Question

Sea f una función lineal para la cual se cumple que:
3f(3) - f(-1) = 10 y 2f(4) + 5f(2) = 1
Si A= < -3,7 ] Hallar f(A)

Answer 1

Primero definamos nuestra función lineal f

                $\boxed{\sf{f(x) = mx + b}}\hspace{40pt} \mathsf{Donde} \hspace{30pt}\begin{array}{ll}\mathsf{\blue{\rightarrow}\ m:Pendiente\ } &\\\\\mathsf{\blue{\rightarrow}\ b:Ordenada\ al\ origen}&\end{array}$

Del problema conocemos que:

                                    $\begin{array}{c}\\\sf{3f(3) - f(-1) = 10}\\\\\sf{3\Big(m(3)+b\Big) - \Big(m(-1)+b\Big) = 10}\\\\\sf{3\Big(3m+b\Big) - \Big(-m+b\Big) = 10}\\\\\sf{9m+3b+ m-b = 10}\\\\\sf{10m+2b = 10}\\\\\boxed{\blue{\sf{5m+b = 5}}}\end{array}$    

   

                                      $\begin{array}{c}\\\sf{2f(4) + 5f(2) = 1}\\\\\sf{2\Big(m(4)+b\Big) + 5\Big(m(2)+b\Big) = 1}\\\\\sf{2\Big(4m+b\Big) + 5\Big(2m+b\Big) = 1}\\\\\sf{8m+2b + 10m+5b = 1}\\\\\boxed{\blue{\sf{18m+7b= 1}}}\end{array}$

Con las dos ecuaciones que tenemos hallamos "m" y "b", por ello multiplicamos por 7 al primer cuadrito y lo restamos con el segundo

                                      $\begin{array}{cccccccccc}\\\sf{35m}&\sf{+}&\sf{7b}&\sf{=}&\sf{35}&\boldsymbol{\sf{-}}\\\sf{18m}&\sf{+}&\sf{7b}&\sf{=}&\sf{1}&\\\rule{19pt}{0.2ex}&\rule{19pt}{0.2ex}&\rule{19pt}{0.2ex}&\rule{19pt}{0.2ex}&\rule{19pt}{0.2ex}&\\\sf{17m}&\sf{+}&\sf{0}&=&\sf{34}\\&&\sf{17m}&=&\sf{34}\\&&\sf{m}&=&\sf{2}\end{array}$

Reemplazamos "m" para determinar "b"

                                                       $\begin{array}{c}\\\sf{5m+b=5}\\\\\sf{5(2)+b=5}\\\\\sf{10+b=5}\\\\\sf{b=-5}\\\\\end{array}$

Nuestra función lineal es:

                                                    $\begin{array}{c}\sf{f(x) = mx + b}\\\\\boxed{\boldsymbol{\sf{f(x) = 2x - 5}}}\end{array}$

Nos piden hallar f(A) si A = < -3,7 ], entonces

                                                 $\begin{array}{c}\sf{-3 < A\leq 7}\\\\\sf{Multiplicamos\ por\ 2}\\\\\sf{-6 < 2A\leq14}\\\\\sf{Restamos\ 5}\\\\\sf{-11 < 2A-5\leq9}\\\\\boxed{\boxed{\boldsymbol{\red{\sf{-11 < f(A)\leq9}}}}}\end{array}$

Rpta. f(A) = <-11,9]

                                                 $\boxed{\sf{{R}}\quad\raisebox{10pt}{ \sf{\red{O}} }\!\!\!\!\raisebox{-10pt}{ \sf{\red{O}} }\quad\raisebox{15pt}{ \sf{{G}} }\!\!\!\!\raisebox{-15pt}{ \sf{{G}} }\quad\raisebox{15pt}{ \sf{\red{H}} }\!\!\!\!\raisebox{-15pt}{ \sf{\red{H}} }\quad\raisebox{10pt}{ \sf{{E}} }\!\!\!\!\raisebox{-10pt}{ \sf{{E}} }\quad\sf{\red{R}}}\hspace{-64.5pt}\rule{10pt}{.2ex}\:\rule{3pt}{1ex}\rule{3pt}{1.5ex}\rule{3pt}{2ex}\rule{3pt}{1.5ex}\rule{3pt}{1ex}\:\rule{10pt}{.2ex}$