Question

f(x)=(6e^(x)−4)/(14e^(x)+6). Find the inverse function and its domain (interval notation)

Answer 1

By f(x)=(6e^(x)−4)/(14e^(x)+6), do you mean\[y=\frac{ 6e^x-4 }{ 14e^6+6 }?\]

Answer 2

oh it's this:
\[f(x)=\frac{ (6e^x-4) }{ (14e^x +6) }\]

Answer 3

@mathmale

Answer 4

thanks for verifying the problem statement. Writing your version as a fraction, as I did, eliminates the need for parentheses.\[y=\frac{ 6e^x-4 }{ 14e^x+6 }\]

Answer 5

You're correct; the problem was given to you labeled f(x). The next step is to replace the "f(x)=" label with "y=" ... which I have done.

Next, interchange x and y: where you see x, write y; where you see y, write x instead. Pls do that now.

Answer 6

\[x=\frac{ 6e^y -4}{ 14e^y+6 }?\]

Answer 7

Yes. Now, please solve this new equation for e^y.

Answer 8

I'm not sure how to do this part... Do I need to change e to ln?? But I know that I can move the denominator to the right hand side:
\[x(14e^y+6)=6e^y-4\]

Answer 9

You're on the right track! However, you still need to solve this equation for e^y. Mult. the quantity within parentheses by x.
Next, group the e^y terms on the left side and the remaining terms on the right side. Solve for e^y.

Answer 10

So like this:
\[6x+4=6e^y-14xe^y\]
\[6x+4=e^y(6-14x)\]

Answer 11

Looks good, and I appreciate your showing your work as you have.
Please divide both sides of this equation by (6-14x) to isolate e^y.
Note that the left side can be reduced. How?

Answer 12

It will like this:
\[\frac{ 6x+4 }{ 6-14x }=e^y\]
\[\frac{(3x+2) }{(3-7x) }=e^y\]

Answer 13

Summary: replace f(x) with y.
2. Interchange x and y
3 Solve for y
4. replace y with \[f ^{-1}(x)=\]

Answer 14

We overlooked one step.
\[\frac{(3x+2) }{(3-7x) }=e^y\]

Answer 15

... is fine, but we need to go all the way in solving for y.
Apply the ln operator to both sides of this most recent question. That will isolate y on the right side.

Answer 16

oohh...
\[\ln \frac{ 3x+2 }{ 3-7x }=(e^y)\ln\]

Answer 17

\[y=\ln(\frac{ 3x+2 }{ 3-7x })\]

Answer 18

Looks good! You may either leave it that way, or write:\[y=\ln (3x+2)-\ln (3-7x)\]

Answer 19

Oh I see... I forgot the properties for "e" and "ln"

Answer 20

You could check by either:
1) substitute this result (right side only) into the original equation, replacing x in the original equation, or
2) substitute the original equation (right side only) into our most recent "answer."

If the end result is merely "x"

Answer 21

...then you have found the correct inverse function of the original f(x).

Answer 22

Oh okay... How would we find the domain (interval notation)??

Answer 23

"ln x" shows up 2x in your latest result.

Answer 24

Recall that the domain of ln x is (0, +inf).
Example: Set 3x+2>0 and solve for x.
Do the same for the second argument:

3-7x>0.

graph both sets. The overall, final domain is the set of all x where the 2 previous sets overlap one another.

Answer 25

So the domain would be:
\[x >-2/3\] and \[x >3/7\]

Answer 26

You must now graph your 2 results. Determine the interval on which these two results overlap. ONLY the overlapping part constitutes the domain for your original inequality.

Answer 27

Glad to have the chance to work with you. I must get off the 'Net now, but if you have further questions about this probelm, just leave a note for me; I will see it in the morning. Good night!

Answer 28

I think I got it. Thank you so much! I learned a lot! :)

Answer 29

But I'm not sure what you mean by overlapping... I think the interval notation is (-2/3, 3/7)
@mathmale