Answer
(a) $g'(a)=\frac{-2}{(2a-1)^2}$
(b) $y=−\frac{2}{9}x-\frac{5}{9}$
$y=−2x-1$
$y=−2x+3$
(c) See graph.
Work Step by Step
(a) Use the formula given in this section, we have:
$$g'(a)=\lim_{h\to 0}\frac{g(a+h)-g(a)}{h}=\lim_{h\to 0}\frac{\frac{1}{2(a+h)-1}-\frac{1}{2a-1} }{h}=\lim_{h\to 0}\frac{2a-1-2a-2h+1 }{h(2a-1)(2a+2h-1)}=\lim_{h\to 0}\frac{-2}{(2a-1)(2a+2h-1)}=\frac{-2}{(2a-1)^2}$$
(b) At $x=-1$, we have $f(-1)=-\frac{1}{3}$ and $f′(-1)=−\frac{2}{9}$ which is the slope of the tangent line. Knowing the slope and a point $(-1,-\frac{1}{3})$ on the line, we can get the equation as $y=−\frac{2}{9}(x+1)-\frac{1}{3}$ or $y=−\frac{2}{9}x-\frac{5}{9}$
At $x=0$, we have $f(0)=-1$ and $f′(0)=−2$ which is the slope of the tangent line. Knowing the slope and a point $(0,-1)$ on the line, we can get the equation as $y=−2x-1$
At $x=1$, we have $f(1)=1$ and $f′(1)=−2$ which is the slope of the tangent line. Knowing the slope and a point $(1,1)$ on the line, we can get the equation as $y=−2(x-1)+1$ or $y=−2x+3$
(c) See graph.