Find the relative minimum point(s) of $f ( x ) = \frac { x ^ { 4 } } { 4 } - x ^ { 3 } - 5 x ^ { 2 } - 10$
A) (-2, f(-2)) and (0, f(0))
B) (2, f(2)) and (-5, f(-5))
C) (-2, f(-2)) and (5, f(5))
D) (0, f(0))
E) none of these

Question

Quizplus · Accepted Answer

To find the relative minimum, we first find the derivative $f'(x) = x^3 - 3x^2 - 10x$ and set it to zero to find critical points. Solving $x^3 - 3x^2 - 10x = 0$ gives $x(x^2 - 3x - 10) = 0$, leading to $x = 0$ or solving $x^2 - 3x - 10 = 0$ gives $x = -2$ and $x = 5$. Using the second derivative test, $f''(x) = 3x^2 - 6x - 10$, we find $f''(-2) > 0$ and $f''(5) > 0$, indicating relative minima at $x = -2$ and $x = 5$.

Find the Relative Minimum Point(s) Of f(x)=x44−x3−5x2−10f ( x ) = \frac { x ^ { 4 } } { 4 } - x ^ { 3 } - 5 x ^ { 2 } - 10f(x)=4x4​−x3−5x2−10

Find the Relative Minimum Point(s) Of $f ( x ) = \frac { x ^ { 4 } } { 4 } - x ^ { 3 } - 5 x ^ { 2 } - 10$