zarkam21:

Help

4 months ago
zarkam21:
4 months ago

Vocaloid:

so when you take the inverse the domain and range are switched (this gets complicated when you consider limitations on domains and ranges)

4 months ago
Vocaloid:

so a) if f(2) = 17 we would expect f(17) = 2 since the domain and range are switched

4 months ago
Vocaloid:

for b) do you remember what you get if you plug an inverse into the original function?

4 months ago
zarkam21:

undefined?

4 months ago
Vocaloid:

that's close, basically if you plug in an inverse into itself you get x so f^(-1) of f(x) = x so if you input 5 you get 5 back

4 months ago
zarkam21:

for a it would be f(17)-1?

4 months ago
zarkam21:

oh i see so it just what the x value is in f(x) is what you are gonna get back

4 months ago
Vocaloid:

ah, no the problem tells you that f(2) = 17 since a) is asking about f^-1 of 17, since we reverse the domain and range for an inverse function, we get f^-1 (17) = 2, so 2 is the solution for a)

4 months ago
Vocaloid:

for b) the inverse and the function both cancel each other out to get x so you get a generic y = x function which just returns whatever input you give it, so 5 is the sol'n for b)

4 months ago
zarkam21:

but for the then that means the f(2)=17 function doesnt matteR?

4 months ago
zarkam21:

for b

4 months ago
Vocaloid:

not really any function and its own inverse will give you x, so whatever f happens to be, then f^-1(f) and f^-1(f) will both equal x

4 months ago
zarkam21:

okay so c would be 5?

4 months ago
Vocaloid:

yeah

4 months ago
zarkam21:

d= f-1(2)=2?

4 months ago
Vocaloid:

that's a good attempt but we only have f^-1 by itself, and if we don't know the original function there's not much we can do here (so "more information needed")

4 months ago
Vocaloid:

same logic with e)

4 months ago
zarkam21:

okay so for e it would be -2?

4 months ago
Vocaloid:

oh there's something I didn't catch before the original function has domain all real numbers and range all positive numbers since we switch the domain + range for the inverse, the inverse has domain all positive numbers and range all real numbers so f^(-1)(-17) would be out of the domain and therefore undefined

4 months ago
zarkam21:

okay so for these three last problems i got one

4 months ago
zarkam21:
4 months ago

zarkam21:

I got x=ln(10^3)/ln^2)

4 months ago
zarkam21:

is this right

4 months ago
Vocaloid:

almost just be careful with your parentheses taking the ln of both sides x = \[\ln_{2} 10^3\] which is ln(10^3)/ln(2)

4 months ago
zarkam21:

got itt. the next would just be x=1 right

4 months ago
Vocaloid:

hm not quite (the logs don't cancel out the way you might think they would) log(log(x)) = 1 taking base 10 on both sides log(x) = 10^1 taking base 10 again x = 10^10

4 months ago
Vocaloid:

wait hold on

4 months ago
zarkam21:

oh so logs keep their 10 value?

4 months ago
zarkam21:

i thought they cancelled out

4 months ago
Vocaloid:

oh duh I was working in base e not base 10

4 months ago
Vocaloid:

but yeah, logs of logs don't cancel out you may be thinking about 10^log(x) which does cancel out

4 months ago
zarkam21:

oh okay so this would be 10^log

4 months ago
Vocaloid:

hm not quite, see this section log(log(x)) = 1 taking base 10 on both sides log(x) = 10^1 taking base 10 again x = 10^10

4 months ago
Vocaloid:

to cancel out a log you have to take the base 10 of both sides

4 months ago
Vocaloid:

I really don't remember how to solve for c) but I can ask someone >>

4 months ago
zarkam21:

i se

4 months ago
zarkam21:

if you can i would appreciate it

4 months ago
Vocaloid:

alright I was like 100% overthinking this you just divide both sides by e^bx to get e^ax/e^bx = C then you just re-write the left side using exponent rules to get e^(ax-bx) = C take the natural log of both sides, then solving for x should be straightforward

4 months ago
zarkam21:

x=ln(c)/a-b

4 months ago
Vocaloid:

yeah just be careful w/ parentheses and capitalization ln(C)/(a-b)

4 months ago
zarkam21:

perfect

4 months ago