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Algebra and Functions Help

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?

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Starting with the inverse,

y=e2x+3y=e^{2x}+3

How would we solve for xx?


The y at the end is supposed to be f^(-1) (x)
(edited 6 years ago)
Reply 3
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emx_eco
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Original post by _gcx
Starting with the inverse,

y=e2x+3y=e^{2x}+3

How would we solve for xx?


I got x = (y-3)/2e

is this correct?
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:smile:
Original post by emx_eco
I got x = (y-3)/2e

is this correct?


It's e2xe^{2x} not 2ex2ex :smile:
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Original post by NGEO


what did you do to get from line 4 to line 5?
Original post by emx_eco
what did you do to get from line 4 to line 5?


Are you familiar with ln⁡\ln?
Reply 8
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emx_eco
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Original post by _gcx
It's e2xe^{2x} not 2ex2ex :smile:


I dont understand how you got that answer, could you please explain
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?
Reply 10
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Original post by _gcx
Are you familiar with ln⁡\ln?


yh, the same as log base e right?
Original post by emx_eco
I dont understand how you got that answer, could you please explain


If we have y=exy = e^x, how would we rearrange to make xx the subject? Think back to C2 logs.
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.
Reply 13
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Original post by _gcx
If we have y=exy = e^x, how would we rearrange to make xx the subject? Think back to C2 logs.


use a log base e on both sides?
Original post by emx_eco
use a log base e on both sides?


Yup, so we'd have log⁡e(y)=x\log_e(y) = x. In C3/C4 we normally write log⁡e\log_e as ln⁡\ln. Is it now clearer what ngeo did to get from e2y=x−3e^{2y} = x-3 to 2y=ln⁡(x−3)2y = \ln(x-3)?
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Original post by _gcx
Yup, so we'd have log⁡e(y)=x\log_e(y) = x. In C3/C4 we normally write log⁡e\log_e as ln⁡\ln. Is it now clearer what ngeo did to get from e2y=x−3e^{2y} = x-3 to 2y=ln⁡(x−3)2y = \ln(x-3)?


ohh I get that part now :smile: So I have the equation (lnx-3)/2 now. How do you find the domain of this?
Original post by emx_eco
ohh I get that part now :smile: So I have the equation (lnx-3)/2 now. How do you find the domain of this?


Where is ln⁡\ln undefined? Can we have the log of 0 or a negative number?
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Original post by _gcx
Where is ln⁡\ln undefined? Can we have the log of 0 or a negative number?


no...?
Original post by emx_eco
no...?


So what's the domain of ln⁡(x)\ln(x) and hence ln⁡(x−3)\ln(x-3)? :smile:
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Original post by _gcx
Where is ln⁡\ln 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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