homework 1 in Calculus 2
homework 1 in Calculus 2
Project description
Write an outline for sections 6.26.4. Work on the solutions to Section 6.1: exercises 21 and 36, Section 6.2: exercises 54 and 89, Section 6.3: exercises 44 and 64, and Section 6.4: exercises 26 and 74.
Submit a report in the comment box of this homework detailing your progress and status toward completion. The end of the report should contain a brief summary stating whether the outlines and exercises were given sufficient effort (whether the correct answer was derived or not).
NOTE: I do not need to see your work – I only want you to report on your homework activity (for each problem, whether it was found to be difficult, obvious, easy/hard, whether you at first had troubles but eventually found the/a solution, etc.) I would like you to document (i.e., report on) your homework experience. You can grade your own work via the solutions I provide. I award 3 pts for your reported work on the outlined sections (3pts per section outline) and 3pts for each exercise problem reported.
Name Math 1920R5X
Date
Outline for Section 7.1, “Inverse Functions”
I. Inverse Functions
A. Representation of functions
1. Table
2. Graph
3. Mathematical expression
B. Definition: A function has an inverse over its domain if it is onetoone, or equally written, 1:1.
C. Definition: A function is 1:1 if it never takes on the same value twice, i.e., it passes the horizontal line test provided we have a graph of the function.
D. Domain/Range of the inverse function: If the domain/range of the function is given by D and R, respectively, then the domain/range of its inverse, written here as Dinv/Rinv is simply Dinv=R and Rinv=D. See page 415 of the text.
E. Mathematical representation of the inverse function: how to find it. If possible, write down y=f(x), and solve for x in terms of y. For consistent notation, it is possible to exchange x and y when finished so that you again have something that looks like See the steps outlined on page 416. ).(1xfy=
F. Graph of an inverse function is found by reflecting the graph of f about the line y=x.
II. The Calculus of Inverse Functions
A. If f is a 1:1, continuous function defined on an interval, then its inverse function is also continuous. This is a theorem found on page 417 and is used as a condition needed for the existence of the derivative of an inverse function. )(1xfy=
B. Theorem about differentiability of the inverse function: the function f must have the properties, 1) f is 1:1, 2) f is differentiable. Having checked for this, and giving its inverse the following notation; , then at some point a, )()(1xfxg=))((1)(‘axaxxgfxgdxd=== (see page 418 of the text, Theorem 7. Here is an example why it is important that the function be 1:1….Let . Note that the function is not 1:1 on this interval. But we will pretend not to notice this. Find 8<<8=x) f ( x2 , x
21)(=xxfdxd. By the theorem, we have xxf2)(‘= and xxfxg==)()(1. So, we have that 22121))((‘1)()2()(211======–agxxxagfxfdxdfdxd and this does not exist (we cannot take the square root of a negative number.
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