Given that (y = x(x + 1)^{2}), calculate the maximum value of y.
-
A.
-2 -
B.
0 -
C.
1 -
D.
2
Correct Answer: Option B
Explanation
To find the maximum value, we can use the second derivative test where, given (f(x)), the second derivative < 0, makes it a maximum value.
(x(x + 1)^{2} = x(x^{2} + 2x + 1) = x^{3} + 2x^{2} + x)
(frac{mathrm d y}{mathrm d x} = 3x^{2} + 4x + 1 = 0)
Solving, we have ( x = frac{-1}{3}) or (-1).
(frac{mathrm d^{2} y}{mathrm d x^{2}} = 6x + 4)
When (x = frac{-1}{3}, frac{mathrm d^{2} y}{mathrm d x^{2}} = 2 > 0)
When (x = -1, frac{mathrm d^{2} y}{mathrm d x^{2}} = -2 < 0)
At maximum value of x being -1, (y = -1(-1 + 1)^{2} = 0)