[tex]\text{f(x)} = \text x^3-3\text x+4[/tex]
item a)
Vamos achar o coeficiente angular da reta tangente, basta derivar a função no ponto (2,6), ou seja :
[tex]\displaystyle \text {m}=\text{f'(x)} \\\\ \text m = \lim_{\displaystyle \text h\to 0 } \ \frac{\text{f(x+h)}-\text{f(x)}}{\text h} \\\\\\ \text m =\lim_{\displaystyle \text h\to 0 } \frac{(\text{x+h})^3-3(\text{x+h})+4-\text x^3+3\text x-4}{\text h} \\\\\\ \text m = \lim_{\displaystyle \text h\to 0 } \frac{\text x^3+\text h^3+3\text{x.h(x+h)}-3\text x-3\text h+4-\text x^3+3\text x-4}{\text h} \\\\\\ \text m =\lim_{\displaystyle \text h\to 0 }\frac{\text h^3+3\text {x.h(x+h)}-3\text h}{\text h}[/tex]
[tex]\displaystyle \text m = \lim_{\text h\to 0} \frac{\text h[\text h^2+3\text{x(h)}-3]}{\text h}\\\\\\ \text m = \lim_{\displaystyle \text h\to 0 } \text h^2+3\text {x(x+h)}-3 \\\\\\ \text m = 0^2+3\text x(\text x+0)-3 \\\\ \text m = 3\text x^2-3 \\\\ \text{Substitiundo o ponto x = 2 } \\\\ \text m = 3.2^2-3 \\\\ \text m = 9[/tex]
Sabemos que o coeficiente angular da reta é a tangente do ângulo de inclinação, então :
[tex]\displaystyle \text{tg }\alpha = 9 \\\\ \underline{\text{Portanto a inclina{\c c}{\~a}o da reta tangente {\'e}}}: \\\\ \huge\boxed{\alpha =\text{arcTg(9)}\ }\checkmark[/tex]
item b)
Equação da reta tangente à f no ponto (2,6) :
[tex]\text y-\text y_\text o=\text m (\text x-\text x_\text o) \\\\ \text y - 6=9(\text x-2) \\\\ \text y-6 = 9\text x-18 \\\\ \underline{\text{Eq da reta tangente {\`a} f}}:\\\\ \huge\boxed{\ 9\text x-\text y -12=0\ }\checkmark[/tex]
item c) Veja a imagem
