[tex]\text f(2\text x+3)=6\text x+11[/tex]
item a)
[tex]\text f(11) = \ ? \\\\ \text 2\text x+3 = 11 \\\\ 2\text x = 8 \\\\ \text x = 4[/tex]
façamos x = 4 :
[tex]\text f(2.4+3) = 6.4+11 \\\\ \huge\boxed{\text f(11) = 35 }\checkmark[/tex]
item b)
[tex]\displaystyle \text {f(x) = ? } \\\\ \text f(2.\text x+3) =6\text x+11 \\\\ \text{fazendo x = k } \\\\ \text f(2\text k+3)=6\text k + 11 \\\\ \text 2\text k + 3 = \text x \\\\ \text k = \frac{\text x-3}{2} \\\\ \text{substituindo o valor na f : } \\\\ \text f\ (\frac{\ 2(\text x-3)}{2}+3 \ ) = \frac{6(\text x-3)}{2}+11 \\\\ \text f\ (\text x-3+3) = 3\text x-9+11 \\\\ \huge\boxed{\text f\ (\text x) = 3\text x+2 }\checkmark[/tex]
item c)
[tex]\text f^{-1}(\text x) = \ ? \ \text{( inversa )}[/tex]
[tex]\text f(\text x) =3\text x+2 \\\\ \text y = 3\text x+2 \\\\ \underline{\text{trocando x por y e isolando y }}: \\\\ \text x = 3\text y+2 \\\\ 3\text y=\text x-2 \\\\ \text y = \displaystyle \frac{\text x-2}{3} \\\\ \underline{\text{Portanto }}: \\\\ \huge\boxed{\ \text f^{-1}(\text x) = \frac{\text x-2 }{3}\ }\checkmark[/tex]
