We have from the fundamental theorem of calculus that:
$$\int_{x_{1}}^{x_{2}}\frac{\mathrm{d}}{\mathrm{d}x}\left(f(x)\right)\:\mathrm{d}x=f(x_{2})-f(x_{1})$$
We also have the formula for an average of a function $f(t)$ between two values $t_{1}$ and $t_{2}$:
$$f_{\text{avg}}=\frac{1}{t_{2}-t_{1}}\int_{t_{1}}^{t_{2}}f(t)\:\mathrm{d}t$$
In your case, we therefore have:
$$g'_{\text{avg}}=\frac{1}{a-1}\int_{1}^{a}\frac{\mathrm{d}g(x)}{\mathrm{d}x}\:\mathrm{d}x=\frac{g(a)-g(1)}{a-1}=\frac{\frac{1}{a}-1}{a-1}=-\frac{1}{a}$$
It appears you are having trouble simplifying the fraction; note that if you multiply top and bottom by $a$ (which is valid because $\frac{a}{a}=1$) you have:
$$\frac{\frac{1}{a}-1}{a-1}=\frac{\frac{a}{a}-a}{a^{2}-a}=\frac{1-a}{a^{2}-a}=\frac{1-a}{a(a-1)}=-\frac{a-1}{a(a-1)}=-\frac{1}{a}$$
The best I can come up with is defining $\lambda(x,y)=\frac{f(y)-f(x)}{y-x}$ and averaging it, using $$f_{AROC}=\frac{1}{\#U}\iint_U\lambda(x,y) dy dx$$
where I use $\#U$ to represent the length, area, volume, size etc of $U$.
For example, when $f: [1,2]\to [2,4] ; f(x)=x^2$,
then $\lambda : [1,2]\times [1,2] \mapsto [2,4]; \lambda(x,y)=x+y$, and $$f_{AROC}=\frac{1}{1\cdot 1}\int_1^2\int_1^2 (x+y) dy dx =3$$
Applying this to $g$, we observe that: $$\lambda_g(x,y)=\begin{cases}\frac{-1}{y-x} & x\geq \frac 12, y< \frac 12 \\ \frac{1}{y-x} & x<\frac12, y\geq \frac12 \\ 0& \text{otherwise} \end{cases}$$
Again the area is $1\cdot 1=1$, and $$g_{AROC}=\bigg[\int_0^\frac 12 \int_\frac 12^1 \frac{1}{y-x}dy dx+\int_\frac12^1 \int_0^\frac 12 \frac{-1}{y-x} dy dx\bigg]=2\ln2$$
As seen here, note both double integrals are equal (but nasty!)
As for $h$, that integral does not converge, as is to be expected, due to the graph's asymptote at $x=0$
Best Answer
You get the values of the function by reading them off the graph. The $x$-coordinate at $D$ is $2.5$, the $y$-coordinate of the graph at that point is $15$, so $f(D)=15$. Therefore, the average rate of change over the interval $DE$ is $$ \frac{f(E)-f(D)}{E-D}=\frac{12.5-15}{4-2.5}=-\frac{5}{3} $$ Do the same thing for the interval $EF$.