Question

can any one find the rangeeeeee of following plseee

Answer 1

solve for f(x)

Answer 2

\[\Large e^x +e^f(x) =e \] \[\Large e^f(x) =e -e^x\] take ln of both sides \[\Large f(x) = \ln (e -e^x)\]

Answer 3

Remember that (e-e^x) > 0 so that should help find the range.

Answer 4

okay

Answer 5

Not quite. There is a maximum value for the range, what happens when x approaches neg infinity?

e -e^x will approach...?
\[\Large \lim_{x \rightarrow - \infty} e -e^x = \]

Answer 6

Yes, we do, to find the max value of y (there's a horizontal asymptote)

\[\Large \lim_{x \rightarrow - \infty} \ln(e -e^x) = \]will approach...

Answer 7

plse explain me more, i mean we can also solve it fro this point
(e-e^x) > 0

Answer 8

*from this point

Answer 9

No, that was just saying that it MUST be greater than zero, that only finds the domain, since for ln(e -e^x) the (e -e^x) must be greater than zero.

Answer 10

k

Answer 11

for range
we put y = ln (e-e^x)

Answer 12

Yes, and we want to find the max and min values of y from that.

Answer 13

k

Answer 14

Correct. \[\Large \lim_{x \rightarrow - \infty} \ln(e -e^x) =\] \[\Large \lim_{x \rightarrow 1} \ln(e -e^x) =\]

Answer 15

well the inside function's limit is e in the first, and lne= ?

And as the inside function approaches 0, ln (e-e^x) will approach...?

Answer 16

Correct, which will give you the range.

Answer 17

So, y=1 is the max value for range. Did you find the min value?

Answer 18

@agent0smith
is it right or wrong

Answer 19

Min value isn't zero... what does lnx approach as x approaches 0?

Answer 20

log 0

Answer 21

Well... yes but what does it actually approach as x approaches 0? You can't just plug in 0, log 0 is undefined.

Answer 22

Correct. Max value is 1, min value is neg infinity.

Answer 23

okay
@agent0smith
thanks bro