Question

Encontre uma equação da reta tangente da curva y = x^3−3x+1 no ponto (2, 3)

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\mathsf{y = x^3 - 3x + 1}$

$\mathsf{P(2;3)}}$

$\mathsf{m = f'(x) = 3x^2 - 3}$

$\mathsf{f'(2) = 3(2)^2 - 3}$

$\mathsf{f'(2) = 3(4) - 3}$

$\mathsf{f'(2) = 12 - 3}$

$\mathsf{f'(2) = 9}$

$\mathsf{y - y_0 = m(x - x_0)}$

$\mathsf{y - 3 = 9(x - 2)}$

$\mathsf{y - 3 = 9x - 18}$

$\boxed{\boxed{\mathsf{9x - y - 15 = 0}}}$

Answer 2

$\Large\boxed{\begin{array}{l}\underline{\rm De~\!\!finic_{\!\!,}\tilde ao\,de\,derivada\,no\,ponto}\\\displaystyle\sf f'(x_0)=\lim_{x \to x_0}\dfrac{f(x)-f(x_0)}{x-x_0}\\\underline{\rm Equac_{\!\!,}\tilde ao\,da\,reta\,tangente}\\\sf y=y_0+f'(x_0)(x-x_0)\end{array}}$

$\Large\boxed{\begin{array}{l}\sf y=x^3-3x+1~~P(2,3)\implies x_0=2~~y_0=3\\\displaystyle\sf f'(2)=\lim_{x \to 2}\dfrac{ x^3-3x+1-3}{x-2}\\\\\displaystyle\sf f'(2)=\lim_{x \to 2}\dfrac{x^3-3x-2}{x-2}\end{array}}$

$\Large\boxed{\begin{array}{l}\sf x^3-3x-2|\underline{x-2~~~~~~~~~~}\\\sf-\underline{x^3+2x^2}~~~~x^2+2x+1\\\sf 2x^2-3x-2\\\sf\underline{-2x^2+4x}\\\sf~~ x-2\\\sf\underline{-x+2}\\\sf~~0\end{array}}$

$\Large\boxed{\begin{array}{l}\displaystyle\sf f'(2)=\lim_{x \to 2}\dfrac{\diagup\!\!\!\!(x-\diagup\!\!\!\!2)\cdot(x^2+2x+1)}{\diagup\!\!\!\!(x-\diagup\!\!\!2)}\\\\\displaystyle\sf f'(2)=\lim_{x \to 2}x^2+2x+1\\\sf f'(2)=2^2+2\cdot2+1\\\sf f'(2)=9\end{array}}$

$\Large\boxed{\begin{array}{l}\sf y =y_0+f'(x_0)\cdot(x-x_0)\\\sf y=3+9\cdot(x-2)\\\sf y=3+9x-18\\\sf y=9x-15\end{array}}$