AP Calculus AB Concavity and the Second Derivative Lesson (Free PDF)

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This concavity lesson for AP Calculus AB covers concavity and the second derivative (Common Core AP CALC AB 5.7, AP CALC AB 5.8, AP CALC AB 5.9), with a free PDF to print.

The second derivative f″ shows concavity: f is concave up where f″ is positive and concave down where f″ is negative. An inflection point is where concavity changes, so f″ changes sign. The Second Derivative Test classifies a critical number c with f′(c) = 0: a positive f″(c) gives a relative minimum and a negative f″(c) gives a relative maximum.

AP Calculus AB · Common Core State Standards: AP CALC AB 5.7, AP CALC AB 5.8, AP CALC AB 5.9

Key facts in this concavity lesson

Key facts
  • A positive f″ means concave up (slope increasing); a negative f″ means concave down (slope decreasing).
  • An inflection point needs a change in the sign of f″, not just f″ = 0.
  • Second Derivative Test: f′(c) = 0 and a positive f″(c) gives a relative minimum; a negative f″(c) gives a maximum.
  • If f″(c) = 0 the test fails; use the First Derivative Test.

What does the second derivative tell you?

The second derivative \(f^{\prime\prime}\) measures how the slope changes. If \(f^{\prime\prime} > 0\), the slope is increasing and the graph is concave up, like a cup. If \(f^{\prime\prime} < 0\), the slope is decreasing and the graph is concave down, like a frown.

-112345-4-22468xyinflection (2, 3)concave downconcave up
The graph bends down before x = 2 and bends up after it; (2, 3) is an inflection point.
Inflection point

An inflection point is a point on the graph where the concavity changes. At such a point \(f^{\prime\prime}=0\) or \(f^{\prime\prime}\) is undefined, and \(f^{\prime\prime}\) changes sign. A zero of \(f^{\prime\prime}\) alone is not enough.

Example. For \(f(x)=x^{3}-6x^{2}+9x+1\), \(f^{\prime\prime}(x)=6x-12\), which is zero at \(x=2\) and changes sign there. The graph is concave down on \((-\infty,2)\), concave up on \((2,\infty)\), with an inflection point at \((2,3)\).

f″−concave down+concave upx = 2
A sign chart of f″ shows where the graph is concave up or down.

What is the second derivative test?

If \(f^{\prime}(c)=0\), check \(f^{\prime\prime}(c)\). A positive value means the graph is a cup at \(c\), so \(f(c)\) is a relative minimum. A negative value gives a relative maximum. If \(f^{\prime\prime}(c)=0\), the test fails and you must use the First Derivative Test.

f′(c) = 0criticalf″(c) > 0minimumf″(c) < 0maximumf″(c) = 0no conclusion
The second derivative test classifies a critical number with one evaluation.

How do f, f′ and f″ connect?

Feature of f Where to read it
Increasing or decreasing \(f^{\prime} > 0\) or \(f^{\prime} < 0\)
Relative extremum \(f^{\prime}\) changes sign
Concave up or down \(f^{\prime\prime} > 0\) or \(f^{\prime\prime} < 0\) (or \(f^{\prime}\) increasing or decreasing)
Inflection point \(f^{\prime\prime}\) changes sign (\(f^{\prime}\) has a local extremum)
-112345-8-6-4-22468101214xyff′f″
f (blue), f′ (green) and f″ (orange) for the same function.

Example (second derivative test). For \(f(x)=x^{3}-12x+3\), \(f^{\prime}=3x^{2}-12=0\) gives \(x=\pm 2\), and \(f^{\prime\prime}=6x\). Since \(f^{\prime\prime}(2)=12\) is positive, \(x=2\) is a relative minimum. Since \(f^{\prime\prime}(-2)=-12\) is negative, \(x=-2\) is a relative maximum. No sign chart is needed, which makes this test fast when \(f^{\prime\prime}\) is simple.

Look at the green graph of \(f^{\prime}\): it has its lowest point at \(x=2\), exactly where the blue graph of \(f\) changes concavity. An inflection point of \(f\) is a place where the slope is at a local extreme.

Concavity lesson: questions and answers

What does concave up mean?

A graph is concave up on an interval if its slope is increasing there, which means f″ is positive. The graph bends like a cup and lies above its tangent lines.

What is an inflection point?

It is a point where the graph changes concavity. The second derivative is zero or undefined there and changes sign.

Does f″(c) = 0 always mean an inflection point?

No. For f(x) = x^4, f″(0) = 0 but f″ is positive on both sides, so the concavity never changes and there is no inflection point.

How does the second derivative test work?

At a critical number c where f′(c) = 0, a positive f″(c) means relative minimum and a negative f″(c) means relative maximum. If f″(c) = 0, the test is inconclusive.

How can you find concavity from the graph of f′?

f is concave up where the graph of f′ is increasing and concave down where the graph of f′ is decreasing. Inflection points of f occur at local extrema of f′.

How to use this concavity lesson

Read the definitions first, then follow the worked method line by line and redo the example on paper without looking. When every step makes sense, move on to the worksheet, then check yourself with the quiz.

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Download the concavity lesson as a PDF with the button below and keep it with your notes.

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AP Calculus AB Concavity and the Second Derivative Lesson

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