Answer:
The maximum acceleration over that interval is 
. 
Step-by-step explanation:
The acceleration of this car is modelled as a function of the variable 
.
Notice that the interval of interest 
 is closed on both ends. In other words, this interval includes both endpoints: 
 and 
. Over this interval, the value of 
 might be maximized when 
 is at the following:
- One of the two endpoints of this interval, where 
 or 
. - A local maximum of 
, where 
 (first derivative of 
 is zero) and 
 (second derivative of 
 is smaller than zero.) 
Start by calculating the value of 
 at the two endpoints: 
.
.
Apply the power rule to find the first and second derivatives of 
:
.
.
Notice that both 
 and 
 are first derivatives of 
 over the interval 
. 
However, among these two zeros, only 
 ensures that the second derivative 
 is smaller than zero (that is: 
.) If the second derivative 
 is non-negative, that zero of 
 would either be an inflection point (if
) or a local minimum (if 
.) 
Therefore 
 would be the only local maximum over the interval 
.
Calculate the value of 
 at this local maximum:
.
Compare these three possible maximum values of 
 over the interval 
. Apparently, 
 would maximize the value of 
. That is: 
 gives the maximum value of 
 over the interval 
. 
However, note that the maximum over this interval exists because 
 is indeed part of the 
 interval. For example, the same 
 would have no maximum over the interval 
 (which does not include 
.)