Question

Which of the following tangent points on g(x)=36/(x - 1) has an instantaneous rate of change of -9?

Answer 1

The statement correct is: a(-3,-9).

Answer 2

$\bf g(x)=\cfrac{36}{x-1}\implies g(x)=36(x-1)^{-1}\qquad thus \\\\\\ \cfrac{dg}{dx}=36\cdot -1(x-1)^{-2}\implies \cfrac{dg}{dx}=-\cfrac{36}{(x-1)^2}\\\\ -------------------------------\\\\ \textit{now, if the instantaneous rate of change is -9, then} \\\\\\ -9-\cfrac{36}{(x-1)^2}\implies (x-1)^2=\cfrac{-36}{-9}\implies (x-1)^2=4 \\\\\\ x-1=\pm\sqrt{4}\implies x-1=\pm 2\quad x= \begin{cases} 3\\ -1 \end{cases}$

well, then we know when x = 3 or -1, the slope is -9, what's "y"?

$\bf x=3\qquad g(x)=\cfrac{36}{(3)-1}\implies g(x)=18\qquad \boxed{(3,18)} \\\\\\ x=-1\qquad g(x)=\cfrac{36}{(-1)-1}\implies g(x)=-18\qquad (3,-18)$