𝑓(𝑥) = 𝑒 −𝑥 1+𝑒−𝑥 . i) Determine whether 𝑓(𝑥) is a one-to-one function.
Let's solve the equation f(x)=f(y)f(x)=f(y)f(x)=f(y) :
e−x1+e−x=e−y1+e−y\frac{e^{-x}}{1+e^{-x}}=\frac{e^{-y}}{1+e^{-y}}1+e−xe−x=1+e−ye−y
1ex+1=1ey+1\frac{1}{e^x+1}=\frac{1}{e^y+1}ex+11=ey+11
ex+1=ey+1e^x+1=e^y+1ex+1=ey+1
ex=eye^x=e^yex=ey
x=yx=yx=y
This means that the function f(x)f(x)f(x) is one-to-one.
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