- Precalculus review worksheet sec. 9.1 to 9.3 how to#
- Precalculus review worksheet sec. 9.1 to 9.3 license#
Interval of downward concavity: (1, 2) (4, 7) (8, ) d. Interval of upward concavity: (, 1) (2, 4) (7, 8) c. We can now answer the final two questions. f 0 f 0 on an interval, then f is increasing on that interval. 2 2 y=f () Solution: First we draw picture 1 for this graph. coordinate(s) of the point(s) of inflection. Intervals over which f is concave upward or concave downward, d. Intervals over which f is increasing/decreasing, b. A Preview of Calculus: The Limit, Derivative, and Integral of a Function 14.1 Finding Limits Using Tables and Graphs 14.2 Algebra Techniques for Finding Limits 14.3 One-sided Limits Continuous Functions 14.4 The Tangent Problem The Derivative 14.5 The Area Problem The Integral Chapter Review Chapter Test Chapter Projects Appendix A. Our resource for Modern Chemistry includes answers to chapter exercises, as well as detailed.
Precalculus review worksheet sec. 9.1 to 9.3 how to#
Below is the graph of the derivative f () of a continuous function f(). At Quizlet, we’re giving you the tools you need to take on any subject Now, with expert-verified solutions from Modern Chemistry 2nd Edition, you’ll learn how to solve your toughest homework problems. For picture 1 we will include the discontinuities.
(Remember 3, -3 are not in the domain of f). This means that this function has no critical numbers, since its derivative is never 0 and it is differentiable at all in its domain. Picture = 0 f () = (2 9) (2) ( 2 9) ( 2 9) 2 = 0 = 2 9 ( 2 9) 2 The last equation has no solution, since it is equivalent to 2 = 9 which has no solution. To find the horizontal asymptotes we evaluate the following. We can find the vertical asymptotes by taking the it as approaches our discontinuities 3 and = 2 9 = 3 and = 3 are vertical asymptotes. Notice that f( ) = f(), so f is an odd function. Sketch the graph of Solution: Here f() = 2 9 y = 2 9 is a rational function, so its domain consists of all so that the denominator is nonzero, i.e. You can solve by substitution (isolating or ), graphically, or by addition. By the time the company produces one unit they are already making profit. This means there is no realistic break-even point. f 0 f is concave downward f is concave upward 1 (1,f(1)) is an inflection point f (0) = 6(0) 6 0ģ E. No, you can either have zero, one, or infinitely many. Picture 2 f () = 0 The critical number of f is = 1. Picture = f () = 0 3( 2 2 3) = 0 3( + 1)( 3) = 0 The critical numbers of f are = 1 and = 3. We see there are no horizontal asymptotes. For horizontal asymptotes we check the its at ±. To find the asymptotes, we first notice that there are no vertical asymptotes. We skip the -intercept since we would have to solve the cubic equation = 0. Here f() = is a polynomial, so its domain is (, ). Sketch the graph of y = Eamples Solution: A. Sketch: Use all of the previous information to sketch y = f().Ģ 1. Concavity and Inflection Points: Draw picture 2 for f. Increase/Decrease and Etrema: Draw picture 1 for f. An oblique asymptote is a an asymptote which is neither a vertical nor a horizontal line. We have already found vertical and horizontal asymptotes of functions. Also find the slant or oblique asymptotes of f. Asymptotes: Find the horizontal and vertical asymptotes of f. Also determine whether f is periodic and what the period P is. Symmetries: Determine whether f is an even function (f( ) = f()), and odd function (f( ) = f()) or neither. Skip finding -intercepts if f() is very complicated. Intercepts: Find the y-intercept (f(0)) and any -intercepts.
Precalculus review worksheet sec. 9.1 to 9.3 license#
Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License license.1 Notes on Curve Sketching The following checklist is a guide to sketching the curve y = f().
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