Question

PLEASE HELP
f(x)=log(x+5)-3
a) domain of f is______
b) graph
c) range of f is___
vertical asymptote is___ or No asymptote
d) Find f^-1, the inverse of f
e) domain of f^-1
range of f^-1
f) Graph f^-1

Answer 1

To start are there any you think you know how to do already?

Answer 2

I know that when I find the domain and range of f(x) it would be the opposite for f^-1(x) I know the graph already but I just dont know how to find the domain and range of vertical asymptote

Answer 3

Okay, for logarithms I believe that the range will be (-INFINITY, INFINITY) every time.
The domain can be found by knowing where a logarithmic function DNE. This would be at log(0) so to find domain, just set the inside equal to 0 and solve for x. Your domain would then be (x, INFINITY).
Then you can almost guess the vertical asymptote from there because since it will head off to negative infinity, but never reach x, you know that that must be the vertical asymptote.

Answer 4

so the range is (-infinity, infinity) the vertical asymptote is (-2),
how would you do the domain? still confused

Answer 5

Careful. Asymptote wouldn't be (-2) because the 3 is outside of the log function. So all that is doing is shifting the graph down on the y-axis.

As for domain, log(x) exists for ALL x > 0. So we have log(x+5) and we know that (x+5) > 0. Solve for x and we find that x > -5, therefore the domain is (-5, INFINITY).

Answer 6

so the vertical asymptote would be -5?

Answer 7

Yep.

Answer 8

how would you get the inverse?

Answer 9

of the equation?

Answer 10

Okay, so here we need to use what we know about inverses.\[f(f^{-1}(x)) = x\]For us, \[f(x) = log(x+5)-3\]so then, combine those two equations and we find that \[f(f^{-1}(x))=log(f^{-1}(x)+5)-3=x\]From here we would use properties of logs to find \[f^{-1}(x)\]So remember that \[\log_{b}(a)=x \] tells us that \[b^{x} = a\]

Answer 11

okay then how would it look like?

Answer 12

Okay, so we plugged in \[f^{-1}(x)\] for x in f(x). Whenever you're given just log without a base it is really \[log_{10}\]. So we know the equation is equal to x. We want the log by itself on one side so we can apply the basic property of logarithms so\[log_{10}(f^{-1}(x) +5) -3 =x\] becomes\[log_{10}(f^{-1}(x) +5) = x+3\]Apply the property of logs we know and we find that \[10^{x+3} = f^{-1}(x) +5\] subtract 5 from each side, and we see that \[f^{-1}(x) = 10^{x+3} - 5\]

Answer 13

okay i can see how you got the answer now! thank you super much!!!

Answer 14

Your welcome! :)

Answer 15

can you help me find the inverse of (5/(x-3))

Answer 16

Sure.

Answer 17

Okay, the key to finding pretty much any inverse function is knowing that \[f(f^{-1}(x))=x\]Suppose that you have the function \[f(x) = x+2\]and you want to find \[f(5)\]How would you do that?

Answer 18

plug in 5 for x

Answer 19

Exactly! And it is precisely the same approach that we will use to find the inverse. If \[f(f^{-1}(x)) =x\] and \[f(x)=\frac{5}{x-3}\]then how could we express \[f(f^{-1}(x))?\]

Answer 20

so it will look like\[\frac{ 5 }{ f ^{-1} -3}\]

Answer 21

to start out with first

Answer 22

Yes. And then that would be equal to what?

Answer 23

\[f ^{-1}=5+3 ???\]

Answer 24

not quite. Remember that \[f(f^{1}(x)) = x\] so when we plugged in \[f^{-1} (x)\] to the function, we know that it will be equal to x. So then we have \[\frac{5}{f^{-1}(x)-3}=x\]Leaving us with just needing to solve for \[f^{-1}(x)\]

Answer 25

so it would be \[f ^{-1}=5x+3\]

Answer 26

Close. Remember that when you have a plus or minus sign with the x you need to be careful how you reorder terms. We would want to treat the bottom as though it were all in parenthesis like so\[\frac{5}{(f^{-1}(x)-3)}=x\]This makes the algebra a bit messy. The way I find best to keep the terms straight is to multiply the denominator by both sides so that you get\[5=x(f^{-1}(x)-3)\]then divide by x giving you \[\frac{5}{x}=f^{-1}(x)-3\]and finally add 3 to both sides to get \[\frac{5}{x}-3=f^{-1}(x)\]

Answer 27

The other way you could solve it is by dividing both sides by 5 first so that you have \[\frac{1}{f^{-1}(x) - 3}=\frac{x}{5}\]And then you invert both fractions so that you have \[f^{-1}(x)-3=\frac{5}{x}\]And finally subtract 3\[f^{-1}(x)=\frac{5}{x}-3\]The reason why I prefer to do it the other way is I feel that when you do it this way you become prone to forgetting to invert both sides so you'll only flip the side you are trying to solve for thereby changing the actual equation.

Answer 28

okay that makes way more sense when i plug in the calculator

Answer 29

okay thank you sooooooooo MUCH!!!!!!!!!!!!!!!

Answer 30

No problem :)

Answer 31

but wouldn't it be +3

Answer 32

oops. yes haha. good catch.

Answer 33

thank you again!