1. Design a polynomial function of degree 4 that has exactly one local minimum at $x = 0$, no local maximum, and is concave up for all $x$.
- A.$f(x) = x^4 + x^2$
- B.$f(x) = x^4 - 4x^2$
- C.$f(x) = -x^4$
- D.$f(x) = x^4 - 2x^2 + 1$
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Answer: $f(x) = x^4 + x^2$
Check each candidate systematically: We need: (1) exactly one local minimum at $x = 0$, (2) no local maximum, (3) concave up for all $x$ (i.e., $f''(x) > 0$ everywhere, or $f''(x) \geq 0$ with equality only at isolated points). Analyze $f(x) = x^4 + x^2$: $f'(x) = 4x^3 + 2x = 2x(2x^2 + 1)$. Since $2x^2 + 1 > 0$ always, $f'(x) = 0$ only at $x = 0$. For $x < 0$: $f'(x) < 0$ (decreasing). For $x > 0$: $f'(x) > 0$ (increasing). So $x = 0$ is the only critical point and it is a local minimum. ✓ No local maximum. ✓ $f''(x) = 12x^2 + 2 \geq 2 > 0$ for all $x$, so $f$ is concave up everywhere. ✓ Why distractors fail: Option B: $f'(x) = 4x^3 - 8x = 4x(x^2-2)$, giving critical points at $x = 0, \pm\sqrt{2}$. Sign analysis shows $x = 0$ is a local max and $x = \pm\sqrt{2}$ are local minima — fails conditions 1 and 2. Option C: $f'(x) = -4x^3$, so $x = 0$ is a local max (not min), and $f''(x) = -12x^2 \leq 0$ (concave down) — fails all conditions. Option D: $f'(x) = 4x^3 - 4x = 4x(x^2-1)$, giving critical points at $x = 0, \pm 1$. At $x = 0$: local max; at $x = \pm 1$: local minima — fails conditions 1 and 2.