Determine a field on the real line that has the following characteristics: Determine a field on the real line that has the following characteristics:
a) At point $-1$ the field is canceled and is a singular repulsive point.
b) In the origin the field is canceled and it is a singular attractor point.
c) At point $1$ the field is canceled and is a singular repulsive point.
I just can think of fields for each of the subsections. Any hint of what could be a field that accomplish all tree? 
 A: Consider the vector field
$X(x) = x^3 - x, \; x \in \Bbb R; \tag 1$
we have
$X(-1) = X(0) = X(1) = 0, \tag 2$
that is, each of $-1$, $0$, $1$ is a critical point, or as our OP Sofia Contreras phrases it, a point where "the field is cancelled.". Note that
$x < -1 \Longrightarrow X(x) < 0, \tag 3$
$-1 < x < 0 \Longrightarrow X(x) > 0, \tag 4$
$0 < x < 1 \Longrightarrow X(x) < 0, \tag 5$
$x > 1 \Longrightarrow X(x) > 0; \tag 6$
it follows that the motion of a particle moving under the influence of the vector field $X(x)$, that is, the motion of which satisfies the ordinary differential equation
$\dot x = X(x) = x^3 - x, \tag 7$
is in the direction of decreasing $x$ under conditions (3) and (5), and in the direction of increasing $x$ in cases (4) and (6), since the direction of motion is given by the sign of $\dot x$, which indicates whether $x(t)$ becomes greater or lesser with $t$; these observations are sufficient to infer that $-1$ and $1$ are repulsive points, whereas $0$ is an attractive fixed point of the flow of $X(x)$
We may also analyze the stability/instability of the three points $-1$, $0$, $1$ by examining the first derivative of $X(x)$ at each of these $x$-values; indeed we find
$X'(x) = 3x^2 - 1, \tag 8$
whence
$X'(-1) = X'(1) = 2, \tag 9$
and
$X'(0) = -1; \tag{10}$
since
$X'(-1) = X'(1) > 0, \tag{11}$
the standard theory shows that $-1$ and $1$ are repellors, whilst
$X'(0) < 0 \tag{12}$
indicates $0$ is an attractive, in accord with our previous discussion.
