Given:
Given the function
[tex]y=3+\frac{3}{x}+\frac{2}{x^2}[/tex]and a point x = 3.
Required: Equation of the line tangent to y at x = 3.
Explanation:
The derivative of a function is he slope of the tangent line of the function at a given point. So, finding the derivative gives the slope of the tangent line.
[tex]y^{\prime}=-\frac{3}{x^2}-\frac{4}{x^3}[/tex]Substitute 3 for x into the derivative.
[tex]\begin{gathered} y^{\prime}|_{x=3}=-\frac{3}{3^2}-\frac{4}{3^3} \\ =-\frac{31}{27} \end{gathered}[/tex]Therefore, the slope of the tangent line is -31/27.
Substitute 3 for x into y.
[tex]\begin{gathered} y|_{x=3}=3+\frac{3}{3}+\frac{2}{3^2} \\ =3+1+\frac{2}{9} \\ =4+\frac{2}{9} \\ =\frac{38}{9} \end{gathered}[/tex](3, 38/9) is the only point on the tangent line where it intersects the original graph.
Plug these coordinates along with slope into the general point-slope form to find the equation.
[tex]\begin{gathered} y-y_1=m(x-x_1) \\ y-\frac{38}{9}=-\frac{31}{27}(x-3) \end{gathered}[/tex]Solving for y will give the equation in slope-intercept form.
[tex]\begin{gathered} y=-\frac{31}{27}(x-3)+\frac{38}{9} \\ =-\frac{31}{27}x+\frac{69}{9} \end{gathered}[/tex]Final Answer: The equation of the tangent line is
[tex]y=-\frac{31}{27}x+\frac{69}{9}[/tex]
