Question

An exponent raised to another exponent...whoa, find the domain

Answer 1

\[f(x)=\frac{ 1-e^x^2 }{ 1-e^1-x^2 }\]

Answer 2

f(x)=(1-e^x^2)/(1-e^1-x^2)

Answer 3

the denominator cant be zero

Answer 4

What is your numerator by the way??
What is the power of e there ??

Answer 5

one minus e to the power of x squared

Answer 6

\[\large f(x) = \frac{1 - e^{x^2}}{1- e^1 - x^2}\]
Right ??

Answer 7

ya except the bottom is 1-e^(1-x^2)

Answer 8

\[\large f(x) = \frac{1 - e^{x^2}}{1- e^{1 - x^2}}\]

Now ??

Answer 9

yup perfect

Answer 10

So, you know denominator can't be 0..

Answer 11

ya how do i solve this tho

Answer 12

do i have to use log

Answer 13

\[1 - e^{1-x^2} \ne 0 \implies e^{1-x^2} \ne 1\]

Answer 14

i hate u

Answer 15

You must..

I hate you too..

Because I don't like boys..

I like Girls..


Ha ha ha...

Answer 16

i dont know how i didnt see that i wanted to solve it somehow

Answer 17

thanks for your help as always

Answer 18

If \(x^2 = 1\)

Then :

\[e^{1-1} = e^0 = 1\]

But we don't need it is equal to 1..

So \(x^2 \ne 1\)

Answer 19

ya i see that now :-)

Answer 20

Got or not ??

Or need more explanation??

Answer 21

\[x \ne \pm 1\]

Answer 22

You can use ln here if you want:

\[(1-x^2) \ln(e) \ne \ln(1) \implies 1 - x^2 \ne 0 \implies x^2 \ne 1\]

Answer 23

thats perfect water, thats the way i was originally thinking of it but im rusty with logs

Answer 24

Thanks..