Answer
$(a)$
Local maximum: $(0,2)$
local minimum: $(2-\sqrt{3},1.866).$
$(b)$
Absolute maximum: none.
Absolute minimum: $(2-\sqrt{3},1.866).$
$(c)$
See graph.
Work Step by Step
$g(x)=\displaystyle \frac{x-1}{x^{2}-1},\quad x\in[0,1)$
$g'(x)=\displaystyle \frac{1(x^{2}-1)-(x-2)(2x)}{(x^{2}-1)^{2}}=\frac{-x^{2}+4x-1}{(x-1)^{2}(x+1)^{2}}=-\frac{x^{2}-4x+1}{(x-1)^{2}(x+1)^{2}}$
$g'(x)$ is undefined for $ x=\pm 1$: not critical points (not in the domain).
$g'(x)=0$ for
$x=\displaystyle \frac{4\pm\sqrt{16-4}}{2}=\frac{4\pm 2\sqrt{3}}{2}=2\pm\sqrt{3}$
In the interval $[0,1)$, x$=2-\sqrt{3}\approx 0.2679$ is the sole critical point.
$\left[\begin{array}{llllll}
interval & [ & (-2,0) & 2-\sqrt{3} & (0,1) & )\\
t & 0 & 0.1 & 1 & 0.5 & 1\\
f'(t) & & -0.89 & & 1.333 & \\
f(t) & 2 & \searrow & 1.866 & \nearrow & ...\\
& & & & &
\end{array}\right]$
$(a)$
Local maximum: $(0,2)$
local minimum: $(2-\sqrt{3},1.866).$
$(b)$
Absolute maximum: none.
Absolute minimum: $(2-\sqrt{3},1.866).$
$(c)$
See graph.
