Concavity of Functions
Over Their Domains
Master how the second derivative reveals whether a curve bends upward or downward, and find inflection points where concavity changes.
AP Calculus BC
What Is Concavity?
Concavity describes the direction a curve bends. Imagine driving along a curved road: concavity tells you whether the road curves like a bowl (opening up) or like a cap (opening down). Formally, it describes how the slope of a function is changing.
Concave Up
The curve lies above its tangent lines. The slope is increasing. Shaped like a bowl: \(\cup\)
Concave Down
The curve lies below its tangent lines. The slope is decreasing. Shaped like a cap: \(\cap\)
Seeing Concavity in a Graph
Why Does Concavity Matter?
Concavity goes beyond just knowing if a function is increasing or decreasing. It tells you how fast it's increasing or decreasing, enabling deeper analysis.
Second Derivative Test
Classify critical points as local maxima or local minima without testing intervals.
Curve Sketching
Produce accurate graphs by knowing where the curve bows up or down.
Optimization
In applied problems, concavity confirms whether a solution is truly a maximum or minimum.
The Concavity Theorem
\( f''(x) \) Controls the Bend
Concavity vs. Increasing/Decreasing
| Increasing/Decreasing | Concavity | |
|---|---|---|
| Determined by | \( f'(x) \) | \( f''(x) \) |
| Tells you | Direction (up/down in value) | Shape of the bend |
| Positive sign | \( f'(x) > 0 \): function rises | \( f''(x) > 0 \): curve bows up |
| Negative sign | \( f'(x) < 0 \): function falls | \( f''(x) < 0 \): curve bows down |
| Special points | Critical points (\( f' = 0 \)) | Inflection points (\( f'' = 0 \)) |
The Four Combinations
Increasing + Concave Up
\( f' > 0,\; f'' > 0 \)
Rising faster and faster. Example: \( e^x \)
Rising faster and faster. Example: \( e^x \)
Increasing + Concave Down
\( f' > 0,\; f'' < 0 \)
Rising but slowing. Example: \( \ln x \)
Rising but slowing. Example: \( \ln x \)
Decreasing + Concave Up
\( f' < 0,\; f'' > 0 \)
Falling but leveling off. Example: \( e^{-x} \)
Falling but leveling off. Example: \( e^{-x} \)
Decreasing + Concave Down
\( f' < 0,\; f'' < 0 \)
Falling faster and faster. Example: \( -x^2 \) for \( x > 0 \)
Falling faster and faster. Example: \( -x^2 \) for \( x > 0 \)
How to Determine Concavity
1
Compute \( f''(x) \). Differentiate \( f(x) \) twice.
2
Find candidates. Solve \( f''(x) = 0 \) and locate where \( f''(x) \) is undefined. These are potential inflection points.
3
Build a sign chart. Pick test values in each interval formed by the candidates. Evaluate \( f''(\text{test}) \).
4
State concavity. Positive \( f'' \) means concave up; negative means concave down.
5
Identify inflection points. Where \( f'' \) changes sign, you have an inflection point at \( (c, f(c)) \).
Example 1: \( f(x) = x^3 - 6x^2 + 9x + 1 \)
1
First derivative: \( f'(x) = 3x^2 - 12x + 9 \)
2
Second derivative: \( f''(x) = 6x - 12 \)
3
Set \( f''(x) = 0 \): \( 6x - 12 = 0 \implies x = 2 \)
4
Sign chart: Test \( x = 0 \): \( f''(0) = -12 < 0 \) (concave down). Test \( x = 3 \): \( f''(3) = 6 > 0 \) (concave up).
Example 1: Graph
Example 2: \( g(x) = x^4 - 4x^3 \)
1
Compute \( g''(x) \): \( g'(x) = 4x^3 - 12x^2 \), so \( g''(x) = 12x^2 - 24x = 12x(x - 2) \).
2
Solve \( g''(x) = 0 \): \( 12x(x-2)=0 \implies x = 0 \) or \( x = 2 \).
