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

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 $f(x) = x^4 - 4x^2 + 1$.First derivative: $f'(x) = 4x^3 - 8x$.Setting $f'(x) = 0$ gives the critical points: $x = 0, x = \sqrt{2}, x = -\sqrt{2}$.Second derivative: $f''(x) = 12x^2 - 8$.Evaluating the second derivative at each critical point:- At $x = 0$, $f''(0) = -8$, which indicates a concave down, hence a relative maximum.- At $x = \sqrt{2}$, $f''(\sqrt{2}) = 12(\sqrt{2})^2 - 8 = 16$, which indicates a concave up, hence a relative minimum.- At $x = -\sqrt{2}$, $f''(-\sqrt{2}) = 16$, which also indicates a concave up, hence another relative minimum.Therefore, the function has a relative maximum at $x = 0$ and two relative minima at $x = \sqrt{2}$ and $x = -\sqrt{2}$.

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 $f ( x ) = x ^ { 4 } - 4 x ^ { 2 } + 1$