-8, inspect...

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The intervals where the graph of the function f ( x ) = ln ( x 2 − 4 x + 5 ) is concave upward and concave downward and the x coordinate of the inflection points.

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To find: The intervals where the graph of the function f(x)=ln(x2−4x+5) is concave upward and concave downward and the x coordinate of the inflection points.

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