a. find the derivative function f’ for the function f. b. determine an equation of the line tangent to the…

Answer:

b. $y = -\frac{3}{2}x - \frac{1}{2}$

Explanation:

Step1: Find $f(a)$

First, substitute $a=-1$ into $f(x)=\frac{2}{3x + 1}$. So $f(-1)=\frac{2}{3\times(-1)+1}=\frac{2}{-3 + 1}=-1$.

Step2: Find the slope of the tangent - line

We know that $f^{\prime}(x)=-\frac{6}{(3x + 1)^2}$, substitute $x = - 1$ into $f^{\prime}(x)$. Then $f^{\prime}(-1)=-\frac{6}{[3\times(-1)+1]^2}=-\frac{6}{(-2)^2}=-\frac{3}{2}$. The slope of the tangent - line $m = f^{\prime}(-1)=-\frac{3}{2}$.

Step3: Use the point - slope form

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(a,f(a))=(-1,-1)$ and $m = -\frac{3}{2}$. Substitute these values into the formula: $y-(-1)=-\frac{3}{2}(x - (-1))$. Simplify the equation: $y + 1=-\frac{3}{2}(x + 1)$. Expand the right - hand side: $y+1=-\frac{3}{2}x-\frac{3}{2}$. Solve for $y$: $y=-\frac{3}{2}x-\frac{3}{2}-1=-\frac{3}{2}x-\frac{1}{2}$.