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# Algebra and Functions Help watch

1. How would you find the inverse of this function :

f(x) = e^(2x) + 3

and how would you work out the domain of its inverse?
2. Starting with the inverse,

How would we solve for ?

3. The y at the end is supposed to be f^(-1) (x)
4. (Original post by _gcx)
Starting with the inverse,

How would we solve for ?
I got x = (y-3)/2e

is this correct?
5. (Original post by NGEO)
x

The y at the end is supposed to be f^(-1) (x)
Best not to post full solutions, it is much preferred to give hints instead. You might be interested in looking at the posting guidelines
6. (Original post by emx_eco)
I got x = (y-3)/2e

is this correct?
It's not
7. (Original post by NGEO)
what did you do to get from line 4 to line 5?
8. (Original post by emx_eco)
what did you do to get from line 4 to line 5?
Are you familiar with ?
9. (Original post by _gcx)
It's not
I dont understand how you got that answer, could you please explain
10. (Original post by NGEO)

The y at the end is supposed to be f^(-1) (x)
Have you seen the guideline (sticky) about not posting full solutions?
11. (Original post by _gcx)
Are you familiar with ?
yh, the same as log base e right?
12. (Original post by emx_eco)
I dont understand how you got that answer, could you please explain
If we have , how would we rearrange to make the subject? Think back to C2 logs.
13. When you find the inverse of a function, essentially what you are doing is reflecting it in the line y=x. In order to find the 'new' equation, you want to rearrange it in terms of x=.... then just swap all the signs around like NGEO did, but did it in the first step instead.
14. (Original post by _gcx)
If we have , how would we rearrange to make the subject? Think back to C2 logs.
use a log base e on both sides?
15. (Original post by emx_eco)
use a log base e on both sides?
Yup, so we'd have . In C3/C4 we normally write as . Is it now clearer what ngeo did to get from to ?
16. (Original post by _gcx)
Yup, so we'd have . In C3/C4 we normally write as . Is it now clearer what ngeo did to get from to ?
ohh I get that part now So I have the equation (lnx-3)/2 now. How do you find the domain of this?
17. (Original post by emx_eco)
ohh I get that part now So I have the equation (lnx-3)/2 now. How do you find the domain of this?
Where is undefined? Can we have the log of 0 or a negative number?
18. (Original post by _gcx)
Where is undefined? Can we have the log of 0 or a negative number?
no...?
19. (Original post by emx_eco)
no...?
So what's the domain of and hence ?
20. (Original post by _gcx)
Where is undefined? Can we have the log of 0 or a negative number?
ooooooh therefore the domain is x > 3. I get it now thank you!!

Do you think you could help me with part b) -

f(x) = e^(2x) + 3 where x is any real number
g(x) = ln(x-1) where x is any real number and x > 1

Find fg(x) and state its range

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