Find the x-values of all relative maxima of the given function. $y = \frac { 1 } { 3 } x ^ { 3 } - 4 x ^ { 2 } + 12 x + 9$
A) $x = 0$
B) $x = 6$
C) $x = 4$
D) $x = 2$
E)no relative maxima

Question

Quizplus · Accepted Answer

To find relative maxima, we first find the derivative: $y' = x^2 - 8x + 12$. Setting $y' = 0$, we solve $x^2 - 8x + 12 = 0$, which factors to $(x-6)(x-2) = 0$. Thus, $x = 6$ and $x = 2$. To determine which is a maximum, we use the second derivative test: $y'' = 2x - 8$. At $x = 2$, $y'' = -4$ (negative, indicating a maximum). At $x = 6$, $y'' = 4$ (positive, indicating a minimum). Therefore, the relative maximum occurs at $x = 2$.

Find the X-Values of All Relative Maxima of the Given y=13x3−4x2+12x+9y = \frac { 1 } { 3 } x ^ { 3 } - 4 x ^ { 2 } + 12 x + 9y=31​x3−4x2+12x+9

Find the X-Values of All Relative Maxima of the Given $y = \frac { 1 } { 3 } x ^ { 3 } - 4 x ^ { 2 } + 12 x + 9$