Respuesta :

konrad509
[tex]f(x)=4x-3\\ y=4x-3\\ 4x=y+3\\ x=\frac{1}{4}y+\frac{3}{4}\\ f^{-1}(x)=\frac{1}{4}x+\frac{3}{4}\\\\ f^{-1} \circ f=\frac{1}{4}(4x-3)+\frac{3}{4}\\ f^{-1} \circ f=\frac{4x}{4}-\frac{3}{4}+\frac{3}{4}\\ f^{-1} \circ f=x[/tex]
kate200468
[tex]f(x)=4x-3\ \ \ \ and\ \ \ D_f=R\\\\y=4x-3\ \ \ \Rightarrow\ \ \ y+3=4x\ /:4\ \ \ \Rightarrow\ \ \ \frac{1}{4}y+ \frac{3}{4}=x\\\\f^{-1}(x)= \frac{1}{4}x+ \frac{3}{4} \ \ \ \ and\ \ \ \ D_{f^{-1}}=R\\\\ (f^{-1}\circ f) (10)=f^{-1}\bigg( f (10)\bigg)=\\=f^{-1}\bigg( 4\cdot10-3\bigg)=f^{-1}(37)=\frac{1}{4}\cdot37+ \frac{3}{4} =\frac{37}{4} +\frac{3}{4} = \frac{40}{4} =10[/tex]