Answer
$ a.\quad f$ is one-to-one
$b.\quad f^{-1}(x)=\displaystyle \frac{3x-1}{2}$
Work Step by Step
$f(x)=\displaystyle \frac{2x+1}{3}=\frac{2}{3}x+\frac{1}{3}$
$ a.\quad$
The function is linear, non-constant.
Its graph is an oblique line that passes the horizontal line test
(It is impossible to draw a horizontal line that intersects a function's graph more than once.)
It is one-to-one and has an inverse.
$ b.\quad$
To find a formula for the inverse,
1. Replace $f(x)$ with $y.$
$y=\displaystyle \frac{2x+1}{3}$
2. Interchange $x$ and $y$. (This gives the inverse function.)
$x=\displaystyle \frac{2y+1}{3}$
3. Solve for $y.$
... multiply with 3,
$ 3x=2y+1\qquad$... subtract 1,
$ 3x-1=2y\qquad$...divide with 2
$\displaystyle \frac{3x-1}{2}=y$
4. Replace $y$ with $f^{-1}(x)$ . (This is inverse function notation.)
$f^{-1}(x)=\displaystyle \frac{3x-1}{2}$
