Calculus AB · Analytical Applications of Differentiation · 14 min read · Updated 2026-05-10

Connecting f, f', f'' qualitatively — AP Calculus AB

AP Calculus AB · Analytical Applications of Differentiation · 14 min read

1. Relating f and f': Increasing/Decreasing and Critical Points ★★☆☆☆ ⏱ 3 min

The core relationship between $f$ and $f'$ comes from the definition of the derivative as the instantaneous slope of $f$ at any point $x$.

If $f'(x) > 0$ for all $x$ in an interval, $f$ is **increasing**: as $x$ increases, $f(x)$ increases
If $f'(x) < 0$ for all $x$ in an interval, $f$ is **decreasing**: as $x$ increases, $f(x)$ decreases

Exam tip: On all AP exam questions asking for intervals of increase/decrease, always use open intervals. AP graders never accept closed intervals for this question type.

2. Relating f and f'': Concavity and Inflection Points ★★☆☆☆ ⏱ 3 min

Concavity describes how the slope of $f$ changes as $x$ increases, so it is determined by the derivative of $f'$, which is $f''$.

If $f''(x) > 0$, $f'$ is increasing on the interval, so $f$ is **concave up** (shaped like a cup $\cup$)
If $f''(x) < 0$, $f'$ is decreasing on the interval, so $f$ is **concave down** (shaped like a cap $\cap$)

Exam tip: Always justify inflection points by explicitly stating that concavity (or the sign of $f''$) changes at the point. AP graders will deduct points if you only state that $f''(c)=0$ with no mention of a sign change.

3. Classifying Local Extrema Qualitatively ★★★☆☆ ⏱ 3 min

Once you have identified the critical points of $f$, you can classify them as local maxima, local minima, or neither using two qualitative tests:

**First Derivative Test**: Works for all critical points: $f'$ changes from positive to negative = local maximum; negative to positive = local minimum; no sign change = no extremum
**Second Derivative Test**: Only for critical points where $f'(c)=0$: $f''(c) < 0$ = local maximum; $f''(c) > 0$ = local minimum; $f''(c)=0$ = test is inconclusive, use the first derivative test

Exam tip: If an FRQ asks you to justify a local extremum, you must explicitly reference the test you use (e.g., "by the second derivative test, $f''(3) < 0$ so $x=3$ is a local maximum"). A bare conclusion earns zero points.

4. Sketching One Graph From Another ★★★☆☆ ⏱ 3 min

A common AP question asks you to sketch the graph of $f$ given the graph of $f'$ (or vice versa), using only qualitative relationships. The process follows three simple steps: mark all key points, divide the x-axis into intervals between key points, then assign the correct increasing/decreasing and concavity to each interval.

📐 Worked Example

The graph of $f'(x)$ is a parabola opening upward with roots at $x=0$ and $x=4$, and $f(0) = 2$. Identify all key features of $f$ (extrema, inflection points) to prepare a sketch.

First, find intervals of increase/decrease for $f$: since $f'(x)$ is an upward opening parabola, it is negative between its roots ($0 < x < 4$) and positive outside ($x < 0$ and $x > 4$). So $f$ increases on $(-\infty, 0)$, decreases on $(0, 4)$, and increases on $(4, \infty)$.
Find extrema of $f$: $f'$ changes from positive to negative at $x=0$, so $f$ has a local maximum at $(0, 2)$. $f'$ changes from negative to positive at $x=4$, so $f$ has a local minimum at $x=4$.
Find concavity and inflection points of $f$: the vertex of the parabola $f'(x)$ is at $x=2$, so $f'(x)$ is decreasing for $x < 2$ and increasing for $x > 2$. This means $f''(x) < 0$ for $x < 2$ and $f''(x) > 0$ for $x > 2$, so $f$ changes concavity at $x=2$, which is an inflection point.
Final key features: local maximum at $(0, 2)$, inflection point at $x=2$, local minimum at $x=4$.

Exam tip: When asked to sketch a graph on AP FRQ, you only need to correctly plot and label all required key features and get the general shape right. You do not need to plot every point to earn full credit.

5. Concept Check ★★★★☆ ⏱ 2 min

📐 Worked Example

The function $f$ is twice differentiable for all real $x$, with first derivative $f'(x) = x(x - 2)^2$. (a) Identify all critical points of $f$. (b) On what intervals is $f$ decreasing? Justify your answer. (c) Classify each critical point as a local maximum, local minimum, or neither. (d) Identify all inflection points of $f$. Justify your answer.

(a) Critical points occur where $f'(x)=0$ or $f'(x)$ is undefined. $f'(x)$ is a polynomial, so defined everywhere. Set $x(x - 2)^2 = 0$, giving critical points at $x=0$ and $x=2$.
(b) $f$ is decreasing when $f'(x) < 0$. $(x-2)^2$ is non-negative for all real $x$, so the sign of $f'(x)$ matches the sign of $x$. $f'(x) < 0$ when $x < 0$, so $f$ is decreasing on $(-\infty, 0)$.
(c) For $x=0$: $f'(x) < 0$ when $x < 0$, and $f'(x) > 0$ when $0 < x < 2$. $f'$ changes from negative to positive, so $x=0$ is a local minimum. For $x=2$: $f'(x) > 0$ when $0 < x < 2$ and $f'(x) > 0$ when $x > 2$, so there is no sign change. $x=2$ is neither a local maximum nor minimum.
(d) To find inflection points, first compute $f''(x)$:
$f'(x) = x^3 - 4x^2 + 4x \implies f''(x) = 3x^2 - 8x + 4 = (3x - 2)(x - 2)$
Set $f''(x)=0$ to get $x = \frac{2}{3}$ and $x=2$. Testing sign: $f''(x) > 0$ for $x < \frac{2}{3}$, $f''(x) < 0$ for $\frac{2}{3} < x < 2$, $f''(x) > 0$ for $x > 2$. $f''$ changes sign at both points, so inflection points at $x = \frac{2}{3}$ and $x=2$.

Common Pitfalls

Why: Students memorize that inflection points occur where $f''=0$, so they assume all such points qualify, forgetting the concavity change requirement.

Why: Students incorrectly assume closed intervals are acceptable because monotonicity can extend to endpoints.

Why: When given a graph of $f'$, students mix up what tells you about increase/decrease of $f$ versus concavity of $f$.

Why: Students assume all critical points are extrema by definition.

Why: Students forget the test is inconclusive, not negative, when $f''(c)=0$.

Why: Students confuse the location of concavity changes with slope changes.

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