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Introduction
1. Limits
2. Derivative basics
3. Advanced differentiation
4. Contextual uses
5. Analytical uses
5.1 Important theorems
5.2 1st derivative test
5.3 2nd derivative test
5.3.1 Concavity
5.3.2 Inflection points
5.4 Graphs & curve sketching
5.5 Optimization
6. Integration
7. Differential equations
8. Applications of integrals
9. Testing details tag
Wrapping up
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5.3.1 Concavity
Achievable AP Calculus AB
5. Analytical uses
5.3. 2nd derivative test
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Concavity

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What you’ll learn

  • How to classify relative extrema using the 2nd derivative test instead of the 1st derivative sign chart
  • Understanding the 2nd derivative as concavity
  • How to find intervals of concavity and identify inflection points

In section 4.4 on linear approximations, it was briefly mentioned that concavity describes whether a graph bends upward or downward.

Once you’ve found a critical point using f′(x), you can often classify it as a relative maximum or minimum using the 2nd derivative test. This lets you use f′′(x) (concavity) instead of building a full sign chart for f′(x).

2nd derivative test for extrema

To classify extrema using this test:

  1. Find the critical points (where f′(x)=0 or is undefined).
  2. Find the 2nd derivative, f′′(x).
  3. Plug each critical point into f′′(x):

  • If f′′(x)>0, then f(x) is concave up (shaped “like a cup”), and the critical point is a relative minimum.

  • If f′′(x)<0, then f(x) is concave down (shaped “like a frown”), and the critical point is a relative maximum.

  • If f′′(x)=0, the 2nd derivative test is inconclusive, so you’ll need to use the 1st derivative test to classify the point.

2nd derivative test
2nd derivative test

Pitfall 1 - inconclusive case: When f′′(c)=0, the test tells you nothing. You must fall back to the 1st derivative test. For example, f(x)=x4 has f′′(0)=0, but x=0 is still a relative minimum (as the 1st derivative sign chart confirms).

Pitfall 2 - inflection points: A point where f′′(c)=0 is not automatically an inflection point. You must verify that f′′ actually changes sign there. If f′′ doesn’t change sign, the concavity doesn’t change and there’s no inflection point.

Here’s why the sign of f′′ tells you the concavity of f:

  • The 1st derivative f′(x) tells you the slope of the tangent line to f(x).
  • The 2nd derivative f′′(x) tells you how those slopes are changing as x increases.

So:

  • If f′′(x)>0, the slopes are increasing (tangent lines get “less negative” and then more positive). That corresponds to a concave-up shape.
  • If f′′(x)<0, the slopes are decreasing. That corresponds to a concave-down shape.

For example, take f(x)=x2.

  • f′(x)=2x
  • f′′(x)=2 (always positive, so the graph is always concave up)

At x=−2, the slope is f′(−2)=−4. At x=−1, the slope is f′(−1)=−2. The slope increased from −4 to −2, which matches f′′(x)>0.

Now compare that to f(x)=−x2.

  • f′(x)=−2x
  • f′′(x)=−2 (always negative, so the graph is always concave down)

As you move left to right, the slopes decrease, matching f′′(x)<0.

Intervals of concavity and inflection points

An inflection point is a point on the graph where the concavity changes - where f′′ switches from positive to negative, or vice versa. Note that f′′(c)=0 is a necessary condition, not a sufficient one: you must confirm that f′′ changes sign at c.

To find intervals of concavity and any inflection points:

  1. Find f′′(x).
  2. Set f′′(x)=0 (and note where it’s undefined) to find candidate points.
  3. Use those candidates to divide the number line into intervals.
  4. Test the sign of f′′ on each interval.
    • Where f′′>0: concave up.
    • Where f′′<0: concave down.
  5. An inflection point occurs at a candidate x=c only if f′′ changes sign there.

Example: Find the intervals of concavity and any inflection points for f(x)=x3−3x2+1.

From the 2nd derivative test section, f′′(x)=6x−6.

Set f′′(x)=0: 6x−6=0, so x=1.

Test the sign of f′′ on each side:

  • For x<1 (e.g., x=0): f′′(0)=−6<0 → concave down.
  • For x>1 (e.g., x=2): f′′(2)=6>0 → concave up.

