The derivative at x = 3 is:
[tex]f^{\prime}(3) = -\frac{7}{11}[/tex]
Using the derivative, the equation for the tangent line is:
[tex]y - 3 = -\frac{7}{11}(x - 3)[/tex]
The function is:
[tex]f(x) = \frac{11x}{2 + x^2}[/tex]
Applying the quotient rule, the derivative is:
[tex]f^{\prime}(x) = \frac{[11x]^{\prime}(2 + x^2) - [(2 + x^2)]^{\prime}(11x)}{(2 + x^2)^2}[/tex]
[tex]f^{\prime}(x) = \frac{22 + 11x^2 - 22x^2}{(2 + x^2)^2}[/tex]
[tex]f^{\prime}(x) = \frac{-11x^2 + 22}{(2 + x^2)^2}[/tex]
At x = 3:
[tex]f^{\prime}(3) = \frac{-11(3)^2 + 22}{(2 + (3)^2)^2} = -\frac{77}{121} = -\frac{7}{11}[/tex]
The equation of a tangent-line, in point-slope form, is:
[tex]y - y_0 = m(x - x_0)[/tex]
In which the slope is [tex]m = f^{\prime}(x_0)[/tex].
Point (3,3), thus [tex]x_0 = 3, y_0 = 3, f^{\prime}(x_0) = f^{\prime}(3) = -\frac{7}{11}[/tex]
Then
[tex]y - y_0 = m(x - x_0)[/tex]
[tex]y - 3 = -\frac{7}{11}(x - 3)[/tex]
A similar problem is given at https://brainly.com/question/22426360
