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  "body": "\n## Proof\n\nWe read off the sign of $f'$ near $c$ from $f''(c)$, then use strict monotonicity on each side.\n\n1. Let $I$ be the open interval around $c$ on which $f$ is differentiable. Assume first that $f''(c) \u003e 0$.\n2. By the definition of the [[second derivative|0Y9D2FJRQ4]] and of the [[derivative|0F1WR9C83K]], $(f'(x) - f'(c))/(x - c) \\to f''(c)$ as $x \\to c$.\n3. Since $f'(c) = 0$, this says $f'(x)/(x - c) \\to f''(c)$ as $x \\to c$.\n4. Apply the definition of the [[limit|0TXTSEMHR5]] with the open interval $V = (0, 2 f''(c))$, which contains $f''(c)$.\n5. This gives an open interval $U$ containing $c$ with $f'(x)/(x - c) \\in V$ for $x \\in U \\cap I$, $x \\ne c$. Choose $\\eta \u003e 0$ with $(c - \\eta, c + \\eta) \\subseteq U \\cap I$. Then $f'(x)/(x - c) \u003e 0$ for $0 \u003c |x - c| \u003c \\eta$.\n6. So $f'(x) \u003c 0$ for $c - \\eta \u003c x \u003c c$, and $f'(x) \u003e 0$ for $c \u003c x \u003c c + \\eta$.\n7. The function $f$ is [[continuous|07ZMX4ANJS]] on $I$, since it is differentiable on $I$.\n8. Let $x \\in (c, c + \\eta)$. On $[c, x]$, $f$ is continuous, and $f' \u003e 0$ on $(c, x)$.\n9. By the [[positive derivative test|07J5X2T1ZX]], $f$ is strictly increasing on $[c, x]$. So $f(c) \u003c f(x)$.\n10. Let $x \\in (c - \\eta, c)$. On $[x, c]$, $-f$ is continuous, and $(-f)' = -f' \u003e 0$ on $(x, c)$ by the [[constant multiple rule|0Z9JB71BS0]].\n11. By the [[positive derivative test|07J5X2T1ZX]], $-f$ is strictly increasing on $[x, c]$. So $-f(x) \u003c -f(c)$, that is, $f(c) \u003c f(x)$.\n12. By steps 9 and 11, $f(x) \u003e f(c)$ for all $x \\ne c$ with $|x - c| \u003c \\eta$. So $f$ has a strict [[local minimum|0NS8BP96QX]] at $c$.\n13. Now assume $f''(c) \u003c 0$. Then $-f$ is differentiable on $I$, $(-f)'(c) = 0$, and $(-f)''(c) = -f''(c) \u003e 0$.\n14. By step 12, $-f$ has a strict local minimum at $c$. So $f(x) \u003c f(c)$ for $x \\ne c$ near $c$, a strict [[local maximum|00999TYPXB]].\n\n## Idea\n\nSince $f'(c) = 0$, the second derivative is the limit of $f'(x)/(x - c)$. If it is positive, $f'$ has the sign of $x - c$ near $c$. So $f$ strictly falls before $c$ and strictly rises after $c$, which makes $c$ a strict minimum.\n",
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