Question #97128
Check whether the given function is odd or even
f(x) = ln|(1-e⁻ˣ)/(1+e⁻ˣ)|

please its urgent answer it.
1
Expert's answer
2019-10-22T08:36:58-0400


We are going to check the given function is Odd (or) Even.



A function f(X)f(X) is said to be Odd function if

f(X)=f(X){f(-X)} = - f(X)

A function f(X)f(X) is said to be Even function if

f(X)=+f(X){f(-X)} = + f(X)

The given function is

f(X)=ln1eX1+eXf(X) = ln | \frac { 1 - e^{-X}} { 1 + e^{-X}}|

Now plug X = - X in the function f(X)f(X):


f(X)=ln1e(X)1+e(X)f(- X) = ln |\frac {1- e^ {-(-X)}} {1 + e^ {-(-X)}}|


f(X)=ln1e(+X)1+e(+X)f(- X) = ln |\frac {1- e^ {(+X)}} {1 + e^ {(+X)}}|


f(X)=ln1e(X)1+e(X)f(- X) = ln |\frac {1- e^ {(X)}} {1 + e^ {(X)}}|

Now, divide the numerator and denominator by e^X



f(X)=ln1eX11eX+1f(- X) = ln |\frac {\frac {1} { e^ {X}} -1} {\frac {1} { e^ {X}} +1}|

f(X)=lneX1eX+1f(- X) = ln |\frac { e^ {-X} -1} { { e^ {-X}} +1}|

f(X)=ln(eX+1)eX+1f(- X) = ln |\frac { -(-e^{-X} +1)} { { e^ {-X}} +1}|


f(X)=ln(eX+1)eX+1f(- X) = ln |\frac { (-e^ {-X} +1)} { { e^ {-X}} +1}|

Since a=a| - a | = |a|,

f(X)=ln(1eX)1+eX=f(X)f(- X) = ln |\frac { (1 -e^ {-X})} { { 1 + e^ {-X}} }| = f(X)

Thus, it is an Even function.


Answer: The given function is Even


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