- Inverse functions- basically looks like the original function after it has been flipped over and rotated 90 degrees
- Logs- have to do with finding exponents. I'd explain but it seems many of my fellow classmates have already done so and quite well if I'm not mistaken.
- Natural logs- same as logs but uses "e" instead of actual numbers (even though e stands for a number...)
- Trigonometric functions- I actually remember this, ha! Sine and cosine are wavy graphs, secant and cosecant are parabolas and have asymptotes, tangent and cotangent graphs look like the graph of x³ and -x³ and have asymptotes. Let's see... oh and those trigonometric identities like secθ=1/cosθ, cscθ=1/sinθ, cotθ=1/tanθ, and those... others (forgot what they were called) the equations, cos²θ+sin²θ=1 and them others... (but that's a little too much innit? I only remember because we used them so much in math analysis with Gapac. On an unrelated note... I can also recite pi to the 17th place... Don't blame me though... I was bored in those few minutes before the bell rang and I noticed that banner/poster thing in her room so I decided Why not memorize pi?! Ha... how fun am I you guys, really? x]
2. What I did not understand completely:
Think I understood most things pretty well except for one or two things... I have to go find them so... found them!
Okay... first thing: finding the inverse equation of f(x). Like in the homework C2 #43, 44 and 45. I got it when Hwang explained it in class but then I tried to do these and... nothing. Completely clueless...
And homework C3 #60. The hell does that mean!? Stupid question annoyed me because it was so confusing...
Yeah I think that's it so... Help? Please?
3. Answered questions: seems most things have already been explained and rather well I might add.
Haha I didn't get #60 either but I do know how to solve #43,44,45. (I had to get help)
ReplyDelete-Its similar to the first equations we did on finding the inverse equation but I understand that having an exponent, it can be tricky.
Okay for 43:
*f(x)= 100/1+2^-x
*x = 100/1+2^-y
*(1+2^-y)x =100/1+2^-y (1+2^-y)
*(1+2^-y)x =100
*(1+2^-y)x/x =100/x
*1+2^-y = 100/x
-1 -1
*2^-y = 100/x - x/x <--- (-1)
*log(2) 2^-y = log(2)100-x/x
*-y = log(2)100-x/x
/-1 /-1
ANSWER: f-(x)= log(2) x/100-x
(log(2) is log base 2 okk)
Okay and thats how you solve 44 as well.
For #45 all you do is solve for the inverse of each function [there is an a) and b) part]
and when you're done, the function f should be its own inverse. Hope I helped!
Yeah, I didn't understand those either and I'm not sure if anyone explained it yet. Oh wait... *looks up* >___>;; uh, nevermind...
ReplyDeletePi!? To the 17th place?!? Damn you!!
Oh yeah? well...I can recite my ABD's backwards!! Ha! x]
I did #60 but im not sure if i did it half way or if i did it wrong but it might help you look at it in a differnt way:
ReplyDeleteso
g(x)= 1/f(x) move f(x) to other side
g(x)f(x)=1
so now you could say:
f(x)=1/g(x)
It doesnt mean that g(x)= f(x)
But you can figure that g(x) has to equal to 1 and f(x) has to equal to 1 so that when multiply they give you 1
Not only does multiplying 1 and 1 give you 1 but multiplying the x (being any number) and 1/x (the reciprical of the number) give you 1
Im not even sure i answered the question but i pointed something out right??