Integral Calculus> hi in numerator it is e raise to power mo...

hi in numerator it is e raise to power modulus of x ...please help ….i posted ….......last time i posted this quetion …..at that time i called e raise to power integral part which is wrong sorry for that....

milind , 10 Years ago

Grade 12

1 Answers

Jitender Singh

Last Activity: 10 Years ago

Ans:
Hello Student,
Please find answer to your question below

$f:R\rightarrow R$
$f(x) = \frac{e^{|x|}-e^{-x}}{e^{x}+e^{-x}}$
$f(x) = \frac{e^{|x|+x}-1}{e^{2x}+1}$
So, range of f(x) varies from -1 to 1 (both exclusive).
But the co-domain is R.
So f(x) is into function. (Function is onto when co-domain equals to range).
For the second part,
$f(0) = \frac{e^{|0|+0}-1}{e^{2(0)}+1}$
$f(0) = \frac{1-1}{e^{2(0)}+1} = 0$
$f(-1) = \frac{e^{|-1|+(-1)}-1}{e^{2(-1)}+1}$
$f(-1) = \frac{e^{1-1}-1}{e^{-2}+1}$
$f(-1) = \frac{1-1}{e^{-2}+1} = 0$
So function takes same velue at two inputs in domain.
So its many to one.