Answer
$g(x)$ is linear, $f(x)$ is not
$g(x)=3x-1$
Work Step by Step
For linear functions,
a change of $\Delta x$ units in results in a change of $\Delta y=m\Delta x$ units in $y$.
Using m$=\displaystyle \frac{\Delta y}{\Delta x} =\displaystyle \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$, on the points defined by the first two columns
for f(x), m should be $\displaystyle \frac{6-2}{3-0}=\frac{4}{3},$that is, $\displaystyle \Delta y=\frac{4}{3}\Delta x$
Observe the last two columns: x changes from 6 to 9, $\Delta x$=3,
but $\displaystyle \Delta y=15-12=3\neq\frac{4}{3}\cdot 3$,
so f is not linear.
Checking g(x), the first two columns suggest that $m=\displaystyle \frac{8-(-1)}{3-0}=\frac{9}{3}=$3
So, $\Delta y=3\cdot\Delta x:$
x changes from $3$ to $5$, $\Delta x=2$, $\Delta y=14-8=6$, OK
x changes from $5$ to $6$, $\Delta x=1$, $\Delta y=17-14=3$, OK
x changes from $6$ to $9$, $\Delta x=3$, $\Delta y=26-17=9$, OK
so, g is linear $(g(x)=mx+b$) and has slope $3$.
The table shows
$b=g(0)=-1$ so
$g(x)=mx+b$
$g(x)=3x-1$
