Inferring f' Behavior from the Sign of f' and f'' on Each Side

Many konkur problems hand you only the signs of $f'$ and $f''$ on the two sides of a point $a$ and ask what $f$ — or its one-sided derivatives at $a$ — must look like. The whole trick is to read two layers of monotonicity off those two signs, then use boundedness to pin down a one-sided limit of $f'$.

The two-layer sign reading

Think of $f'$ as a function in its own right. On an interval $I$:

The first sign controls $f$; the second sign controls $f'$. So '$f' < 0 < f''$' means $f$ is decreasing while its slope is rising toward $0$ (a decreasing-but-flattening, convex piece), and '$f'' < 0 < f'$' means $f$ is increasing while its slope is falling (an increasing-but-flattening, concave piece).

Squeezing out a one-sided limit of f'

Apply the Monotone Convergence Theorem to $f'$, not to $f$. Suppose on $(b, a)$ we have $f'' > 0$ (so $f'$ is increasing) and $f' < 0$ (so $f'$ is bounded above by $0$). An increasing function bounded above has a finite limit at the right end:

$L_- = \lim_{x \to a^-} f'(x) = \sup_{x<a} f'(x) \le 0.$

The bound that makes $f'$ monotone comes from $f''$; the bound that makes it bounded comes from $f'$. Mirror image on the right: $f'$ decreasing ($f''<0$) and positive ($f'>0$) gives a finite (or $+\infty$) limit

$R_+ = \lim_{x \to a^+} f'(x) = \sup_{x>a} f'(x) > 0,$

and because $f'$ is strictly decreasing the inequality $R_+ > f'(x) > 0$ is strict for every interior $x$.

From limit of f' to the one-sided derivative

The missing link is the derivative-limit theorem (a clean corollary of the Mean Value Theorem):

If $f$ is continuous at $a$ and differentiable just to one side, and $\lim_{x\to a^-} f'(x) = L$ exists (finite or $\pm\infty$), then the left-hand derivative equals that limit: $f'_-(a) = L$.

So a derivative cannot have a jump or oscillatory discontinuity from a one-sided monotone limit — the one-sided derivative at $a$ is that one-sided limit. This is exactly what lets you conclude things like '$f'_-(a) \le 0 \le f'_+(a)$, so $a$ is a local-min-type corner', or compare the two one-sided slopes to decide differentiability at $a$.

How to apply it (4 moves)

1. Translate signs. Convert each '$f'$ sign' to monotonicity of $f$, and each '$f''$ sign' to monotonicity of $f'$.
2. Bound $f'$. Use the sign of $f'$ itself ($f'<0$ or $f'>0$) as the bound $0$ that the monotone $f'$ approaches.
3. Take the one-sided limit. Monotone $+$ bounded $\Rightarrow$ $\lim_{x\to a^\pm} f'(x)$ exists (or is $\pm\infty$); it equals the relevant sup/inf.
4. Identify the one-sided derivative. By the derivative-limit theorem this limit is $f'_\pm(a)$; compare $f'_-(a)$ and $f'_+(a)$ to settle differentiability or the corner type at $a$.

Mini-example

Let $f$ be continuous on $\mathbb{R}$ with $f' < 0 < f''$ for $x < 0$ and $f'' < 0 < f'$ for $x > 0$. Left of $0$: $f'$ increasing and $< 0$, so $f'_-(0) = \lim_{x\to0^-}f'(x) = \sup_{x<0}f'(x) \le 0$. Right of $0$: $f'$ decreasing and $> 0$, so $f'_+(0) = \lim_{x\to0^+}f'(x) = \sup_{x>0}f'(x) > 0$. Hence $f'_-(0) \le 0 < f'_+(0)$: the two one-sided slopes differ, so $f$ need not be differentiable at $0$, and $0$ is a downward corner (local minimum of $f$). A concrete witness is $f(x) = \sqrt{|x|+1} - 1$ near $0$-style functions, or simply any V-shaped-but-curved profile matching the signs.

Exam trick: Keep the two derivatives in separate lanes — the sign of $f'$ tells you about $f$ (rising/falling), the sign of $f''$ tells you about $f'$ (rising/falling). The single most useful consequence is that '$f'$ monotone and one-signed near $a$' forces $\lim_{x\to a^\pm} f'$ to exist and to equal the one-sided derivative $f'_\pm(a)$ — so you may read those slopes straight off the supremum/infimum, never needing a formula for $f$.