Question

Preciso de ajuda, derivada;

Answer 1

Explicação passo-a-passo:

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7)Temos a seguinte função:

$\boxed{f(x)=\sqrt{x-2}}$

Calculando o ponto em que x=3:

$f(3)=\sqrt{3-2}\\\\f(3)=\sqrt{1} \\\\f(3)=1\\\\\boxed{P(3,1)}$

Calculando a derivada da função f:

$f(x)=(x-2)^{\frac{1}{2} } \\\\\text{Usando a Regra da Cadeia:}\\\\f'(x)=\dfrac{1}{2} .(x-2)^{-\frac{1}{2} }. 1\\\\\boxed{f'(x)=\dfrac{1}{2.\sqrt{x-2} } }$

Calculando o coeficiente angular da reta tangente r a função:

$m_{r} =f'(3)\\\\m_{r} =\dfrac{1}{2.\sqrt{3-2} }\\\\\boxed{m_{r}=\dfrac{1}{2}}$

A reta normal s é perpendicular a reta tangente r:

$m_{s}.m_{r}=-1\\\\m_{s}.\dfrac{1}{2}=-1\\\\\boxed{m_{2}=-2}$

Calculando a equação da reta norma que passa pelo ponto P:

$y-y_{0} =m.(x-x_{0} )\\\\y-1=-2.(x-3)\\\\y-1=-2x+6\\\\\boxed{\boxed{y=-2x+7}}$

8)Temos a seguinte função:

$\boxed{f(x)=2sen(x)+x}$

Calculando a derivada:

$\boxed{f'(x)=2cos(x)+1}$

Temos a seguinte reta r:

$r:y-x=-1\\\\r:y=x-1\\\\\implies \boxed{m_{r}=1}$

Como as retas são paralelas elas tem o mesmo coeficiente angular:

$m=m_{r} \\\\\boxed{m=1}$

Calculando o valor de x que terá esse coeficiente angular:

$f'(x)=m\\\\f'(x)=1\\\\2.\cos(x)+1=1\\\\2.\cos(x)=0\\\\\cos(x)=0\\\\\boxed{x=\dfrac{\pi }{2}\ \ ou \ \ x=\dfrac{3\pi }{2} }$

Calculando o ponto em que x=π/2:

$f\bigg(\dfrac{\pi }{2}\bigg)=2.sen\bigg(\dfrac{\pi }{2}\bigg) +\dfrac{\pi }{2} \\\\f\bigg(\dfrac{\pi }{2}\bigg)=2+\dfrac{\pi }{2} \\\\\boxed{P_{1}\bigg(\dfrac{\pi }{2}\ ,\ 2+ \dfrac{\pi }{2}\bigg) }$

Calculando o ponto em que x=3π/2:

$f\bigg(\dfrac{3\pi }{2}\bigg)=2.sen\bigg(\dfrac{3\pi }{2}\bigg) +\dfrac{3\pi }{2} \\\\f\bigg(\dfrac{3\pi }{2}\bigg)=-2+\dfrac{3\pi }{2} \\\\\boxed{P_{2}\bigg(\dfrac{3\pi }{2}\ ,\ -2+ \dfrac{3\pi }{2}\bigg) }$

Calculando as equações das retas tangente:

$y-y_{1}=m.(x-x_{1})\\\\y-2-\dfrac{\pi }{2} =1.\bigg(x-\dfrac{\pi }{2} \bigg)\\\\y-2-\dfrac{\pi }{2} =x-\dfrac{\pi }{2} \\\\\boxed{\boxed{y=x+2}}\\\\y-y_{2}=m.(x-x_{2})\\\\y+2-\dfrac{3\pi }{2} =1.\bigg(x-\dfrac{3\pi }{2} \bigg)\\\\y+2-\dfrac{3\pi }{2} =x-\dfrac{3\pi }{2} \\\\\boxed{\boxed{y=x-2}}$