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  "body": "\n## Proof\n\nWe apply the monotonicity test on each side of $c$.\n\n1. Assume $f'(x) \\le 0$ for $c - \\delta \u003c x \u003c c$ and $f'(x) \\ge 0$ for $c \u003c x \u003c c + \\delta$.\n2. The function $f$ is [[continuous|07ZMX4ANJS]] at each $x \\ne c$, since it is differentiable there. It is continuous at $c$ by hypothesis.\n3. Let $x \\in (c, c + \\delta)$. Then $f$ is continuous on $[c, x]$ and differentiable on $(c, x)$ with $f' \\ge 0$ there.\n4. By the [[nonnegative derivative test|026BJBQNGJ]] on the interval $[c, x]$, $f$ is increasing there. So $f(c) \\le f(x)$.\n5. Let $x \\in (c - \\delta, c)$. The function $-f$ is continuous on $[x, c]$ and differentiable on $(x, c)$.\n6. By the [[constant multiple rule|0Z9JB71BS0]], $(-f)' = -f' \\ge 0$ on $(x, c)$.\n7. By the [[nonnegative derivative test|026BJBQNGJ]], $-f$ is increasing on $[x, c]$. So $-f(x) \\le -f(c)$, that is, $f(c) \\le f(x)$.\n8. Steps 4 and 7 give $f(x) \\ge f(c)$ for all $x$ with $|x - c| \u003c \\delta$. So $f$ has a [[local minimum|0NS8BP96QX]] at $c$.\n9. Now assume the reversed inequalities: $f' \\ge 0$ on $(c - \\delta, c)$ and $f' \\le 0$ on $(c, c + \\delta)$.\n10. Then $-f$ satisfies the hypotheses of step 1, since $(-f)' = -f'$. By step 8, $-f$ has a local minimum at $c$.\n11. So $-f(x) \\ge -f(c)$ near $c$, that is, $f(x) \\le f(c)$. Thus $f$ has a [[local maximum|00999TYPXB]] at $c$.\n\n## Idea\n\nA function that does not increase up to $c$ and does not decrease after $c$ is lowest at $c$. The mean value theorem, through the monotonicity test, turns the sign conditions on $f'$ into these monotonicity facts. Continuity at $c$ lets the test include the point $c$ itself.\n",
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