Help

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

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

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

8 months agoundefined?

8 months agothat'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

8 months agofor a it would be f(17)-1?

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

8 months agoah, 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)

8 months agofor 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)

8 months agobut for the then that means the f(2)=17 function doesnt matteR?

8 months agofor b

8 months agonot 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

8 months agookay so c would be 5?

8 months agoyeah

8 months agod= f-1(2)=2?

8 months agothat'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")

8 months agosame logic with e)

8 months agookay so for e it would be -2?

8 months agooh 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

8 months agookay so for these three last problems i got one

8 months agoI got x=ln(10^3)/ln^2)

8 months agois this right

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

8 months agogot itt. the next would just be x=1 right

8 months agohm 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

8 months agowait hold on

8 months agooh so logs keep their 10 value?

8 months agoi thought they cancelled out

8 months agooh duh I was working in base e not base 10

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

8 months agooh okay so this would be 10^log

8 months agohm 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

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

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

8 months agoi se

8 months agoif you can i would appreciate it

8 months agoalright 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

8 months agox=ln(c)/a-b

8 months agoyeah just be careful w/ parentheses and capitalization ln(C)/(a-b)

8 months agoperfect

8 months ago