Question

escreva em cada caso a lei........​

Answer 1

$\boxed{\begin{array}{l}\bf 1)~Escreva~em~cada~caso\\\bf~a~lei~de~formac_{\!\!,}\tilde ao~da~func_{\!\! ,}\tilde ao~\\\bf afim~f:\mathbb{R}\longrightarrow\mathbb{R}~tal~que,~f(x)=a\cdot x+b\\\rm a)~Se~f(-1)=3~e~f(2)=0~ou~seja\\\rm dados~os~pontos~A(-1,3)~e~B(2,0)\\\rm b)~se~f(1)=-\dfrac{1}{10}~e~f(-2)=\dfrac{7}{5}~ou~seja\\\rm dado~os~pontos~A\bigg(1,-\dfrac{1}{10}\bigg)~e~B\bigg(\!\!-2,\dfrac{7}{5}\bigg)\\\rm Encontre~a~e~b~e~escreva~a~func_{\!\!,}\tilde ao\end{array}}$

$\boxed{\begin{array}{l}\underline{\rm soluc_{\!\!,}\tilde ao\!:}\\\rm a)~\sf f(-1)=a\cdot(-1)+b\\\sf-a+b=3\\\sf f(2)=a\cdot 2+b\\\sf 2a+b=0\\\begin{cases}\sf-a+b=3\\\sf2a+b=0\end{cases}\\\rm multiplicando~a~1^a~equac_{\!\!,}\tilde ao~por~-1~temos:\\\begin{cases}\sf a-b-=-3\\\sf 2a+b=0\end{cases}\\\rm adicionando~algebricamente~as~equac_{\!\!,}\tilde oes~teremos\\+\underline{\begin{cases}\sf a-\diagdown\!\!\!\!b=-3\\\sf 2a+\diagdown\!\!\!\!\!b=0\end{cases}}\\\sf 3a=-3\\\sf a=-\dfrac{3}{3}\end{array}}$

$\Large\boxed{\begin{array}{l}\sf a=-1\\\rm substituindo~a~na~2^a~equac_{\!\!,}\tilde ao~temos\\\sf 2a+b=0\implies b=-2a\\\sf b=-2\cdot(-1)\\\sf b=2\\\sf portanto~a~lei~de~formac_{\!\!,}\tilde ao~ser\acute a\\\huge\boxed{\boxed{\boxed{\boxed{\sf f(x)=-x+2}}}}\end{array}}$

$\boxed{\begin{array}{l}\rm b)~\sf f(1)=a\cdot1+b\\\sf a+b=-\dfrac{1}{10}\\\sf f(-2)=a\cdot(-2)+b\\\sf -2a+b=\dfrac{7}{5}\\\begin{cases}\sf a+b=-\dfrac{1}{10}\\\\\sf -2a+b=\dfrac{7}{5}\end{cases}\\\rm multiplicando~a~1^a~equac_{\!\!,}\tilde ao~por~2~temos:\\+\underline{\begin{cases}\sf\diagdown\!\!\!\!\!2a+2b=-\dfrac{1}{5}\\\\\\\sf-\diagdown\!\!\!\!\!2a+b=\dfrac{7}{5}\end{cases}}\end{array}}$

$\large\boxed{\begin{array}{l}\sf3b=\dfrac{6}{5}\\\sf 15b=6\\\sf b=\dfrac{6\div3}{15\div3}=\dfrac{2}{5}\\\rm substituindo~na~1^a~equac_{\!\!,}\tilde ao~temos\\\sf a+b=-\dfrac{1}{10}\cdot10\\\sf 10a+10b=-1\\\sf 10a+\diagdown\!\!\!\!\!10\cdot\dfrac{2}{\diagdown\!\!\!\!5}=-1\\\sf 10a+4=-1\\\sf 10a=-1-4\\\sf 10a=-5\\\sf a=-\dfrac{5\div5}{10\div5}\\\sf a=-\dfrac{1}{2}\\\rm portanto~a~lei~de~formac_{\!\!,}\tilde ao~ser\acute a\\\huge\boxed{\boxed{\boxed{\boxed{\sf f(x)=-\dfrac{1}{2}x+\dfrac{2}{5}}}}}\end{array}}$