9514 1404 393
Answer:
Step-by-step explanation:
The extrema will be at the ends of the interval or at a critical point within the interval.
The derivative of the function is ...
f'(x) = 3x² -4x -4 = (x -2)(3x +2)
It is zero at x=-2/3 and at x=2. Only the latter critical point is in the interval. Since the leading coefficient of this cubic is positive, the right-most critical point is a local minimum. The coordinates of interest in this interval are ...
f(0) = 2
f(2) = ((2 -2)(2) -4)(2) +2 = -8 +2 = -6
f(3) = ((3 -2)(3) -4)(3) +2 = -3 +2 = -1
The absolute maximum on the interval is f(0) = 2.
The absolute minimum on the interval is f(2) = -6.
S(v)=<span>∛v
if v = 64 then
s(v) </span><span>∛64 = 4
answer
</span><span>b. s >= 4</span>
If you plot the coordinates on a piece of graph paper your answer would come out too (1,-8)
The final steps is to review your results.
This means that you have to check if the results meet the original requirements or statements.
Ideally, you should try to solve the same problem by a second method, and/or you should substitute the results into the given relationships given in the problem statement to chek coherence of your results.
Then the answer is option b.