a. for the following function, find f(a)\nb. determine an equation of the line tangent to the graph of f at…

Explanation:

Step1: Use the quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. For $f(x)=\frac{4}{3x + 1}$, let $u = 4$ and $v=3x + 1$. Then $u'=0$ and $v'=3$. $f'(x)=\frac{0\times(3x + 1)-4\times3}{(3x + 1)^{2}}=\frac{-12}{(3x + 1)^{2}}$

Step2: Find $f'(a)$ when $a = 2$

Substitute $x = 2$ into $f'(x)$. $f'(2)=\frac{-12}{(3\times2+1)^{2}}=\frac{-12}{(6 + 1)^{2}}=\frac{-12}{49}$

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

$f(2)=\frac{4}{3\times2+1}=\frac{4}{7}$

Step4: Use the point - slope form for the tangent line

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(a,f(a))=(2,\frac{4}{7})$ and $m = f'(2)=-\frac{12}{49}$. $y-\frac{4}{7}=-\frac{12}{49}(x - 2)$ $y-\frac{4}{7}=-\frac{12}{49}x+\frac{24}{49}$ $y=-\frac{12}{49}x+\frac{24}{49}+\frac{4}{7}$ $y=-\frac{12}{49}x+\frac{24 + 28}{49}$ $y=-\frac{12}{49}x+\frac{52}{49}$

Answer:

a. $f'(a)=-\frac{12}{49}$ b. $y =-\frac{12}{49}x+\frac{52}{49}$