Respuesta :

konrad509
[tex]f(x)=x\left(3-\dfrac{4}{x+7}\right)=3x-\dfrac{4x}{x+7}=\dfrac{3x(x+7)}{x+7}-\dfrac{4x}{x+7}=\\=\dfrac{3x^2+21x-4x}{x+7}= \dfrac{3x^2+17x}{x+7}\\ f'(x)=\dfrac{(6x+17)(x+7)-(3x^2+17x)}{(x+7)^2}\\ f'(x)=\dfrac{6x^2+17x+42x+119-3x^2-17x}{(x+7)^2}\\ f'(x)=\dfrac{119+42x+3x^2}{(x+7)^2}\Rightarrow \text{A}[/tex]

Otras preguntas