Question

prove que função inversa f(x)=x, sabendo que f(x) = ax+b

Answer 1

$\boxed{\begin{array}{l}\sf f(x)=ax+b\\\sf ax=f(x)-b\\\sf x=\dfrac{f(x)-b}{a}\\\sf f^{-1}(x)=\dfrac{x-b}{a}\\\underline{\rm demonstrac_{\!\!,}\tilde ao}\\\\\sf f(x)\circ f^{-1}(x)=f[f^{-1}(x)]=a\cdot f^{-1}(x)+b\\\\\sf f[f^{-1}(x)]=\backslash\!\!\!a\cdot\bigg(\dfrac{x-b}{\backslash\!\!\!a}\bigg)+b\\\\\sf f[f^{-1}(x)]=x-\backslash\!\!\!b+\backslash\!\!\!b=x\\\\\sf f^{-1}(x)\circ f(x)=f^{-1}[f(x)]\\\\\sf f^{-1}[f(x)]=\dfrac{f(x)-b}{a}\\\\\sf f^{-1}[f(x)]=\dfrac{ax+\backslash\!\!\!b-\backslash\!\!\!b}{a}\\\\\sf f^{-1}[f(x)]=\dfrac{\backslash\!\!\!ax}{\backslash\!\!\!a}\\\\\sf f^{-1}[f(x)]=x\end{array}}$