Since f′′ changes sign at x=1 (from negative to positive), there is an inflection point at x=1. The y-value is f(1)=1−3+1=−1, so the inflection point is (1,−1).

Examples

Classify the extrema of

f(x)=x3−3x2+1

Solution

(spoiler)

The 1st derivative is

f′(x)=3x2−6x

To find the critical points, set equal to 0:

3x2−6x=0

3x(x−2)=0

x=0,2

These are the two critical points.

Now use the 2nd derivative test. The 2nd derivative is

f′′(x)=6x−6

Plug in each critical point:

f′′(0)=6(0)−6=−6

f′′(2)=6(2)−6=6

  • Since f′′(0)<0, f(x) is concave down at x=0, so x=0 is a relative max.

  • Since f′′(2)>0, f(x) is concave up at x=2, so x=2 is a relative min.

Classify the extrema of

f(x)=xe−x

Solution

(spoiler)

The 1st derivative is

f′(x)=xe−x(−1)+e−x(1)

=−e−x(x−1)

To find the critical points, solve

−e−x(x−1)=0

x=1

The 2nd derivative is

f′′(x)=−e−x(1)+(x−1)(e−x)

=−e−x(1−(x−1))

=−e−x(−x+2)

Classify the critical point x=1:

f′′(1)=−e−1(−1+2)=−e1​

Since f′′(1)<0, f(x) is concave down at x=1, so x=1 is a relative max.

Challenge problem

A curve with derivative

dxdy​=y+5x2−4x+3​

has y=−2 as a tangent line. At what point is the line tangent to the curve, and is it a relative maximum, a relative minimum, or neither?

Solution

(spoiler)

The curve meets y=−2 at the point of tangency (a,−2).

Because y=−2 is a horizontal tangent line, its slope is 0, so at the tangency point we must have

dxdy​=0.

From

dxdy​=y+5x2−4x+3​,

the derivative is zero when the numerator is zero:

x2−4x+3=0

(x−1)(x−3)=0

x=1,3

So (1,−2) and (3,−2) are both points where the tangent line could be horizontal. (And the derivative is defined there because y+5=−2+5=3=0.)

Now classify each point using the 2nd derivative.

The 2nd derivative is

dx2d2y​=(y+5)2(y+5)(2x−4)−(x2−4x+3)(dxdy​)​

For (1,−2):

dx2d2y​=(−2+5)2(−2+5)(2⋅1−4)−(12−4⋅1+3)(0)​

=9(3)(−2)​

=−32​

Since the 2nd derivative is negative, the curve is concave down at (1,−2), so (1,−2) is a relative max.

For (3,−2):

dx2d2y​=(−2+5)2(−2+5)(2⋅3−4)−(32−4⋅3+3)(0)​

=9(3)(2)​

=32​

Since the 2nd derivative is positive, the curve is concave up at (3,−2), so (3,−2) is a relative min.

2nd derivative test

  • Classifies critical points using f′′(x) (concavity)
  • Steps:
    • Find critical points: f′(x)=0 or undefined
    • Compute f′′(x) at each critical point
  • Interpretation:
    • f′′(x)>0: concave up, relative minimum
    • f′′(x)<0: concave down, relative maximum
    • f′′(x)=0: test is inconclusive, use 1st derivative test

Concavity and derivatives

  • f′(x): slope of tangent line
  • f′′(x): rate of change of slope (concavity)
    • f′′(x)>0: slopes increase, graph bends upward
    • f′′(x)<0: slopes decrease, graph bends downward

Worked examples

  • f(x)=x3−3x2+1
    • Critical points: x=0,2
    • f′′(0)=−6 (max), f′′(2)=6 (min)
  • f(x)=x3+1x2−1​
    • Critical points: x=0,2
    • f′′(0)=2 (min), f′′(2)<0 (max)
  • f(x)=cos(x)−sec(x) on (0,2π)
    • Critical point: x=π
    • f′′(π)=2 (min)
  • f(x)=xe−x
    • Critical point: x=1
    • f′′(1)<0 (max)
  • f(x)=ln(x2+1)
    • Critical point: x=0
    • f′′(0)=2 (min)

