Calculus: particle farthest left when f(x) is at absolute min... why?

Question

On a given interval, why is the "particle farthest to the left" when the function is at its absolute minimum? Why not at the point at the negative-most end of the given interval (i.e. at x=0 on interval [0,2])?

The question given was: function of motion of particle f(x)=e^(-x)sin(x) for [0,2pi], at what point is the particle farthest to the left?

Answer 1

OK I tried to figure out what you mean.
I guess you are given the equation of the trajectory of a particle. And then you are asked to compute its absolute minimum to find some point which is farthest.
better write the problem as it is stated.
edit: i'm thinking that x is here the time, and f is the function of motion in rapport with time.
Now, I see this motion as an oscillation around the origin, it goes in positive direction and returns to 0 and goes to negative direction and goes back. You can think the time as a parameter along the curve. Think about an ant that walks on the curve of f, a curve that looks like a series of loops.

Alternatively, we can think about the usual way we graph the sinus along the x-axis. In this case, the "farthest to the left" is the lowest point on graph attained by f.