Explanation Definition used: If f ′ ( x ) > 0 on an interval ( a , b ) , then the graph of f ( x ) increases on that interval and if f ′ ( x ) < 0 then f ( x ) decreases on that interval. Concavity: For the interval ( a , b ) , if f ″ ( x ) > 0 , then the function f is concave upward and if f ″ ( x ) < 0 , then f is concave downward. If a graph approaches a line that is neither horizontal nor vertical as x approaches ∞ and − ∞ , then that line is called an oblique asymptote. Calculation: Consider the function f ( x ) = x 2 − 3 x + 2 x 2 − 4 x + 4 . The function f ( x ) = x 2 − 3 x + 2 x 2 − 4 x + 4 can be simplified as follows. f ( x ) = x 2 − 3 x + 2 x 2 − 4 x + 4 = ( x − 2 ) ( x − 1 ) ( x − 2 ) ( x − 2 ) = ( x − 1 ) ( x − 2 ) Thus, the function f ( x ) = x 2 − 3 x + 2 x 2 − 4 x + 4 can be rewritten as f ( x ) = x − 1 x − 2 . The domain of the function is obtained as follows. The function f ( x ) = x 2 − 3 x + 2 x 2 − 4 x + 4 is not defined when the denominator becomes zero. Hence equate the denominator to zero to obtain the undefined points. x 2 − 4 x + 4 = 0 ( x − 2 ) ( x − 2 ) = 0 x = 2 Therefore, the domain of f ( x ) = x 2 − 3 x + 2 x 2 − 4 x + 4 is all real numbers except 2. For rational functions, the vertical asymptote are the undefined points of the simplified function. The function is not defined at x = 2 . So, the vertical asymptote is x = 2 . For any rational function with the same degree of numerator and denominator, the horizontal asymptote is at y = a m x m b n x n , where m and n are leading coefficients. In the function f ( x ) = x − 1 x − 2 , the horizontal asymptote is y = x x = 1 . Therefore, the horizontal asymptote is y = 1 . The x -intercept of f ( x ) = x − 1 x − 2 can be found by substituting y = f ( x ) = 0 . x − 1 x − 2 = 0 x − 1 = 0 x = 1 Therefore, the x- intercept is x = 1 . The y -intercept of f ( x ) = x − 1 x − 2 can be found by substituting x = 0 in f ( x ) . f ( 0 ) = y = 0 − 1 0 − 2 y = − 1 − 2 y = 1 2 Therefore, the y- intercept is y = 1 2 . Differentiate f ( x ) = x − 1 x − 2 with respect to x on both sides, f ′ ( x ) = d d x ( x − 1 x − 2 ) = ( x − 2 ) ( 1 ) − ( x − 1 ) ( 1 ) ( x − 2 ) 2 = x − 2 − x + 1 ( x − 2 ) 2 = − 1 ( x − 2 ) 2 Therefore, f ′ ( x ) = − 1 ( x − 2 ) 2 . There exists partition numbers where the function f ′ ( x ) = 0 or where f ′ ( x ) is not defined. The function f ′ ( x ) = − 1 ( x − 2 ) 2 is not defined when the denominator is equal to 0

Pearson eText for College Mathematics for Business, Economics, Life Sciences, and Social Sciences -- Instant Access (Pearson+)

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ISBN: 9780137553341

Author: Raymond Barnett, Michael Ziegler

Publisher: PEARSON+

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Chapter 11.4, Problem 74E

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To explain: The pertinent information obtained by applying the graphing strategy and also to sketch the graph of the function f(x)=x2−3x+2x2−4x+4.

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Chapter 11.4, Problem 73E

Chapter 11.4, Problem 75E

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Chapter 11 Solutions

Pearson eText for College Mathematics for Business, Economics, Life Sciences, and Social Sciences -- Instant Access (Pearson+)

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 pertinent information obtained by applying the graphing strategy and also to sketch the graph of the function f ( x ) = x 2 − 3 x + 2 x 2 − 4 x + 4 .

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To explain: The pertinent information obtained by applying the graphing strategy and also to sketch the graph of the function f(x)=x2−3x+2x2−4x+4.

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Chapter 11 Solutions