3
Sign chart:
\( x = -1 \): \( g''(-1)=12(-1)(-3)=36 > 0 \) (concave up)
\( x = 1 \): \( g''(1)=12(1)(-1)=-12 < 0 \) (concave down)
\( x = 3 \): \( g''(3)=12(3)(1)=36 > 0 \) (concave up)
\( x = -1 \): \( g''(-1)=12(-1)(-3)=36 > 0 \) (concave up)
\( x = 1 \): \( g''(1)=12(1)(-1)=-12 < 0 \) (concave down)
\( x = 3 \): \( g''(3)=12(3)(1)=36 > 0 \) (concave up)
Example 3: \( h(x) = \frac{x}{x^2 + 1} \)
1
Quotient rule for \( h'(x) \): \( h'(x) = \dfrac{(x^2+1) - x(2x)}{(x^2+1)^2} = \dfrac{1 - x^2}{(x^2+1)^2} \)
2
Second derivative \( h''(x) \): After differentiation and simplification: \( h''(x) = \dfrac{2x(x^2 - 3)}{(x^2+1)^3} \)
3
Solve \( h''(x) = 0 \): \( 2x(x^2-3)=0 \implies x=0,\; x=\pm\sqrt{3} \). Three inflection point candidates.
The Second Derivative Test
Common Mistakes
✕ \( f''(c)=0 \) means inflection
\( f''(c)=0 \) alone does not guarantee an inflection point. You must verify that \( f'' \) changes sign at \( c \). For example, \( f(x)=x^4 \) has \( f''(0)=0 \) but no inflection there.
✕ Confusing \( f' \) with \( f'' \)
Concavity uses the second derivative, not the first. \( f'(x) > 0 \) means the function is increasing, not concave up. Always differentiate twice.
✕ Ignoring domain restrictions
If \( f''(x) \) is undefined at some \( x = c \) within the domain, that point is also a candidate for an inflection point. Don't only solve \( f'' = 0 \).
Summary Reference Table
| Condition | Meaning | Graph Behavior |
|---|---|---|
| \( f''(x) > 0 \) | Concave Up | Curve opens upward (\(\cup\)) |
| \( f''(x) < 0 \) | Concave Down | Curve opens downward (\(\cap\)) |
| \( f''(c) = 0 \) and sign change | Inflection Point | Concavity switches at \( x = c \) |
| \( f'(c)=0,\; f''(c)>0 \) | Local Minimum | Bowl shape at critical point |
| \( f'(c)=0,\; f''(c)<0 \) | Local Maximum | Cap shape at critical point |
| \( f'(c)=0,\; f''(c)=0 \) | Inconclusive | Use first derivative test |
Question 1 of 5
On which interval is \( f(x) = x^3 - 3x^2 + 2 \) concave up?
\( f''(x) = 6x - 6 = 6(x-1) \). Setting \( f''(x) > 0 \) gives \( x > 1 \), so the function is concave up on \( (1, \infty) \).
Question 2 of 5
What is the \( x \)-coordinate of the inflection point of \( g(x) = x^4 - 12x^2 \)?
\( g''(x) = 12x^2 - 24 = 12(x^2 - 2) \). Setting \( g''(x) = 0 \) gives \( x^2 = 2 \), so \( x = \pm\sqrt{2} \). Checking sign changes confirms both are inflection points.
Question 3 of 5
If \( f''(3) = 0 \) and \( f''(x) < 0 \) for \( x < 3 \) and \( f''(x) < 0 \) for \( x > 3 \), what can you conclude about \( x = 3 \)?
Since \( f'' \) does not change sign at \( x = 3 \) (it is negative on both sides), \( x = 3 \) is not an inflection point. The function remains concave down through this point.
Question 4 of 5
Given \( f'(2)=0 \) and \( f''(2) = -5 \), what does the Second Derivative Test tell you about \( x = 2 \)?
Since \( f'(2) = 0 \) (critical point) and \( f''(2) = -5 < 0 \) (concave down), the Second Derivative Test confirms a local maximum at \( x = 2 \).
Question 5 of 5
For \( f(x) = e^{-x^2} \), find where the function is concave up.
\( f'(x) = -2x\,e^{-x^2} \) and \( f''(x) = (4x^2 - 2)e^{-x^2} \). Since \( e^{-x^2} > 0 \) always, the sign of \( f'' \) depends on \( 4x^2 - 2 \). Setting \( 4x^2 - 2 > 0 \) gives \( x^2 > \frac{1}{2} \), i.e., \( |x| > \frac{1}{\sqrt{2}} \). So \( f \) is concave up on \( \left(-\infty, -\frac{1}{\sqrt{2}}\right) \cup \left(\frac{1}{\sqrt{2}}, \infty\right) \).