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  "body": "\n## Proof\n\nWe first treat the case $f'(a) \u003c y \u003c f'(b)$, and find an interior minimum of a shifted function.\n\n1. Assume $f'(a) \u003c y \u003c f'(b)$. Here $f'(a)$ is the right derivative and $f'(b)$ is the left derivative, as [[one-sided derivatives|0HXEX0DPRA]].\n2. Define $g \\colon [a, b] \\to \\R$ by $g(x) = f(x) - y x$. Then $g$ is differentiable on $(a, b)$ with $g'(x) = f'(x) - y$.\n3. In the same way, $g$ has right derivative $g'(a) = f'(a) - y \u003c 0$ and left derivative $g'(b) = f'(b) - y \u003e 0$.\n4. The function $g$ is [[continuous|07ZMX4ANJS]] at each point of $(a, b)$, since it is differentiable there.\n5. For $a \u003c t \\le b$ write $g(t) - g(a) = (t - a) q(t)$, where $q(t) = (g(t) - g(a))/(t - a)$. As $t \\to a^+$, $q(t) \\to g'(a)$ and $t - a \\to 0$. So $g(t) \\to g(a)$.\n6. In the same way $g(t) \\to g(b)$ as $t \\to b^-$. With step 4, $g$ is [[continuous|0P27W9RW0G]] on $[a, b]$.\n7. By the [[extreme value theorem|0WS1A8E2P9]], there is $c \\in [a, b]$ with $g(c) \\le g(t)$ for all $t \\in [a, b]$.\n8. Since $q(t) \\to g'(a) \u003c 0$ as $t \\to a^+$, the definition of the [[right limit|0909PWZR62]] gives $\\delta \u003e 0$ with $q(t) \u003c 0$ for $a \u003c t \u003c a + \\delta$, $t \\le b$.\n9. For such $t$, $g(t) - g(a) = (t - a) q(t) \u003c 0$. So $g(t) \u003c g(a)$, and $c \\ne a$.\n10. Likewise, the quotient $(g(t) - g(b))/(t - b)$ tends to $g'(b) \u003e 0$ as $t \\to b^-$. So it is positive for $t \u003c b$ close to $b$.\n11. For such $t$ we have $t - b \u003c 0$, so $g(t) - g(b) \u003c 0$. Thus $g(t) \u003c g(b)$, and $c \\ne b$.\n12. So $c \\in (a, b)$. The restriction of $g$ to the open interval $(a, b)$ has a [[local minimum|0NS8BP96QX]] at $c$, and it is differentiable at $c$.\n13. By [[Fermat's theorem|0PCVBRV9VJ]], $g'(c) = 0$. So $f'(c) = y$.\n14. Now assume $f'(a) \u003e y \u003e f'(b)$. The function $-f$ satisfies the hypotheses, with $(-f)'(a) \u003c -y \u003c (-f)'(b)$.\n15. By steps 1 to 13 applied to $-f$ and $-y$, there is $c \\in (a, b)$ with $-f'(c) = -y$. So $f'(c) = y$.\n\n## Idea\n\nSubtracting the line $y x$ turns the problem into finding a zero of a derivative. The signs of $g'$ at the ends give values of $g$ below $g(a)$ just right of $a$. They also give values below $g(b)$ just left of $b$. So the minimum of $g$ is inside, and Fermat's theorem makes the derivative zero there.\n",
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