The acceleration, in meters per second per second, of a race car is modeled by A(t)=t^3−15/2t^2+12t+10, where t is measured in seconds. What is the car’s maximum acceleration on the time interval 0≤t≤6 ?

Notice that the interval of interest [tex]0 \le t \le 6[/tex] is closed on both ends. In other words, this interval includes both endpoints: [tex]t = 0[/tex] and [tex]t= 6[/tex]. Over this interval, the value of [tex]A(t)[/tex] might be maximized when [tex]t[/tex] is at the following:

One of the two endpoints of this interval, where [tex]t = 0[/tex] or [tex]t = 6[/tex].
A local maximum of [tex]A(t)[/tex], where [tex]A^\prime(t) = 0[/tex] (first derivative of [tex]A(t)\![/tex] is zero) and [tex]A^{\prime\prime}(t) <0[/tex] (second derivative of [tex]\! A(t)[/tex] is smaller than zero.)

Start by calculating the value of [tex]A(t)[/tex] at the two endpoints:

[tex]A(0) = 10[/tex].
[tex]A(6) = 28[/tex].

Apply the power rule to find the first and second derivatives of [tex]A(t)[/tex]:

[tex]\begin{aligned} A^{\prime}(t) &= 3\, t^{2} - 15\, t + 12 \\ &= 3\, (t - 1) \, (t + 4)\end{aligned}[/tex].

[tex]\displaystyle A^{\prime\prime}(t) = 6\, t - 15[/tex].

Notice that both [tex]t = 1[/tex] and [tex]t = 4[/tex] are first derivatives of [tex]A^{\prime}(t)[/tex] over the interval [tex]0 \le t \le 6[/tex].

However, among these two zeros, only [tex]t = 1\![/tex] ensures that the second derivative [tex]A^{\prime\prime}(t)[/tex] is smaller than zero (that is: [tex]A^{\prime\prime}(1) < 0[/tex].) If the second derivative [tex]A^{\prime\prime}(t)\![/tex] is non-negative, that zero of [tex]A^{\prime}(t)[/tex] would either be an inflection point (if[tex]A^{\prime\prime}(t) = 0[/tex]) or a local minimum (if [tex]A^{\prime\prime}(t) > 0[/tex].)

Therefore [tex]\! t = 1[/tex] would be the only local maximum over the interval [tex]0 \le t \le 6\![/tex].

Calculate the value of [tex]A(t)[/tex] at this local maximum:

[tex]A(1) = 15.5[/tex].

Compare these three possible maximum values of [tex]A(t)[/tex] over the interval [tex]0 \le t \le 6[/tex]. Apparently, [tex]t = 6[/tex] would maximize the value of [tex]A(t)\![/tex]. That is: [tex]A(6) = 28[/tex] gives the maximum value of [tex]\! A(t)[/tex] over the interval [tex]0 \le t \le 6\![/tex].

However, note that the maximum over this interval exists because [tex]t = 6\![/tex] is indeed part of the [tex]0 \le t \le 6[/tex] interval. For example, the same [tex]A(t)[/tex] would have no maximum over the interval [tex]0 \le t < 6[/tex] (which does not include [tex]t = 6[/tex].)