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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....

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.... 

Question Image
Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
6 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.

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