sendo F(x)=3x-1 Determine :
. Fof ⁻¹(X)
Nooel:
X^-1 ?
Soluções para a tarefa
Respondido por
4
Caso tenha problemas para visualizar a resposta abaixo pelo aplicativo, experimente abrir pelo navegador: https://brainly.com.br/tarefa/8064314
Seja f a função definida da seguinte maneira:
![\begin{array}{llll} \mathsf{f}:&\mathbb{R}&\to&\mathbb{R}\\ &\mathsf{x}&\mapsto&\mathsf{3x-1} \end{array} \begin{array}{llll} \mathsf{f}:&\mathbb{R}&\to&\mathbb{R}\\ &\mathsf{x}&\mapsto&\mathsf{3x-1} \end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bllll%7D+%5Cmathsf%7Bf%7D%3A%26amp%3B%5Cmathbb%7BR%7D%26amp%3B%5Cto%26amp%3B%5Cmathbb%7BR%7D%5C%5C+%26amp%3B%5Cmathsf%7Bx%7D%26amp%3B%5Cmapsto%26amp%3B%5Cmathsf%7B3x-1%7D+%5Cend%7Barray%7D)
![\mathsf{f\circ f^{-1}(x)=f\big[f^{-1}(x)\big]=x} \mathsf{f\circ f^{-1}(x)=f\big[f^{-1}(x)\big]=x}](https://tex.z-dn.net/?f=%5Cmathsf%7Bf%5Ccirc+f%5E%7B-1%7D%28x%29%3Df%5Cbig%5Bf%5E%7B-1%7D%28x%29%5Cbig%5D%3Dx%7D)
A composição de uma função com sua inversa resulta na função identidade.
Observe o exemplo em questão:
![\mathsf{f\circ f^{-1}(x)=f\big[f^{-1}(x)\big]}\\\\ \mathsf{x=f\big[f^{-1}(x)\big]}\\\\ \mathsf{x=3\,f^{-1}(x)-1}\\\\ \mathsf{3\,f^{-1}(x)=x+1}\\\\ \mathsf{f^{-1}(x)=\dfrac{x+1}{3}\qquad\quad\checkmark} \mathsf{f\circ f^{-1}(x)=f\big[f^{-1}(x)\big]}\\\\ \mathsf{x=f\big[f^{-1}(x)\big]}\\\\ \mathsf{x=3\,f^{-1}(x)-1}\\\\ \mathsf{3\,f^{-1}(x)=x+1}\\\\ \mathsf{f^{-1}(x)=\dfrac{x+1}{3}\qquad\quad\checkmark}](https://tex.z-dn.net/?f=%5Cmathsf%7Bf%5Ccirc+f%5E%7B-1%7D%28x%29%3Df%5Cbig%5Bf%5E%7B-1%7D%28x%29%5Cbig%5D%7D%5C%5C%5C%5C+%5Cmathsf%7Bx%3Df%5Cbig%5Bf%5E%7B-1%7D%28x%29%5Cbig%5D%7D%5C%5C%5C%5C+%5Cmathsf%7Bx%3D3%5C%2Cf%5E%7B-1%7D%28x%29-1%7D%5C%5C%5C%5C+%5Cmathsf%7B3%5C%2Cf%5E%7B-1%7D%28x%29%3Dx%2B1%7D%5C%5C%5C%5C+%5Cmathsf%7Bf%5E%7B-1%7D%28x%29%3D%5Cdfrac%7Bx%2B1%7D%7B3%7D%5Cqquad%5Cquad%5Ccheckmark%7D)
Verificando as composições:
•![\mathsf{f\circ f^{-1}(x)} \mathsf{f\circ f^{-1}(x)}](https://tex.z-dn.net/?f=%5Cmathsf%7Bf%5Ccirc+f%5E%7B-1%7D%28x%29%7D)
