The second derivative test reveals that $f ( x ) = x ^ { 4 } - 4 x ^ { 2 } + 1$ has
A) a relative minimum only.
B) a relative maximum and two relative minima.
C) neither a relative maximum nor a relative minimum.
D) a relative maximum only.

Question

Quizplus · Accepted Answer

To determine the nature of critical points using the second derivative test, we first find the first and second derivatives of $f(x) = x^4 - 4x^2 + 1$. The first derivative $f'(x) = 4x^3 - 8x$, and setting $f'(x) = 0$ gives the critical points $x = 0, \pm \sqrt{2}$. The second derivative $f''(x) = 12x^2 - 8$. Evaluating $f''(x)$ at the critical points: $f''(0) = -8 < 0$ indicates a relative maximum at $x = 0$, and $f''(\pm \sqrt{2}) = 16 > 0$ indicates relative minima at $x = \pm \sqrt{2}$.

The Second Derivative Test Reveals That f(x)=x4−4x2+1f ( x ) = x ^ { 4 } - 4 x ^ { 2 } + 1f(x)=x4−4x2+1

The Second Derivative Test Reveals That $f ( x ) = x ^ { 4 } - 4 x ^ { 2 } + 1$