Challenge problem

  • Given dxdy​=y+5x2−4x+3​, y=−2 tangent line
    • Points of tangency: (1,−2) (max), (3,−2) (min)
    • Classified using 2nd derivative at each point

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Concavity

What you’ll learn

  • How to classify relative extrema using the 2nd derivative test instead of the 1st derivative sign chart
  • Understanding the 2nd derivative as concavity
  • How to find intervals of concavity and identify inflection points

In section 4.4 on linear approximations, it was briefly mentioned that concavity describes whether a graph bends upward or downward.

Once you’ve found a critical point using f′(x), you can often classify it as a relative maximum or minimum using the 2nd derivative test. This lets you use f′′(x) (concavity) instead of building a full sign chart for f′(x).

2nd derivative test for extrema

To classify extrema using this test:

  1. Find the critical points (where f′(x)=0 or is undefined).
  2. Find the 2nd derivative, f′′(x).
  3. Plug each critical point into f′′(x):

  • If f′′(x)>0, then f(x) is concave up (shaped “like a cup”), and the critical point is a relative minimum.

  • If f′′(x)<0, then f(x) is concave down (shaped “like a frown”), and the critical point is a relative maximum.

  • If f′′(x)=0, the 2nd derivative test is inconclusive, so you’ll need to use the 1st derivative test to classify the point.

Pitfall 1 - inconclusive case: When f′′(c)=0, the test tells you nothing. You must fall back to the 1st derivative test. For example, f(x)=x4 has f′′(0)=0, but x=0 is still a relative minimum (as the 1st derivative sign chart confirms).

Pitfall 2 - inflection points: A point where f′′(c)=0 is not automatically an inflection point. You must verify that f′′ actually changes sign there. If f′′ doesn’t change sign, the concavity doesn’t change and there’s no inflection point.

Here’s why the sign of f′′ tells you the concavity of f:

  • The 1st derivative f′(x) tells you the slope of the tangent line to f(x).
  • The 2nd derivative f′′(x) tells you how those slopes are changing as x increases.

So:

  • If f′′(x)>0, the slopes are increasing (tangent lines get “less negative” and then more positive). That corresponds to a concave-up shape.
  • If f′′(x)<0, the slopes are decreasing. That corresponds to a concave-down shape.

For example, take f(x)=x2.

  • f′(x)=2x
  • f′′(x)=2 (always positive, so the graph is always concave up)

At x=−2, the slope is f′(−2)=−4. At x=−1, the slope is f′(−1)=−2. The slope increased from −4 to −2, which matches f′′(x)>0.

Now compare that to f(x)=−x2.

  • f′(x)=−2x
  • f′′(x)=−2 (always negative, so the graph is always concave down)

As you move left to right, the slopes decrease, matching f′′(x)<0.

Intervals of concavity and inflection points

An inflection point is a point on the graph where the concavity changes - where f′′ switches from positive to negative, or vice versa. Note that f′′(c)=0 is a necessary condition, not a sufficient one: you must confirm that f′′ changes sign at c.

To find intervals of concavity and any inflection points:

  1. Find f′′(x).
  2. Set f′′(x)=0 (and note where it’s undefined) to find candidate points.
  3. Use those candidates to divide the number line into intervals.
  4. Test the sign of f′′ on each interval.
    • Where f′′>0: concave up.
    • Where f′′<0: concave down.
  5. An inflection point occurs at a candidate x=c only if f′′ changes sign there.

Example: Find the intervals of concavity and any inflection points for f(x)=x3−3x2+1.

From the 2nd derivative test section, f′′(x)=6x−6.

Set f′′(x)=0: 6x−6=0, so x=1.

Test the sign of f′′ on each side:

  • For x<1 (e.g., x=0): f′′(0)=−6<0 → concave down.
  • For x>1 (e.g., x=2): f′′(2)=6>0 → concave up.

Since f′′ changes sign at x=1 (from negative to positive), there is an inflection point at x=1. The y-value is f(1)=1−3+1=−1, so the inflection point is (1,−1).

