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

Question

Quizplus · Accepted Answer

The second derivative test involves finding the first and second derivatives of the function, then using these to determine the nature of critical points. For $f(x) = x^4 - 4x^2 + 1$, the first derivative $f'(x) = 4x^3 - 8x$. Setting $f'(x) = 0$ gives critical points at $x = 0, \pm \sqrt{2}$. The second derivative $f''(x) = 12x^2 - 8$. Evaluating $f''(x)$ at the critical points, we find $f''(0) = -8 < 0$ (indicating a relative maximum at $x = 0$), and $f''(\pm \sqrt{2}) = 16 > 0$ (indicating 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 $f ( x ) = x ^ { 4 } - 4 x ^ { 2 } + 1$