Question

Help

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

undefined?

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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)

Answer 9

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)

Answer 10

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

Answer 11

for b

Answer 12

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

Answer 13

okay so c would be 5?

Answer 14

yeah

Answer 15

d= f-1(2)=2?

Answer 16

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")

Answer 17

same logic with e)

Answer 18

okay so for e it would be -2?

Answer 19

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

Answer 20

okay so for these three last problems i got one

Answer 21

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

Answer 22

is this right

Answer 23

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)

Answer 24

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

Answer 25

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

Answer 26

wait hold on

Answer 27

oh so logs keep their 10 value?

Answer 28

i thought they cancelled out

Answer 29

oh duh I was working in base e not base 10

Answer 30

but yeah, logs of logs don't cancel out

you may be thinking about

10^log(x)
which does cancel out

Answer 31

oh okay so this would be 10^log

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

i se

Answer 36

if you can i would appreciate it

Answer 37

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

Answer 38

x=ln(c)/a-b

Answer 39

yeah just be careful w/ parentheses and capitalization

ln(C)/(a-b)

Answer 40

perfect