Examples

Classify the extrema of

f(x)=x3−3x2+1

Solution

(spoiler)

The 1st derivative is

f′(x)=3x2−6x

To find the critical points, set equal to 0:

3x2−6x=0

3x(x−2)=0

x=0,2

These are the two critical points.

Now use the 2nd derivative test. The 2nd derivative is

f′′(x)=6x−6

Plug in each critical point:

f′′(0)=6(0)−6=−6

f′′(2)=6(2)−6=6

  • Since f′′(0)<0, f(x) is concave down at x=0, so x=0 is a relative max.

  • Since f′′(2)>0, f(x) is concave up at x=2, so x=2 is a relative min.

Classify the extrema of

f(x)=xe−x

Solution

(spoiler)

The 1st derivative is

f′(x)=xe−x(−1)+e−x(1)

=−e−x(x−1)

To find the critical points, solve

−e−x(x−1)=0

x=1

The 2nd derivative is

f′′(x)=−e−x(1)+(x−1)(e−x)

=−e−x(1−(x−1))

=−e−x(−x+2)

Classify the critical point x=1:

f′′(1)=−e−1(−1+2)=−e1​

Since f′′(1)<0, f(x) is concave down at x=1, so x=1 is a relative max.

Challenge problem

A curve with derivative

dxdy​=y+5x2−4x+3​

has y=−2 as a tangent line. At what point is the line tangent to the curve, and is it a relative maximum, a relative minimum, or neither?

Solution

(spoiler)

The curve meets y=−2 at the point of tangency (a,−2).

Because y=−2 is a horizontal tangent line, its slope is 0, so at the tangency point we must have

dxdy​=0.

From

dxdy​=y+5x2−4x+3​,

the derivative is zero when the numerator is zero:

x2−4x+3=0

(x−1)(x−3)=0

x=1,3

So (1,−2) and (3,−2) are both points where the tangent line could be horizontal. (And the derivative is defined there because y+5=−2+5=3=0.)

Now classify each point using the 2nd derivative.

The 2nd derivative is

dx2d2y​=(y+5)2(y+5)(2x−4)−(x2−4x+3)(dxdy​)​

For (1,−2):

dx2d2y​=(−2+5)2(−2+5)(2⋅1−4)−(12−4⋅1+3)(0)​

=9(3)(−2)​

=−32​

Since the 2nd derivative is negative, the curve is concave down at (1,−2), so (1,−2) is a relative max.

For (3,−2):

dx2d2y​=(−2+5)2(−2+5)(2⋅3−4)−(32−4⋅3+3)(0)​

=9(3)(2)​

=32​

Since the 2nd derivative is positive, the curve is concave up at (3,−2), so (3,−2) is a relative min.

Key points

2nd derivative test

  • Classifies critical points using f′′(x) (concavity)
  • Steps:
    • Find critical points: f′(x)=0 or undefined
    • Compute f′′(x) at each critical point
  • Interpretation:
    • f′′(x)>0: concave up, relative minimum
    • f′′(x)<0: concave down, relative maximum
    • f′′(x)=0: test is inconclusive, use 1st derivative test

Concavity and derivatives

  • f′(x): slope of tangent line
  • f′′(x): rate of change of slope (concavity)
    • f′′(x)>0: slopes increase, graph bends upward
    • f′′(x)<0: slopes decrease, graph bends downward

Worked examples

  • f(x)=x3−3x2+1
    • Critical points: x=0,2
    • f′′(0)=−6 (max), f′′(2)=6 (min)
  • f(x)=x3+1x2−1​
    • Critical points: x=0,2
    • f′′(0)=2 (min), f′′(2)<0 (max)
  • f(x)=cos(x)−sec(x) on (0,2π)
    • Critical point: x=π
    • f′′(π)=2 (min)
  • f(x)=xe−x
    • Critical point: x=1
    • f′′(1)<0 (max)
  • f(x)=ln(x2+1)
    • Critical point: x=0
    • f′′(0)=2 (min)

Challenge problem

  • Given dxdy​=y+5x2−4x+3​, y=−2 tangent line
    • Points of tangency: (1,−2) (max), (3,−2) (min)
    • Classified using 2nd derivative at each point