Question

Determine a equação da reta tangente ao gráfico de f(x) = ³√x
, no ponto de abscissa 8​

Answer 1

$\displaystyle \text{f(x)}=\sqrt[3]{\text x} \\\\\ \underline{\text{Reta tangente {\`a} f(x)}}: \\\\ \text{y}=\text m.\text x+\text n \\\\ \text{onde}: \\\\ \text m =\text{f'(x)} \\\\ \text{Derivando a f(x)} : \\\\ \text m = [\sqrt[3]{\text x}]'\to \text m = [\displaystyle \text x^{\displaystyle (\frac{1}{3})}]' \\\\ \text m = \frac{1}{3}\text x^{(\frac{1}{3}-1)} \to \text m=\frac{1}{3\sqrt[3]{\text x^2}}$

$\displaystyle \underline{\text{Substituindo o ponto de abscissa 8} }: \\\\ \text m = \frac{1}{3\sqrt[3]{8^2}} \to \text m = \frac{1}{3\sqrt[3]{64}} \\\\\\ \text m =\frac{1}{3.4} \to \boxed{\text m =\frac{1}{12}} \\\\\\ \underline{\text{Equa{\c c}{\~a}o da reta tangente}} : \\\\ \text y = \frac{\text x}{12}+\text n$

Vamos substituir o valor de x = 8 na f(x) para achar o valor de y :

$\text{f(8)}=\sqrt[3]{8} = 2$

Daí na reta tangente :

$\displaystyle 2=\frac{8}{12}+\text n \to \text n = 2-\frac{8}{12} \\\\\\ \text n = \frac{24-8}{12} \to \boxed{\text n = \frac{16}{12} }$

Portanto a equação da reta Tangente é :

$\displaystyle \text y = \frac{\text x+16}{12} \\\\\\ 12\text y =\text x+16 \\\\\\ \huge\boxed{\ \text x-12\text y+16=0\ }\checkmark$

Answer 2

✅ Tendo finalizado os cálculos, concluímos que a equação da reta tangente ao gráfico da referida função é:

           $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf t: y = \frac{1}{12}x + \frac{16}{12}\:\:\:}}\end{gathered} }$

Sejam os dados:

                       $\Large\begin{cases} f(x) = \sqrt[3]{x}\\x = 8\end{cases}$

Organizando a equação, temos:

                        $\Large\begin{cases} f(x) = x^{\frac{1}{3}}\\x = 8\end{cases}$

Para calcular a equação da reta tangente ao gráfico da referida função pelo ponto de abscissa "-2" devemos utilizar a equação da reta em sua forma "ponto/declividade", ou seja:

$\Large\displaystyle\begin{gathered} \bf I\end{gathered}$              $\Large\displaystyle\begin{gathered} y - y_{T} = m_{t}\cdot(x- x_{T})\end{gathered}$

Sabendo que:

$\Large\displaystyle\begin{gathered} \bf II\end{gathered}$                        $\Large\displaystyle\begin{gathered} y_{T} = f(x_{T})\end{gathered}$

Além disso, sabemos também que o coeficiente angular da reta é numericamente igual à derivada primeira da função no ponto de abscissa especificada, ou seja:

$\Large\displaystyle\begin{gathered} \bf III\end{gathered}$                      $\Large\displaystyle\begin{gathered} m_{t} = f'(x_{T})\end{gathered}$

Substituindo "II" e "III" na equação "I", temos:

$\Large\displaystyle\begin{gathered} \bf IV\end{gathered}$        $\Large\displaystyle\begin{gathered} y - f(x_{T}) = f'(x_{T})\cdot(x - x_{T})\end{gathered}$

Substituindo os dados na equação "IV", temos:

            $\Large\displaystyle\text{ \begin{gathered} y - \left[8^{\frac{1}{3}}\right] = \left[\frac{1}{3}\cdot8^{\frac{1}{3} - 1}\right]\cdot\left[x - 8\right]\end{gathered} }$

          $\Large\displaystyle\text{ \begin{gathered} y - \left[\sqrt[3]{8}\right] =\left[\frac{1}{3}\cdot8^{\frac{1 - 3}{3}}\right]\cdot\left[x - 8\right] \end{gathered} }$

                   $\Large\displaystyle\text{ \begin{gathered} y - 2 = \left[\frac{1}{3}\cdot8^{-\frac{2}{3}}\right]\cdot\left[x - 8\right]\end{gathered} }$

                   $\Large\displaystyle\text{ \begin{gathered} y - 2 = \left[\frac{1}{3}\cdot\frac{1}{8^{\frac{2}{3}}}\right]\cdot\left[x - 8\right]\end{gathered} }$

                   $\Large\displaystyle\text{ \begin{gathered} y - 2 = \frac{1}{3\sqrt[3]{8^{2}}}\cdot\left[x - 8\right]\end{gathered} }$

                   $\Large\displaystyle\text{ \begin{gathered} y - 2 = \frac{1}{3\sqrt[3]{64}}\cdot\left[x - 8\right]\end{gathered} }$

                   $\Large\displaystyle\begin{gathered} y - 2 = \frac{1}{3\cdot4}\cdot\left[x - 8\right]\end{gathered}$

                   $\Large\displaystyle\begin{gathered} y - 2 = \frac{1}{12}\cdot\left[x - 8\right]\end{gathered}$

                   $\Large\displaystyle\begin{gathered} y - 2 = \frac{x}{12} - \frac{8}{12}\end{gathered}$

                           $\Large\displaystyle\begin{gathered} y = \frac{x}{12} - \frac{8}{12} + 2\end{gathered}$

                           $\Large\displaystyle\begin{gathered} y = \frac{x - 8 + 24}{12} \end{gathered}$

                           $\Large\displaystyle\begin{gathered} y = \frac{x + 16}{12}\end{gathered}$

                           $\Large\displaystyle\begin{gathered} y = \frac{1}{12}x + \frac{16}{12}\end{gathered}$

✅ Portanto, a equação da reta tangente é:

                       $\Large\displaystyle\begin{gathered} t: y = \frac{1}{12}x + \frac{16}{12}\end{gathered}$

$\LARGE\displaystyle\text{ \begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Bons \:estudos!!\:\:\:Boa\: sorte!!\:\:\:}}}\end{gathered} }$

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$\Large\displaystyle\text{ \begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Observe \:o\:Gr\acute{a}fico!!\:\:\:}}}\end{gathered} }$