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

Answer:

a.

$f'(x)=-\frac{24}{(4x + 1)^2}$

b.

$y=-3x - 6$

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^2}$. For $f(x)=\frac{6}{4x + 1}$, let $u = 6$ and $v=4x + 1$.

Step2: Find $u'$ and $v'$

$u'=0$ (since the derivative of a constant is 0) and $v' = 4$ (derivative of $4x+1$ with respect to $x$ is 4).

Step3: Apply quotient - rule

$f'(x)=\frac{0\times(4x + 1)-6\times4}{(4x + 1)^2}=-\frac{24}{(4x + 1)^2}$.

Step4: Find $f(a)$ and $f'(a)$ when $a=-1$

First, $f(-1)=\frac{6}{4\times(-1)+1}=\frac{6}{-3}=-2$. Second, $f'(-1)=-\frac{24}{[4\times(-1)+1]^2}=-\frac{24}{(-3)^2}=-\frac{24}{9}=-\frac{8}{3}=- 3$.

Step5: Use point - slope form for tangent line

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(a,f(a))=(-1,-2)$ and $m = f'(a)=-3$. Substituting these values, we get $y+2=-3(x + 1)$. Simplifying gives $y=-3x-3 - 2=-3x - 6$.