cube root (x^3) = cube root (1) x = 1. This is our critical point
2) Points where f ' does not exist.
We know f ' (x) = (2-2x^3) / (x^3+2)^2
You cannot divide by 0 ever so f ' does not exist where the denominator equals 0
(x^3 + 2)^2 = 0. Take square roots of both sides sqrt((x^3+2)^2) = sqrt(0) x^3 + 2 = 0. Add -2 to both sides. -2 + x^3 + 2 = -2 + 0 x^3 = -2. Take cube roots of both sides. cube root (x^3) = cube root (-2) x = cube root (-2). This is where f ' doesnt exist. However, it is not in our interval [0,2]. Thus, we can ignore this point.
3) End points of the domain.
The domain was clearly stated as [0, 2]. The end points are 0 and 2.
Therefore, our only options are: 0, 1, 2.
Check the intervals
[0, 1] and [1, 2]. Pick an x value in each interval and determine its sign.
In [0, 1]. Check 1/2. f ' (1/2) = (7/4) / (17/8)^2 which is positive.
In [1, 2]. Check 3/2. f ' (3/2) = (-19/4) / (43/8)^2 which is negative.
Therefore, f is increasing on [0, 1] and decreasing on [1, 2] and 1 is a local maximum.
f (0) = 0 f (1) = 1/3 f (2) = 1/5
Therefore, 0 is a local and absoulte minimum. 1 is a local and absolute maximum. Finally, 2 is a local minimum. </span><span>Thunderclan89</span>
150 is the amount of interest earned for one year because the formula of PRT is, in this case 5000 times 0.03 times one which gets you the answer of 150.