Question

Dada a funcao f(x)=-3x²+×-4,sua taxa se variação no intervalo -1

Answer 1

$\Large\boxed{\begin{array}{l}\underline{ ~~\rm D~\!\!efinic_{\!\!,}\tilde ao~de~derivada~no~ponto}\\\huge\boxed{\boxed{\boxed{\boxed{\displaystyle\sf f'(a)=\lim_{x \to a}\dfrac{f(x)-f(a)}{x-a}}}}}\end{array}}$

$\large\boxed{\begin{array}{l}\sf f(-1)=-3\cdot(-1)^2+(-1)-4\\\sf f(-1)=-3-1-4\\\sf f(-1)=-8\\\displaystyle\sf f'(-1)=\lim_{x \to -1}\dfrac{-3x^2+x-4-(-8)}{x-(-1)}\\\\\displaystyle\sf f'(-1)=\lim_{x \to -1}=\dfrac{-3x^2+x-4+8}{x+1}\\\\\displaystyle\sf f'(-1)=\lim_{x \to -1}\dfrac{-3x^2+x+4}{x+1}\end{array}}$

$\boxed{\begin{array}{l}\sf-3x^2+x+4=0\\\sf\Delta=b^2-4ac\\\sf\Delta=1^2-4\cdot(-3)\cdot4\\\sf\Delta=1+48\\\sf\Delta=49\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\sf x=\dfrac{-1\pm\sqrt{49}}{2\cdot(-3)}\\\\\sf x=\dfrac{-1\pm7}{-6}\begin{cases}\sf x_1=\dfrac{-1+7}{-6}=-\dfrac{6}{6}=-1\\\\\sf x_2=\dfrac{-1-7}{-6}=\dfrac{-8}{-6}=\dfrac{8\div2}{6\div2}=\dfrac{4}{3}\end{cases}\\\sf -3x^2+x+4=-3\cdot(x-[-1])\cdot\bigg(x-\dfrac{4}{3}\bigg)\\\sf -3x^2+x+4=(x+1)\cdot(-3x+4)\end{array}}$

$\Large\boxed{\begin{array}{l}\displaystyle\sf f'(-1)=\lim_{x \to -1}=\dfrac{\diagup\!\!\!\!\!\!(x+\diagup\!\!\!\!\!\!1)\cdot(-3x+4)}{\diagup\!\!\!\!\!\!(x+\diagup\!\!\!\!\!\!\!1)}\\\\\displaystyle\sf f'(-1)=\lim_{x \to-1}-3x+4\\\sf f'(-1)=-3\cdot(-1)+4=3+4=7\end{array}}$