Question #220217

š‘“(š‘„) = š‘’ āˆ’š‘„ 1+š‘’āˆ’š‘„ . i) Determine whether š‘“(š‘„) is a one-to-one function. 


Expert's answer

Let's solve the equation f(x)=f(y)f(x)=f(y) :

eāˆ’x1+eāˆ’x=eāˆ’y1+eāˆ’y\frac{e^{-x}}{1+e^{-x}}=\frac{e^{-y}}{1+e^{-y}}

1ex+1=1ey+1\frac{1}{e^x+1}=\frac{1}{e^y+1}

ex+1=ey+1e^x+1=e^y+1

ex=eye^x=e^y

x=yx=y

This means that the function f(x)f(x) is one-to-one.


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