<span>The function can only change from increasing to decreasing, and visa-versa at those points where the slope of the function is 0. And the slope of the function is determined by the first derivative of the function. So let's calculate the first derivative.
f(t) = (t^3 + 3t^2)^3
f'(t) = d/dt[ (t^3 + 3t^2)^3 ]
f'(t) = 3(t^3 + 3t^2)^2 * d/dt[ t^3 + 3t^2 ]
f'(t) = 3(d/dt[ t^3 ] + 3 * d/dt[ t^2 ])(t^3 + 3t^2)^2
f'(t) = 3(3t^2 + 3 * 2t)(t^3 + 3t^2)^2
f'(t) = 3(3t^2 + 6t)(t^3 + 3t^2)^2
Simplify
f'(t) = 3(3t^2 + 6t)(t^3 + 3t^2)^2
f'(t) = 3 * 3t(t + 2)(t^3 + 3t^2)^2
f'(t) = 9t(t + 2)(t^2(t + 3))^2
f'(t) = 9t(t + 2)t^4(t + 3)^2
f'(t) = 9t^5(t + 2)(t + 3)^2
And looking at the function, it becomes obvious that the roots (or inflection points) are at t = 0, t = -2, and t = -3.
Now the only places where f(t) can switch directions is at those 3 inflection points. And at exactly those inflection points the curve is neither increasing, nor decreasing.
If the slope of the function is positive, then its value is increasing, and if the slope is negative, then the function is decreasing. So all we need to do is calculate the value of the first derivative for any value between each inflection point plus one value smaller than the smallest inflection point and another value higher than the highest inflection point.
Range from [-infinity, -3)
f'(-4) = 18432
Since the value is positive, the function is increasing from [-infinity, -3)
Range from (-3, -2)
f'(-2.5) = 30.51758
Since the value is positive, the function is increasing from (-3, -2)
Range from (-2, 0)
f(-1) = -36
Since the value is negative, the function is decreasing from (-2, 0)
Range from (0, infinity)
f(1) = 64
Since the value is positive, the function is increasing from (0, +infinity)
To summarize:
increasing from [-infinity, -3)
increasing from (-3,-2)
decreasing from (-2,0)
increasing from (0,infinity]</span>
