Question

aplicando a definição, calcule:

a) a derivada da função f(x) = x² + x no ponto de abscissa x = 3:
b) a derivada da função f(x) = x² - 5x + 6 no ponto x = 1.

me ajudem por favor

Answer 1

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Definição de derivada no ponto

$\huge\boxed{\boxed{\boxed{\boxed{\displaystyle\sf f'(p)=\lim_{ x \to p}\dfrac{f(x)-f(p)}{x-p}}}}}$

$\large\boxed{\begin{array}{l}\rm a)~\sf f(x)=x^2+x\\\sf f(3)=3^2+3=9+3=12\\\displaystyle\sf f'(3)=\lim_{x \to 3}\dfrac{x^2+x-12}{x-3}\\\sf x^2+x-12=(x+4)\cdot(x-3)\\\displaystyle\sf f'(3)=\lim_{x \to 3}\dfrac{(x+4)\cdot\diagup\!\!\!\!(x-\diagup\!\!\!\!3)}{\diagup\!\!\!\!\!(x-\diagup\!\!\!\!3)}\\\displaystyle\sf f'(3)=\lim_{x \to 3}x+4\\\sf f'(3)=3+4\\\sf f'(3)=7\end{array}}$x

$\large\boxed{\begin{array}{l}\rm b)~\sf f(x)=x^2-5x+6\\\sf f(1)=1^2-5\cdot1+6=2\\\displaystyle\sf f'(1)=\lim_{x \to 1}\dfrac{x^2-5x+6-2}{x-1}\\\displaystyle\sf f'(1)=\lim_{x \to 1}\dfrac{ x^2-5x+4}{x-1}\\\sf x^2-5x+4=(x-4!\!\!)\cdot(x-1)\\\displaystyle\sf f'(1)=\lim_{x \to 1}\dfrac{(x-4)\cdot\diagup\!\!\!\!(x-\diagup\!\!\!1)}{\diagup\!\!\!\!\!(x-\diagup\!\!\!1)}\\\displaystyle\sf f'(1)=\lim_{x \to 1}x-4\\\sf f'(1)=1-4\\\sf f'(1)=-3\end{array}}$