Which way it bends is a question for f″ AB & BC

On an interval, the following are equivalent:

1. $f$ is concave up.
2. $f'' > 0$ there.
3. $f'$ is increasing there.

Statements 2 and 3 are the same statement, because $f''$ is the derivative of $f'$ and Topic 5.3 already established that a positive derivative means increase. Statement 3 is the one worth keeping in your head, because it says what concavity means in plain terms: the slopes are getting bigger. They can be getting bigger while staying negative, which is a falling curve leveling out, and that is concave up.

Reverse every sign for concave down: $f'' < 0$, and $f'$ decreasing.

A fourth description is geometric and is what people are actually picturing. On an interval where $f$ is concave up, the graph lies above every one of its tangent lines, touching each only at the point of tangency. Concave down puts the graph below its tangents. This is the version that makes the linear approximation over-and-under rule from Topic 4.5 fall out immediately: a tangent line estimate under a concave up curve is an underestimate.

"Rising, so it curves upward" is the single most common sentence in this topic and it is not an argument. Four functions settle it:

1. $f(x) = e^{x}$: increasing, and concave up. $f' = e^{x} > 0$ and $f'' = e^{x} > 0$.
2. $f(x) = \sqrt{x}$: increasing, and concave down. $f' = \frac{1}{2\sqrt{x}} > 0$ and $f'' = -\frac{1}{4}x^{-3/2} < 0$. It climbs forever and flattens as it goes.
3. $f(x) = e^{-x}$: decreasing, and concave up. $f' = -e^{-x} < 0$ and $f'' = e^{-x} > 0$. It falls toward the axis, and the fall gets gentler.
4. $f(x) = -x^{2}$ for $x > 0$: decreasing, and concave down. $f' = -2x < 0$ and $f'' = -2 < 0$. It falls, and the fall gets steeper.

The pattern behind the four: direction is whether the slope is positive or negative, and concavity is whether the slope is growing or shrinking. Those are independent questions about the same number. The third example is the one to remember, because "decreasing" and "concave up" sound contradictory and describe the ordinary behaviour of every decaying quantity there is.

Identical procedure to Topic 5.3, one derivative higher:

1. Compute $f''$.
2. Mark every point where $f'' = 0$ or $f''$ fails to exist.
3. Mark every point not in the domain of $f$.
4. Test the sign of $f''$ in each resulting interval.

For $f(x) = x^{3} - 3x^{2} - 9x + 5$: $f'(x) = 3x^{2} - 6x - 9$ and $f''(x) = 6x - 6$, which is zero at $x = 1$, negative to its left, positive to its right. So $f$ is concave down on $(-\infty, 1)$ and concave up on $(1, \infty)$.

Compare that with the same function's direction chart, which was split at $-1$ and $3$. Different derivative, different marks, different answer. A chart built at $-1$ and $3$ answers a question about rising and falling and says nothing about bending.

Step 3 matters as much here as it did before. $f(x) = \frac{1}{x}$ has $f''(x) = \frac{2}{x^{3}}$, negative for $x < 0$ and positive for $x > 0$. The concavity does change across $x = 0$, and there is no inflection point there, because $x = 0$ is not in the domain and an inflection point has to be a point on the graph.

When the picture is a graph of $f'$, concavity is read from the shape of that curve rather than its position:

1. Where the graph of $f'$ is rising, $f'$ is increasing, so $f$ is concave up.
2. Where the graph of $f'$ is falling, $f$ is concave down.
3. Where the graph of $f'$ has a relative extremum, $f'$ changes from increasing to decreasing or back, so $f$ has an inflection point.

Above and below the axis is the direction question; rising and falling is the concavity question. On a graph of $f'$ those two readings use completely different features of the same curve, and swapping them is how a correct picture produces a wrong answer.

An inflection point of $f$ is a point on the graph of $f$ where the concavity changes. Two things follow that will be worked properly in Topic 5.7. It requires a genuine change, so $f'' = 0$ on its own proves nothing: $f(x) = x^{4}$ has $f''(0) = 0$ and is concave up on both sides. And it does not require $f''$ to exist: $f(x) = x^{1/3}$ has an inflection point at the origin, where $f''$ blows up and the sign flips.