Find the x coordinates of all relative extreme points of $f ( x ) = \frac { 1 } { 2 } x ^ { 4 } - \frac { 2 } { 3 } x ^ { 3 } - 6 x ^ { 2 } - 100$
A) x = -2, 0, 3
B) x = 0
C) x = -3, 0, 2
D) x = -3, 2
E) x = -2, 3

Question

Quizplus · Accepted Answer

To find the x-coordinates of all relative extreme points, we first find the derivative of the function and set it equal to zero to find critical points. The derivative of $f(x) = \frac{1}{2}x^4 - \frac{2}{3}x^3 - 6x^2 - 100$ is $f'(x) = 2x^3 - 2x^2 - 12x$. Factoring out $2x$, we get $2x(x^2 - x - 6)$. Further factoring gives $2x(x-3)(x+2)$. Setting this equal to zero gives $x = 0, 3, -2$, which are the x-coordinates where the slope of the tangent is zero, indicating potential relative extrema.

Find the X Coordinates of All Relative Extreme Points Of f