![=\mathsf{f\big[f^{-1}(x)\big]}\\\\\\ =\mathsf{3\,f^{-1}(x)-1}\\\\\\ =\mathsf{\diagup\!\!\!\! 3\cdot \left(\dfrac{x+1}{\diagup\!\!\!\! 3}\right)-1}\\\\\\ =\mathsf{x+\diagup\!\!\!\! 1-\diagup\!\!\!\! 1}\\\\ =\mathsf{x\qquad\quad\checkmark} =\mathsf{f\big[f^{-1}(x)\big]}\\\\\\ =\mathsf{3\,f^{-1}(x)-1}\\\\\\ =\mathsf{\diagup\!\!\!\! 3\cdot \left(\dfrac{x+1}{\diagup\!\!\!\! 3}\right)-1}\\\\\\ =\mathsf{x+\diagup\!\!\!\! 1-\diagup\!\!\!\! 1}\\\\ =\mathsf{x\qquad\quad\checkmark}](https://tex.z-dn.net/?f=%3D%5Cmathsf%7Bf%5Cbig%5Bf%5E%7B-1%7D%28x%29%5Cbig%5D%7D%5C%5C%5C%5C%5C%5C+%3D%5Cmathsf%7B3%5C%2Cf%5E%7B-1%7D%28x%29-1%7D%5C%5C%5C%5C%5C%5C+%3D%5Cmathsf%7B%5Cdiagup%5C%21%5C%21%5C%21%5C%21+3%5Ccdot+%5Cleft%28%5Cdfrac%7Bx%2B1%7D%7B%5Cdiagup%5C%21%5C%21%5C%21%5C%21+3%7D%5Cright%29-1%7D%5C%5C%5C%5C%5C%5C+%3D%5Cmathsf%7Bx%2B%5Cdiagup%5C%21%5C%21%5C%21%5C%21+1-%5Cdiagup%5C%21%5C%21%5C%21%5C%21+1%7D%5C%5C%5C%5C+%3D%5Cmathsf%7Bx%5Cqquad%5Cquad%5Ccheckmark%7D)
•![\mathsf{f^{-1}\circ f(x)} \mathsf{f^{-1}\circ f(x)}](https://tex.z-dn.net/?f=%5Cmathsf%7Bf%5E%7B-1%7D%5Ccirc+f%28x%29%7D)
![=\mathsf{f^{-1}\big[f(x)\big]}\\\\ =\mathsf{\dfrac{f(x)+1}{3}}\\\\\\ =\mathsf{\dfrac{(3x-1)+1}{3}}\\\\\\ =\mathsf{\dfrac{3x}{3}}\\\\\\ =\mathsf{x\qquad\quad\checkmark} =\mathsf{f^{-1}\big[f(x)\big]}\\\\ =\mathsf{\dfrac{f(x)+1}{3}}\\\\\\ =\mathsf{\dfrac{(3x-1)+1}{3}}\\\\\\ =\mathsf{\dfrac{3x}{3}}\\\\\\ =\mathsf{x\qquad\quad\checkmark}](https://tex.z-dn.net/?f=%3D%5Cmathsf%7Bf%5E%7B-1%7D%5Cbig%5Bf%28x%29%5Cbig%5D%7D%5C%5C%5C%5C+%3D%5Cmathsf%7B%5Cdfrac%7Bf%28x%29%2B1%7D%7B3%7D%7D%5C%5C%5C%5C%5C%5C+%3D%5Cmathsf%7B%5Cdfrac%7B%283x-1%29%2B1%7D%7B3%7D%7D%5C%5C%5C%5C%5C%5C+%3D%5Cmathsf%7B%5Cdfrac%7B3x%7D%7B3%7D%7D%5C%5C%5C%5C%5C%5C+%3D%5Cmathsf%7Bx%5Cqquad%5Cquad%5Ccheckmark%7D)
Dúvidas? Comente.
Bons estudos! :-)
Seja f a função definida da seguinte maneira:
A composição de uma função com sua inversa resulta na função identidade.
Observe o exemplo em questão:
Verificando as composições:
•
•
Dúvidas? Comente.
Bons estudos! :-)
Perguntas interessantes