Question

f(x)=radical(8-e^(2x)). Find domain of f? inverse of f? and domain of inverse of f?

Answer 1

\[f(x)=\sqrt{8-e^{2x}}\]
you have to be sure that
\[8-e^{2x}\geq 0\] meaning
\[e^{2x}\leq 8\]
\[2x\leq \ln(8)\]
\[x\leq \frac{\ln(8)}{2}\]

Answer 2

if you want to be fancy you can write it as
\[x\leq \frac{3}{2}\ln(2)\]

Answer 3

to find the inverse maybe write
\[x=\sqrt{8-e^{2y}}\] and solve for y via
\[x^2=8-e^{2y}\]
\[x^2-8=-e^{2y}\]
\[e^{2y}=8-x^2\]
\[2y=\ln(8-x^2)\]
\[y=\frac{1}{2}\ln(8-x^2)\]
\[y=\ln(\sqrt{8-x^2})\]

Answer 4

what about domain? and to enter it there are two boxes with a comma in between

Answer 5

you have to make sure that
\[8-x^2>0\] because you cannot take the square root of a negative number and you cannot take the log of 0

Answer 6

oh hold on now i have to think. the range of f is non - negative, so you cannot use a negative number for the inverse

Answer 7

satellite what are you thinking about?

Answer 8

so domain is
\[[0,\sqrt{8})\] or
\[[0,2\sqrt{2})\]

Answer 9

oh i just had to make sure the answer made sense

Answer 10

if you just see
\[\frac{1}{2}\ln(8-x^2)\] you would say domain is
\[(-\sqrt{8},\sqrt{8})\] but not in this case

Answer 11

\[f(x)=\sqrt{8-e^{2x}}\]
domain of f is \[8-e^{2x} \ge 0 \]

\[8 \ge e^{2x}\]
natural log of both sides
\[\ln(8) \ge \ln(e^{2x})\]
\[\ln(8) \ge 2x \ln(e)\]
\[\ln(8) \ge 2x\]
\[\frac{\ln(8)}{2} \ge x\]

Answer 12

yeah that is the first thing i wrote

Answer 13

then i found the inverse. but i had to think for a moment about the domain of the inverse

Answer 14

\[y=\sqrt{8-e^{2x}}\]
square both sides
\[y^2=8-e^{2x}\]
subtract 8 on both sides
\[y^2-8=-e^{2x}\]
multiply negative one on both sides
\[-y^2+8=e^{2x}\]
take natural log of both sides
\[\ln(-y^2+8)=\ln(e^{2x})\]
\[\ln(8-y^2)=2x \ln(e)\]
\[\ln(8-y^2)=2x\]
divide both sides by 2
\[\frac{\ln(8-y^2)}{2}=x\]
\[f^{-1}(x)=\frac{\ln(8-x^2)}{2}\]

Answer 15

yeah i did that too!

Answer 16

\[8-x^2>0\] is the domain

Answer 17

that is where the mistake is!

Answer 18

why?

Answer 19

i wrote it out up top. what you wrote i mean. but the domain of
\[f^{-1}(x)=\frac{1}{2}\ln(8-x^2)=\ln(\sqrt{8-x^2})\] is in fact
\[[0,\sqrt{8})\]

Answer 20

because it is defined as the inverse of
\[f(x)=\sqrt{8-e^{2x}}\] and the range of this function does not include negative numbers. so the domain of the inverse does not
that is why i had to take a moment to think

Answer 21

yes you are right! :)

Answer 22

imagine!

Answer 23

so we would say the domain of f inverse is
8-x^2>0 with also remembering that we must also satisfy x>=0

Answer 24

remember... \cup
also \cap
\[\color{red}{\text{red}}\cap \color{blue}{\text{blue}}\]

Answer 25

yeah domain is
\[[0,2\sqrt{2})\]

Answer 26

first we find that \[(-\sqrt{8},\sqrt{8})\] from 8-x^2>0 but then we remember that we must have x>=0 so we only actually have \[[0,\sqrt{8})\]

Answer 27

and yes \[\sqrt{8}=\sqrt{4 \cdot 2}=2\sqrt{2}\]

Answer 28

okay but what about the domain of f? admist all this i got lost a little

Answer 29

the domain of f is the first thing i found

Answer 30

ln(8)/2? how do i write that in two boxes with a comma in between

Answer 31

are you saying how you write an interval notation?
(-inf,ln(8)/2]