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

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The discussion centers on the analysis of the function s(t) = t^3 - 3t to determine the intervals where the object is speeding up. The first derivative, s'(t) = 3t^2 - 6, reveals critical points at t = ±√2. The second derivative, s''(t) = 6t, indicates that the object is speeding up when t > 0, as this is when acceleration is positive. A correction was noted regarding the derivative of 3t, which is 3, not 6, but this does not affect the intervals where the object is speeding up.

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UrbanXrisis
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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?
 
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UrbanXrisis said:
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. :confused:

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

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