Mensagem não lida por joaopcarv » Sex 09 Abr, 2021 15:58
Mensagem não lida
por joaopcarv » Sex 09 Abr, 2021 15:58
Seja [tex3]\mathsf{\alpha_{(x)}}[/tex3]
a inclinação da reta tangente a um ponto de [tex3]\mathsf{f(x) \ = \ e^{\sqrt{x^2 \ - \ 3\cdot x}}.}[/tex3]
Temos que:
[tex3]\mathsf{\alpha_{(x)} \ = \ \dfrac{df(x)}{dx}}[/tex3]
[tex3]\mathsf{\alpha_{(x)} \ = \ \dfrac{d(e^u)}{du} \cdot \dfrac{du}{dx}}[/tex3]
[tex3]\mathsf{\alpha_{(x)} \ = \ e^{\sqrt{x^2 \ - \ 3\cdot x}} \cdot \ \dfrac{1}{2\ \cdot \big(\sqrt{x^2 \ - \ 3\cdot x}\big)} \cdot \ (2\cdot x \ - \ 3)}[/tex3]
Em particular, no ponto [tex3]\mathsf{P \ = \ (-1, f(-1)):}[/tex3]
[tex3]\mathsf{\alpha_{(-1)} \ = \ e^{\sqrt{(-1)^2 \ - \ 3\cdot (-1)}} \cdot \dfrac{1}{2\ \cdot \big(\sqrt{(-1)^2 \ - \ 3\cdot (-1)}\big)} \cdot (2\cdot (-1) \ - \ 3)}[/tex3]
[tex3]\mathsf{\alpha_{(-1)} \ = \ \dfrac{-5\cdot e^2}{4}}[/tex3]
Logo, usando a inclinação e o ponto dado, temos [tex3]\mathsf{L_1:}[/tex3]
[tex3]\mathsf{\dfrac{-5 \cdot e^2}{4} \ = \ \dfrac{\ \ y \ - \ \overbrace{e^2}^{f(-1)}}{x \ + \ 1}}[/tex3]
[tex3]\boxed{\mathsf{L_1: \ y \ = \ \dfrac{-5\cdot e^2 \cdot x \ - e^2}{4}}}[/tex3]
Faremos o mesmo método para [tex3]\mathsf{g(x) \ = \ \dfrac{df(x)}{dx} \ = \ e^{\sqrt{x^2 \ - \ 3\cdot x}} \cdot \dfrac{1}{2\ \cdot \big(\sqrt{x^2 \ - \ 3\cdot x}\big)} \cdot (2\cdot x \ - \ 3).}[/tex3]
Usando a derivação do produto expandida (Para essa derivação, derive cada função do produto e multiplique pelas outras e some cada parcela):
[tex3]\mathsf{\dfrac{dg(x)}{dx} \ = \ \dfrac{1}{2} \cdot \Bigg(\dfrac{e^{\sqrt{x^2 \ - \ 3\cdot x}} \ \cdot (2\cdot x \ - \ 3)^2}{2\cdot \big(x^2 \ - \ 3\cdot x\big)} \ - \ \dfrac{e^{\sqrt{x^2 \ - \ 3\cdot x}} \ \cdot (2\cdot x \ - \ 3)^2}{2 \cdot \big(\sqrt{x^2 \ - 3\cdot x}\big)^3} \ + \ \dfrac{2 \cdot e^{\sqrt{x^2 \ - \ 3\cdot x}}}{\big(\sqrt{x^2 \ - \ 3\cdot x}\big)}\Bigg)}[/tex3]
Seja então [tex3]\mathsf{\beta{(x)}}[/tex3]
a inclinação da reta tangente a um ponto de [tex3]\mathsf{g(x).}[/tex3]
Para [tex3]\mathsf{Q \ = \ (-1, g(-1)):}[/tex3]
[tex3]\mathsf{\beta_{(-1)} \ = \ \dfrac{1}{2} \cdot \Bigg(\dfrac{e^{\sqrt{x^2 \ - \ 3\cdot x}} \ \cdot (2\cdot x \ - \ 3)^2}{2\cdot \big(x^2 \ - \ 3\cdot x\big)} \ - \ \dfrac{e^{\sqrt{x^2 \ - \ 3\cdot x}} \ \cdot (2\cdot x \ - \ 3)^2}{2 \cdot \big(\sqrt{x^2 \ - 3\cdot x}\big)^3} \ + \ \dfrac{2 \cdot e^{\sqrt{x^2 \ - \ 3\cdot x}}}{\big(\sqrt{x^2 \ - \ 3\cdot x}\big)}\Bigg) \Bigg|_{x \ = \ -1}}[/tex3]
[tex3]\mathsf{\beta_{(-1)} \ = \ \dfrac{e^2 \cdot \Big(\frac{25}{8} \ - \ \frac{25}{16} \ + \ 1\Big)}{2}}[/tex3]
[tex3]\mathsf{\beta_{(-1)} \ = \ \dfrac{41 \cdot e^2}{32}}[/tex3]
Sendo [tex3]\mathsf{g(-1) \ = \ e^{\sqrt{x^2 \ - \ 3\cdot x}} \cdot \dfrac{1}{2\ \cdot \big(\sqrt{x^2 \ - \ 3\cdot x}\big)} \cdot (2\cdot x \ - \ 3) \ \Big|_{x \ = \ -1} \ = \ \dfrac{-5 \cdot e^2}{4}}[/tex3]
, temos [tex3]\mathsf{L_2:}[/tex3]
[tex3]\mathsf{\dfrac{41 \cdot e^2}{32} \ = \ \dfrac{\ \ y \ + \ \frac{5 \cdot e^2}{4}}{x \ + \ 1}}[/tex3]
[tex3]\boxed{\mathsf{L_2: \ y \ = \ \dfrac{41\cdot e^2 \cdot x \ + e^2}{32}}}[/tex3]
Intersecção: [tex3]\mathsf{y_{L_1} \ = \ y_{L_2}:}[/tex3]
[tex3]\mathsf{\dfrac{-5\cdot e^2 \cdot x \ -e^2}{4} \ = \ \dfrac{41\cdot e^2 \cdot x \ + e^2}{32}}[/tex3]
[tex3]\mathsf{-9 \cdot \cancel{e^2} \ = \ 81 \cdot \cancel{e^2} \cdot x}[/tex3]
[tex3]\boxed{\boxed{\mathsf{x \ = \ -\dfrac{1}{9}}}}[/tex3]
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