Question

Find Range of y.

Answer 1

\[y=\frac{e^{x}-e^{-|x|}}{e^{x}+e^{|x|}}\]

Answer 2

Can u graph the function

Answer 3

no

Answer 4

this looks like hyperbolic tangent function

Answer 5

also,
the local maxima nad minima can be found from the derivatives.
then check for the absolute maxima and minima using limits.

Answer 6

\[
y=\frac{e^x-e^{-|x|}}{e^x+e^{-|x|}}=\frac{e^{x+|x|}-1}{e^{x+|x|}+1}\\
y=\cases{\frac{e^{2x}-1}{e^{2x}+1} \forall\;x\ge0\\
0\qquad\quad\forall\;x<0}
\]

Answer 7

we notice that the minima is
y=0

and a horizontal asymptote at
y=1

conclusion:
\[\Large y\in[0,1]\]

Answer 8

@electrokid y lies in [0,1/2)

Answer 9

not 1/2
but "1"

Answer 10

that is your range

Answer 11

i am telling the ans.

Answer 12

the given ans.

Answer 13

ooh wait..
I took the function worong

Correct the denominator and folow the same steps

Answer 14

no

Answer 15

\[
y=\cases{\frac{e^x-e^{-x}}{2e^x}\;\forall\;x\ge0\\
0\qquad\quad\forall x<0}
\]

Answer 16

ok then

Answer 17

and then it is pretty simple to get \[\Large y\in\left[1,{1\over2}\right)\]

Answer 18

ans is [0,1/2)

Answer 19

my typo..

Answer 20

it is 0<=y<1/2

Answer 21

@electrokid

Answer 22

you should then click on the best answer

Answer 23

I have done like this but not getting the ans.

Answer 24

you do not need any of that

Answer 25

we don't know calculus by now.

Answer 26

so help me with my sol.

Answer 27

no need of any calculus
\[
y=\cases{\frac{e^x-e^{-x}}{2e^x}={1\over2}-{e^{-2x}\over 2}\;\forall\;x\ge0\\
0\qquad\quad\forall x<0}
\]
from this, we know that y minimum = 0
then
for x>=0, we notice that it is continuously increasing function

\[y_0={1\over2}-{e^0\over2}=0\\
y_\infty={1\over2}-{e^{-\infty}\over2}={1\over2}
\]