The function $g(x)$ is continuous and differentiable at all points in $\mathbb{R}$ as $\var{c} \notin I$.

On differentiating we see that
$$\begin{eqnarray*}g'(x)&=&\simplify{2*x/(x-{c})^({a}/{b}) - {a}*x^2/({b}(x-{c})^({a+b}/{b}))}\\ &=&\simplify{({2*b}x*(x-{c})-{a}*x^2)/({b}(x-{c})^({a+b}/{b}))}\\ &=&\simplify{(x*({2*b-a}x-{2*b*c}))/({b}(x-{c})^({a+b}/{b}))} \end{eqnarray*}$$

Stationary Points.

The stationary points are given by solving $g'(x)=0$.

$\displaystyle g'(x)=0 \Rightarrow \simplify{x*({2*b-a}x-{2*b*c})=0} \Rightarrow x=0 \mbox{ or } x=\simplify[std]{{2*b*c}/{2*b-a}}$

We see that $\displaystyle x=\simplify[std]{{2*b*c}/{2*b-a}}$ is the only stationary point in $I$.

The second derivative can be found by applying the quotient rule to the derivative of $g(x)$ and we obtain:

Using the quotient rule for differentiation we see that
$$g''(x)=\simplify[std]{({a^2-3*a*b+2*b^2}*x^2+{4*b*c*(a-b)}*x+{2*c^2*b^2})/({b^2}(x-{c})^({a+2*b}/{b}))}$$

Hence $$r(x)=\simplify[std]{({a^2-3*a*b+2*b^2}*x^2+{4*b*c*(a-b)}*x+{2*c^2*b^2})}$$

The nature of the stationary points are determined by evaluating $g''(x)$ at the stationary points.

There is only one stationary point $\displaystyle  x=\simplify[std]{{2*b*c}/{2*b-a}}$ in $I$ and at that point we have:
$$g''\left(\simplify[std]{{2*b*c}/{2*b-a}}\right)=\var{valsd} \gt 0$$

Hence this point is a local minimum.

Evaluating at end points of the interval.

The values of $g$ at the endpoints are:

$g(\var{l})=\var{valbegin}$ and $g(\var{m})=\var{valend}$, both to 3 decimal places.

Global Maximum and Minimum

Global Maximum: Since $g$ does not have a local maximum in the interval $I$, it must take a global maximum value at one of the end points of the interval.

From the values found for $g(\var{l})$ and $g(\var{m})$ found above, we see that $x=\var{xma}$ is the global maximum for $g$ in the interval $I$.

Global Minimum: The local minimum of $g$ given by $\displaystyle x=\simplify[std]{{2*b*c}/{2*b-a}} \in I$ is the only local minimum and must be a global minimum in $I$.

Note that the global minimum value for $g$ on $I$ is:

$$g\left(\simplify[std]{{2*b*c}/{2*b-a}}\right)=\var{valmin}$$