Unit 5, one level down AB & BC

The Fundamental Theorem gives $g'(x) = f(x)$, and differentiating again gives $g''(x) = f'(x)$. Those two equations are the entire topic. Substitute them into the Unit 5 rules and read off:

1. $g$ is increasing where $g' > 0$, so where $f$ is above the axis.
2. $g$ is decreasing where $f$ is below the axis.
3. $g$ has a critical point where $f$ crosses the axis: a maximum where $f$ goes from positive to negative, a minimum where it goes from negative to positive.
4. $g$ is concave up where $g'' > 0$, so where $f$ is increasing.
5. $g$ has an inflection point where $f$ turns around, at a maximum or minimum of $f$.

Every one of these reads a sign or a turn of $f$. None of them reads the size of $f$, and that is where the errors come from. "$f$ is large here, so $g$ must peak here" confuses fast with far. A large positive $f$ means $g$ is climbing steeply, which is a reason for $g$ to keep going up, not to stop.

The same slot confusion produces "$g$ has an inflection point where $f$ crosses zero". Zeros of $f$ are $g$'s critical points, because $f$ is $g'$. Inflection points of $g$ need $g'' = f'$ to change sign, which happens where $f$ has a peak or a valley.

Let the graph of $f$ on $[0, 10]$ consist of straight segments through $(0,0)$, $(2,4)$, $(4,0)$, $(6,-3)$, $(8,0)$ and $(10,4)$, and let $g(x) = \int_{0}^{x} f(t)\,dt$. The three regions have areas $8$, $6$ and $4$, with the middle one below the axis.

Direction. $f$ is positive on $(0, 4)$, negative on $(4, 8)$, positive on $(8, 10)$. So $g$ rises, falls, rises.

Critical points. $f$ crosses from positive to negative at $x = 4$, so $g$ has a maximum there. It crosses from negative to positive at $x = 8$, so $g$ has a minimum there. The zero at $x = 0$ is an endpoint and $f$ does not change sign across it.

Values. Accumulate the signed areas:

$$g(0) = 0, \quad g(4) = 8, \quad g(8) = 8 - 6 = 2, \quad g(10) = 2 + 4 = 6.$$

So the absolute maximum of $g$ on $[0, 10]$ is $8$ at $x = 4$, and the absolute minimum is $0$ at $x = 0$. Note that the candidates test from Topic 5.5 is doing the work here, endpoints included: $x = 8$ is a local minimum with value $2$, and it loses to the left endpoint.

Concavity. $f$ increases on $(0, 2)$, decreases on $(2, 6)$, increases on $(6, 10)$. So $g$ is concave up, then concave down, then concave up, with inflection points at $x = 2$ and $x = 6$. Neither of those is where $f$ is zero, and neither is where $g$ has an extremum.

Locating $g$'s features is a question about $f$'s sign; evaluating $g$ is a question about $f$'s area, and the area is signed. Regions below the axis subtract, which is Topic 6.1's rule and does not stop applying because the notation got shorter.

In the worked example, $g(10) = 6$. The total geometric area enclosed by $f$ and the axis is $8 + 6 + 4 = 18$, and that number answers a different question. Both appear on exams and the wording is the only signal.

Three consequences worth writing down.

1. $g$ can be negative even where $f$ is positive, if enough negative area came before. The current sign of $f$ says which way $g$ is moving, never where $g$ is.
2. $g$ returning to zero does not mean nothing happened. It means the positive and negative areas balanced, and the total area is the sum of two nonzero pieces.
3. Geometry does the arithmetic. AP graphs are built from segments, semicircles and quarter circles precisely so the areas come out exactly. A semicircle of radius $2$ contributes $\frac{1}{2}\pi(2)^{2} = 2\pi$, not $4\pi$ and not $8$.

These problems come with a lot of things to confuse, so it pays to name the object before computing anything.

1. A location or a value? "Where does $g$ attain its maximum" wants $x = 4$. "What is the maximum value of $g$" wants $8$. Topic 5.5's distinction, unchanged.
2. $g$, $f$, or $f'$? A question about the maximum of $g$ looks at where $f$ crosses zero. A question about the maximum of $f$ looks at the graph directly. They are almost never at the same place, and here they are at $x = 4$ and $x = 2$.
3. Net or total? Signed area for $g$'s value; unsigned for total area or total distance.

Justifications carry marks of their own, and the accepted form names the feature and where it is. "$g$ has a maximum at $x = 4$ because $f$ changes from positive to negative there" earns it. "Because the graph is highest there" does not, since the graph shown is $f$ and its height is $g'$.

One last check that catches sign errors cheaply: $g$ is increasing on $(0,4)$, so $g(4)$ must exceed $g(0)$. Any computed pair violating a direction you already established is arithmetic to redo, not a result to report.