Single Variable Calculus - unit2 review - Mathematics, Study notes of Mathematics

Download Single Variable Calculus - unit2 review - Mathematics and more Study notes Mathematics in PDF only on Docsity!

MIT OpenCourseWare http://ocw.mit.edu

18.01 Single Variable Calculus

Fall 2006 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.

x 18.01 UNIT 2 REVIEW; Fall 2007 The central theme of Unit 2 is that knowledge of f �^ (and sometimes f ��) tells us something about f itself. This is even true of our first topic, approximation. For instance, knowing that f (x) = e satisfies f (0) = 1 and f �(0) = 1, we can say ex^ � 1 + x provided x � 0 The linear function 1 + x is much simpler than e^ x , so f (0) and f �(0) give us a (very) simplified picture of our function, useful only near near 0. For more detail, use the quadratic approximation, e^ x � 1 + x + x^2 / 2 provided x � 0 (still only works well near 0) The second and third practice exams are actual tests from previous years. The exam this year is similar to the one from 2006 posted at our site. It has 6 questions covering the following topics. (No Newton’s method, but there is a seventh, extra credit problem.)

1. Linear and/or quadratic approximations
2. Sketch a graph y = f (x)
3. Max/min
4. Related rates
5. Find antiderivatives and solve a differential equation by separating variables
6. Mean value theorem. Remarks.
7. Recall that linear [and quadratic] approximation is f (x) � f (a) + f �(a)(x − a) [+(f ��(a)/2)(x − a)^2 ]
8. You should expect to graph a function y = f (x), where f (x) is a rational function (ratio of polynomials). Warnings: a) When asked to label the critical point on the graph, find and mark the point (a, b). In lecture we called x = a the critical point and y = b the critical value, and this is what is used in 18.02, and elsewhere. But for this exam (and this is just an inconsistency in language that you will have to tolerate) the words “critical point” refer to the point on the graph (a, b), not the number a and the point on the x-axis. The same applies to inflection points. b) y = 1/(x − 1) is decreasing on the intervals −≈ < x < 1 and 1 < x < ≈, but it is not decreasing on the interval −≈ < x < ≈. Draw the graph to see. You cannot just use the fact that y�^ = − 1 /(x − 1)^2 < 0 because there is a point in the middle at which y is not differentiable — and not even continuous. So the mean value theorem does not apply. c) Similarly, y = 1/(x − 1)^2 is concave up on −≈ < x < 1 and 1 < x < ≈, but it is not concave up on the interval −≈ < x < ≈. Here y��^ = 6/(x − 1)^4 > 0, but there is a singularity in the middle. Plot the graph yourself to see. 1