Question

determine a equação da reta tangente ao grafico de y=3x/x+2 no ponto (1,1)

Answer 1

$\mathtt{\dfrac{dy}{dx}=\dfrac{3(x+2)-3x.2}{{(x+2)}^{2}}=\dfrac{-3x+6}{{(x+2)}^{2}}}\\\\\mathtt{\dfrac{dy}{dx}\big|_{x=1}=\dfrac{-3.1+6}{({1+2)}^{2}}=\dfrac{1}{3}}$

Equação da reta tangente

$\boxed{\boxed{\mathtt{y=y_{0}+f'(x_{0})(x-x_{0})}}}$

$\mathtt{y=1+\dfrac{1}{3}(x-1)}\\\mathtt{y=1+\dfrac{1}{3}x-\dfrac{1}{3}}}\\\\\huge\boxed{\boxed{\mathtt{y=\dfrac{1}{3}x+\dfrac{2}{3}}}}$