A function f is defined on the set of integers as: f(x) = 2x+1 for -1 <= x < 0, -3x+1 for 0 <= x < 3, -2x-1 for 3 <= x < 6 . a) Write the type of function. b) Find the domain of f. c) Find the range o

The function is defined by different expressions over different intervals. Each expression is a linear polynomial.
This type of function is called a piecewise linear function.

For $-1 \le x < 0$, the only integer is $x = -1$.
For $0 \le x < 3$, the integers are $x = 0, 1, 2$.
For $3 \le x < 6$, the integers are $x = 3, 4, 5$.

The domain of $f$ is the set of all these integers.
Domain: $\boxed{\{-1, 0, 1, 2, 3, 4, 5\}}$

For $x = -1$: $f(-1) = 2(-1) + 1 = -2 + 1 = -1$.
For $x = 0$: $f(0) = -3(0) + 1 = 1$.
For $x = 1$: $f(1) = -3(1) + 1 = -3 + 1 = -2$.
For $x = 2$: $f(2) = -3(2) + 1 = -6 + 1 = -5$.
For $x = 3$: $f(3) = -2(3) - 1 = -6 - 1 = -7$.
For $x = 4$: $f(4) = -2(4) - 1 = -8 - 1 = -9$.
For $x = 5$: $f(5) = -2(5) - 1 = -10 - 1 = -11$.

The range of $f$ is the set of these values.
Range: $\boxed{\{-11, -9, -7, -5, -2, -1, 1\}}$

From part (c), the calculated range values are $\{-1, 1, -2, -5, -7, -9, -11\}$. All these values are unique. Each input $x$ from the domain maps to a unique output $f(x)$.
Therefore, the function is one-to-one.

The problem states that the function is defined on the set of integers. Typically, if the codomain is not explicitly stated, it is assumed to be the set of all integers, $\mathbb{Z}$.
The range of $f$ is $\{-11, -9, -7, -5, -2, -1, 1\}$. This set is a proper subset of the set of all integers $\mathbb{Z}$ (e.g., 0 is an integer but not in the range).
Since the range is not equal to the codomain ($\mathbb{Z}$), the function is not onto.