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Sagot :
The definition of the inverse function is
[tex]\begin{gathered} f(f^{-1}(x))=x \\ \text{and} \\ f^{-1}(f(y))=y \end{gathered}[/tex]In our case,
[tex]f(x)=2x-1[/tex]Then,
[tex]\begin{gathered} f^{-1}(f(x))=x \\ \Rightarrow f^{-1}(2x-1)=x \\ \Rightarrow f^{-1}(x)=\frac{x+1}{2} \end{gathered}[/tex]We need to verify this result using the other equality as shown below
[tex]\begin{gathered} f^{-1}(x)=\frac{x+1}{2} \\ \Rightarrow f(f^{-1}(x))=f(\frac{x+1}{2})=2(\frac{x+1}{2})-1=x+1-1=x \\ \Rightarrow f(f^{-1}(x))=x \end{gathered}[/tex]Therefore,
[tex]\Rightarrow f^{-1}(x)=\frac{x+1}{2}[/tex]The inverse function is f^-1(x)=(x+1)/2.
We say that a relation is a function if, for x in the domain of f, there is only one value of f(x).
In our case, notice that for any value of x, there is only one value of (x+1)/2=x/2+1/2.
The function is indeed a function, it is a straight line on the plane that is not parallel to the y-axis.
The inverse f^-1(x) is indeed a function
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