SOLUTION:
Step 1:
In this question, we are given the following:
Step 2:
The details of the solution are as follows:
[tex]\begin{gathered} Given\text{ the function: y =6 + }\frac{6}{x}+\frac{3}{x^2} \\ Differentiating\text{ the function wwith respect to x, we have that:} \\ \frac{dy}{dx}=\text{ }\frac{-6}{x^2}-\frac{6}{x^3} \end{gathered}[/tex]
[tex]\begin{gathered} \frac{dy}{dx}=\text{ }\frac{-6}{x^2}-\frac{6}{x^3} \\ put\text{ x = 3, we have that:} \end{gathered}[/tex][tex]\frac{dy}{dx}=\text{ m = }\frac{-6}{3^2}-\frac{6}{3^3}=\frac{-6}{9}-\frac{6}{27}=\frac{-18-6}{27}=\frac{-24}{27}=\frac{-8}{9}[/tex][tex]\begin{gathered} put\text{ the value of x = 3 in } \\ \text{y = 6 +}\frac{6}{x}+\text{ }\frac{3}{x^2}\text{, we have that:} \\ y_1\text{= 6 + }\frac{6}{3}+\frac{3}{3^2^}=6\text{ + 2 + }\frac{1}{3}=\text{ 8 }\frac{1}{3}=\frac{25}{3} \end{gathered}[/tex]
Now, we have that:
[tex]\begin{gathered} (x_1,\text{ y}_1)\text{ = \lparen 3, }\frac{25}{3}) \\ and\text{ m = }\frac{-8}{9} \end{gathered}[/tex][tex]\begin{gathered} Using\text{ y -y}_1=\text{ m \lparen x - x}_1) \\ y\text{ -}\frac{25}{3}=\frac{-8}{9}(\text{ x - 3\rparen} \end{gathered}[/tex][tex]\begin{gathered} Multiplying\text{ through by 9, we have that:} \\ 9\text{y -75 = - 8 \lparen x- 3\rparen} \\ 9y\text{ -75 = -8x + 24} \\ 9y\text{ + 8x = 24 + 75} \\ \text{8x + 9y = 99} \end{gathered}[/tex]
The final answer is:
[tex]8x+\text{ 9y = 99}[/tex]
