estimate the instantaneous rate of change of $g(x)=\frac{4}{x - 3}$ at the point $x=-3$. your answer should…

Explanation:

Step1: Recall derivative formula

The derivative of $y = \frac{u}{v}$ is $y'=\frac{u'v - uv'}{v^{2}}$ by the quotient - rule. Here, $u = 4$, $u'=0$, $v=x - 3$, $v' = 1$.

Step2: Apply quotient - rule

$g'(x)=\frac{0\times(x - 3)-4\times1}{(x - 3)^{2}}=-\frac{4}{(x - 3)^{2}}$.

Step3: Evaluate at $x=-3$

Substitute $x = - 3$ into $g'(x)$: $g'(-3)=-\frac{4}{(-3 - 3)^{2}}=-\frac{4}{36}=-\frac{1}{9}\approx - 0.111$.

Answer:

$-0.111$