homework 1 in Calculus 2

homework 1 in Calculus 2Project description
Write an outline for sections 6.2-6.4. 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 1920-R5X
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 one-to-one, 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 221|21))((1|)()2()(211-====-=-=agxxxagfxfdxdfdxd and this does not exist (we cannot take the square root of a negative number.