Wednesday, August 27, 2008

Chapter 1, section 6

10 comments:

David said...

Can someone help with sec 1.6 #2a. What is it asking or how would you answer it...

fBm said...

2a is asking about a function like the one in the left hand diagram of Figure 1 on page 60. These are nice schematics of domains and ranges. Pour through pages 60 and 61. What 2a is specifically asking for is mostly in the pink boxes. Also, as an example, think of f(x) = x^2, where x>=0 ("x greater than or equal to 0"). This function is 1-to-1. Graph f(x) and think about it's domain and range and the domain/range of it's inverse.

Hope that helps. Let me know if you have any more questions on this one or others.

-Jesse

Anonymous said...

Anyone up for a quick study session tonight before the quiz? I'm going to the library at seven, third floor. Text me @ 406-672-3040if interrested

Anonymous said...

Isaac,
I'm interested in the study session but that other side of silence deal is tonight at seven. Interested for another time.

Anonymous said...

I'm having trouble finding the inverse for the function in 16 1.6...

kaitlin.reuss said...

I am having trouble with section 1.6 questions 66 & 68. Any help?

fBm said...

for 16a, either graph f(x) and figure it out from the graph, or "see" that f^-1(3) means that some x makes f(x) = 3. Replace f(x) with the 3+x^2... expression and do some algebra. I'll give more of a hint later. (Gotta head home.)

fBm said...

Here's a graph of f(x) for number 16. here

fBm said...

A hint for 66 (68 is similar):

tan(arcsin(x)) = ?

1) arcsin(x) = arcsin(x/1) = an angle, A, in a triangle with vertical leg x and hypotenuse 1. Remember that arcsin takes the ratio of "opposite/adjacent" and gives the _angle_ necessary for a triangle with that ratio. Draw a right triangle with vertical leg of length x and hypotenuse of length 1. What is the length of the horizontal (eg. "adjacent") leg? Once you have that, you will triangle with angle A (the arcsin(x)), and tan(A) = opp/adj.

[The answer is tan(arcsin(x)) = x/sqrt(1-x^2) -- but make sure you know how you get there. :-) ]

-Jesse

Steve said...

Could someone explain how to do #9 in chap.1 review section?