Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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#30) find the absolute extremum, if any, given by the second derivative test for each function
In Problems 27–42, find the absolute extremum, if any, given by the second derivative test for each function. 27. \( f(x) = x^2 - 4x + 4 \) 28. \( f(x) = x^2 + 2x + 1 \) 29. \( f(x) = -x^2 - 2x + 5 \) 30. \( f(x) = -x^2 + 6x + 1 \) 31. \( f(x) = x^3 - 3 \) 32. \( f(x) = 6 - x^3 \) 33. \( f(x) = x^4 - 7 \) 34. \( f(x) = 8 - x^4 \) 35. \( f(x) = x + \frac{4}{x} \) 36. \( f(x) = x + \frac{9}{x} \) 37. \( f(x) = \frac{-3}{x^2 + 2} \) 38. \( f(x) = \frac{2}{x^2 + 3} \) 39. \( f(x) = \frac{1 - x}{x^2 - 4} \) 40. \( f(x) = \frac{x - 1}{x^2 - 1} \) 41. \( f(x) = \frac{-x^2}{x^2 + 4} \) 42. \( f(x) = \frac{x^2}{x^2 + 1} \)
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Transcribed Image Text:In Problems 27–42, find the absolute extremum, if any, given by the second derivative test for each function. 27. \( f(x) = x^2 - 4x + 4 \) 28. \( f(x) = x^2 + 2x + 1 \) 29. \( f(x) = -x^2 - 2x + 5 \) 30. \( f(x) = -x^2 + 6x + 1 \) 31. \( f(x) = x^3 - 3 \) 32. \( f(x) = 6 - x^3 \) 33. \( f(x) = x^4 - 7 \) 34. \( f(x) = 8 - x^4 \) 35. \( f(x) = x + \frac{4}{x} \) 36. \( f(x) = x + \frac{9}{x} \) 37. \( f(x) = \frac{-3}{x^2 + 2} \) 38. \( f(x) = \frac{2}{x^2 + 3} \) 39. \( f(x) = \frac{1 - x}{x^2 - 4} \) 40. \( f(x) = \frac{x - 1}{x^2 - 1} \) 41. \( f(x) = \frac{-x^2}{x^2 + 4} \) 42. \( f(x) = \frac{x^2}{x^2 + 1} \)
### Example 30: Quadratic Function The given function is: \[ f(x) = -x^2 + 6x + 1 \] #### Key Characteristics: - **Type of Function**: This is a quadratic function, which is a polynomial of degree 2. - **General Form**: The standard form of a quadratic function is \( f(x) = ax^2 + bx + c \) where \( a = -1 \), \( b = 6 \), and \( c = 1 \). #### Graphical Representation: - **Parabola Orientation**: Since the coefficient of \( x^2 \) is negative (\( a = -1 \)), the parabola opens downward. - **Vertex**: The vertex of the parabola can be found using the formula \( x = -\frac{b}{2a} \). - **Axis of Symmetry**: This vertical line passes through the vertex, with an equation \( x = -\frac{b}{2a} \). - **Y-Intercept**: The point where the graph intersects the y-axis, found by evaluating \( f(0) = c \). #### Additional Analysis: - **Vertex Form**: The function can be rewritten in vertex form to easily identify the vertex. - **Roots**: To find the x-intercepts (roots), solve \( f(x) = 0 \). This function is important for understanding the characteristics of quadratic equations in algebra, including their graphical behavior and real-world applications.
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Transcribed Image Text:### Example 30: Quadratic Function The given function is: \[ f(x) = -x^2 + 6x + 1 \] #### Key Characteristics: - **Type of Function**: This is a quadratic function, which is a polynomial of degree 2. - **General Form**: The standard form of a quadratic function is \( f(x) = ax^2 + bx + c \) where \( a = -1 \), \( b = 6 \), and \( c = 1 \). #### Graphical Representation: - **Parabola Orientation**: Since the coefficient of \( x^2 \) is negative (\( a = -1 \)), the parabola opens downward. - **Vertex**: The vertex of the parabola can be found using the formula \( x = -\frac{b}{2a} \). - **Axis of Symmetry**: This vertical line passes through the vertex, with an equation \( x = -\frac{b}{2a} \). - **Y-Intercept**: The point where the graph intersects the y-axis, found by evaluating \( f(0) = c \). #### Additional Analysis: - **Vertex Form**: The function can be rewritten in vertex form to easily identify the vertex. - **Roots**: To find the x-intercepts (roots), solve \( f(x) = 0 \). This function is important for understanding the characteristics of quadratic equations in algebra, including their graphical behavior and real-world applications.
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