Final answer:
To determine the points where the tangent line is horizontal, we would need to find the derivative f'(x) of the function f(x) and solve for f'(x) = 0. However, the specific function is not provided in the question. With provided kinematics context, horizontal tangent lines correspond to constant velocities indicating zero acceleration.
Step-by-step explanation:
To find the derivative f'(x) and the values of x where the tangent line is horizontal, we need to understand that a horizontal tangent line occurs when the slope (derivative) of the function at that point is zero. We do not have the specific function f(x) provided in the question, but if we had it, we would take its derivative and solve for f'(x) = 0 to find the x-values where the tangent is horizontal. As for the given information, it seems to be part of a Physics problem involving position, velocity, and time, which hints at kinematics, but without a specific function or context, we cannot directly apply this information to finding a derivative.
For example, if given a kinematic equation such as s(t) = s_0 + v_0t + (1/2)at^2, the velocity would be the first derivative v(t) = s'(t) and the acceleration would be the second derivative a(t) = v'(t). For a horizontal velocity graph over a time interval, this implies that acceleration (slope) is zero, and hence, the velocity is constant during that interval.
For a graph represented as a horizontal line over an interval [0, 20] for f(x), it means f'(x) is 0 for all x in that interval, indicating that the tangent line is horizontal everywhere on that interval.