Analyzing s(t)=t^3-3t: When is It Speeding Up?

[UrbanXrisis]

s(t)=t^3-3t

which interval(s) of t is this object speeding up?

s'(t)=3x^2-6
0=3x^2-6
x=sqrt(2) and

...+...-...-...+
inf...-sqrt(2)...0...sqrt(2) ...inf

s''(t)=6x
0=6x
x=0

...-...+...
inf...0...inf


inc (-sqrt(2),0) V (sqrt(2),inf)

correct?

 

[Galileo]

s(t)=t^3-3t

which interval(s) of t is this object speeding up?

s'(t)=3x^2-6
0=3x^2-6
x=sqrt(2) and

Why did you change your t's to x's.

The object is speeding up when s''(t) is positive.
6t>0 iff t>0

 

[BobG]

You made one mistake. Derivative of 3t is 3, not 6. That changes the interval where the velocity is positive or negative. Fortunately, that doesn't change your answer as to when acceleration is negative or positive (both 3 and 6 have a derivative of 0).

What's the question actually asking. You asked when the object was speeding up (i.e. - acceleration was positive). Your final answer gave an incorrect answer as to when the object was moving forward (velocity positive).