a. for the following function, find f(a). f(x)=\frac{3}{3x + 1}, a = 2 b. determine an equation of the line…

Explanation:

Step1: Use quotient - rule for differentiation

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. For $f(x)=\frac{3}{3x + 1}=\frac{u}{v}$, where $u = 3$, $u'=0$, $v=3x + 1$, and $v'=3$. [ \begin{align*} f'(x)&=\frac{0\times(3x + 1)-3\times3}{(3x + 1)^{2}}\ &=\frac{- 9}{(3x + 1)^{2}} \end{align*} ]

Step2: Evaluate $f'(a)$ at $a = 2$

Substitute $x = 2$ into $f'(x)$. [ \begin{align*} f'(2)&=\frac{-9}{(3\times2 + 1)^{2}}\ &=\frac{-9}{(6 + 1)^{2}}\ &=\frac{-9}{49} \end{align*} ]

Step3: Find $f(a)$ at $a = 2$

[ \begin{align*} f(2)&=\frac{3}{3\times2+1}\ &=\frac{3}{7} \end{align*} ]

Step4: Use point - slope form for tangent line

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(a,f(a))=(2,\frac{3}{7})$ and $m = f'(2)=-\frac{9}{49}$. [ \begin{align*} y-\frac{3}{7}&=-\frac{9}{49}(x - 2)\ y-\frac{3}{7}&=-\frac{9}{49}x+\frac{18}{49}\ y&=-\frac{9}{49}x+\frac{18}{49}+\frac{3}{7}\ y&=-\frac{9}{49}x+\frac{18 + 21}{49}\ y&=-\frac{9}{49}x+\frac{39}{49} \end{align*} ]

Answer:

a. $f'(2)=-\frac{9}{49}$ b. $y =-\frac{9}{49}x+\frac{39}{49}$