:
f(x) = x^3 + x
find its velocity at a time unit t = 5
I might be wrong here...but I think we need to know something about the time from the information given in the question.
Does "x" stand for time in the formula? Or are you given another equation that tells us the x or y coordinate at any point in time?
If this is the position function, the velocity function is the derivative of this
f'(x) = 3x^2 + 1
Assuming that you meant x = 5, we have that the velocity at that time =
f'(5) = 3(5)^2 + 1 = 76 [ units per second ?? ]
