for this position function discuss the motion along a horizontal line.
Discuss the instantaneous velocity, acceleration, intervals when the particle in motion is moving to the right and left, intervals when the particle in motion is speeding up and slowing down.
how would I do this? Im also suppose to sketch the motion of the particle at its appropriate position above the line of motion.
f(x) = x^3/3-2x^2 + 3x +8
Thanks
(x^3/3) - 2x^2 + 3x +8
is it ok to multiply this whole function by 3 to get
x^3 -6x^2 + 9x +24? and then do the first and second derivative?
the (x^3/3) is giving me problems, can someone be kind enough to tell me how to find the derivative of this fraction?
THanks
No, that is now a different function. You could factor out the 1/3 and write it as (1/3)(x^3- 6x^2+ 9x+ 24) and then differentiate it. You need to know that the derivative of af(x) (a times the function f(x)) is af'(x) (a times the derivative of f).
The derivative of x^3 is 3x^2 so the derivative of x^3/3= (3x^2)/3= x^2.the (x^3/3) is giving me problems, can someone be kind enough to tell me how to find the derivative of this fraction?
THanks
alright thanks, now the next step, original problem is
f(x) = x^3/3-2x^2 + 3x +8
first derivative: x^2 - 4x + 3 .... x = 3 x = 1
second derivative: 2x - 4 .... x = 2
s(0) = 8
s(1) = 9 1/3
s(2) = 8 2/3
s(3) = 8
so now
time velocity acceleration speed
0 < x <1
1< x < 2
2< x < 3
x> 3
how would I find out the velocity acceleration speed?