Concavity and Points of Inflection

The second derivative reveals which way the graph of a function bends. If the second derivative is positive, the graph is concave up (holds water); if negative, it is concave down (spills water). A point where the concavity changes direction is called a point of inflection.

Condition	Conclusion
f''(x)>0 on $I$	Graph is concave up on $I$
f''(x)<0 on $I$	Graph is concave down on $I$
Concavity changes at $x_{0}$ and $f$ continuous there	$(x_{0}, f (x_{0}))$ is an inflection point

Concavity

The sign of the derivative tells us whether a function is increasing or decreasing, but it does not tell us which way the graph bends. Two increasing functions can look very different depending on whether their graphs bend upward or downward.

Two panels: left shows an increasing function that is concave up (bends upward), right shows an increasing function that is concave down (bends downward) — (a) Increasing and concave up. (b) Increasing and concave down.

When a graph is concave up on an interval, the slope of the tangent line is increasing from left to right. When it is concave down, the slope is decreasing.

Graph of a concave up function with tangent lines showing increasing slopes — Concave up: slopes are increasing (the derivative f' is increasing).

Graph of a concave down function with tangent lines showing decreasing slopes — Concave down: slopes are decreasing (the derivative f' is decreasing).

Informally:

A function is concave up on $I$ if its graph over $I$ "holds water."
A function is concave down on $I$ if its graph "spills water."

Assume $f$ is a differentiable function on an interval $I$ .

(a) The graph of $f$ is concave up if f' is increasing on $I$ .

(b) The graph of $f$ is concave down if f' is decreasing on $I$ .

Since f'' is the derivative of f', the Increasing/Decreasing Test applied to f' gives us:

Let $f$ be a function whose second derivative exists at each point of an interval $I$ .

(a) If f''(x)>0 for every $x$ in $I$ , then the graph of $f$ is concave up on $I$ .

(b) If f''(x)<0 for every $x$ in $I$ , then the graph of $f$ is concave down on $I$ .

Examples

Determine where the following functions are concave up and where they are concave down.

(a) $f (x) = 2 x^{2} - 5 x - 7$

(b) $f (x) = x^{3}$

Solution

**(a)** f'(x)=4x-5\Rightarrow f''(x)=4. Since f''(x)=4>0 for all

x

, the function is concave up on

(- \infty, + \infty)

Graph of y = 2x^2 - 5x - 7, a parabola opening upward (concave up everywhere) — Graph of $y = 2 x^{2} - 5 x - 7$ . Always concave up.

**(b)** f'(x)=3x^2\Rightarrow f''(x)=6x. f''(x)<0 when

x < 0

, so

f

is concave down on

(- \infty, 0)

. f''(x)>0 when

x > 0

, so

f

is concave up on

(0, + \infty)

Graph of y = x^3 showing concave down on the left and concave up on the right — Graph of $y = x^{3}$ . Concave down on $(- \infty, 0)$ and concave up on $(0, + \infty)$ .

**(c)** f'(x)=3x^2-6x\Rightarrow f''(x)=6x-6=6(x-1). f''(x)<0 when

x < 1

, so

f

is concave down on

(- \infty, 1)

. f''(x)>0 when

x > 1

, so

f

is concave up on

(1, + \infty)

Graph of y = x^3 - 3x^2 + 1 showing the change of concavity at x = 1 — Graph of $y = x^{3} - 3 x^{2} + 1$ . The concavity changes from downward to upward at $x = 1$ .

Determine where $y = e^{- x^{2}}$ is concave up and where it is concave down.

Solution

Let

u = - x^{2}

: y'=e^{-x^2}(-2x)=-2xe^{-x^2}. Using the product rule with

v = - 2 x

and

w = e^{- x^{2}}

: \begin{aligned} y'' &= (-2)e^{-x^2}+(-2x)(-2xe^{-x^2}) \\ &= -2e^{-x^2}+4x^2e^{-x^2} \\ &= 4e^{-x^2}\!\left(x^2-\frac{1}{2}\right). \end{aligned} Since

e^{- x^{2}} > 0

, the sign of y'' is determined by:

x^{2} - \frac{1}{2} = (x - \frac{1}{\sqrt{2}}) (x + \frac{1}{\sqrt{2}}) .

Sign diagram for y'' showing negative between -1/sqrt(2) and 1/sqrt(2) and positive outside

The graph is concave down on

(- \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})

and concave up on

(- \infty, - \frac{1}{\sqrt{2}}) \cup (\frac{1}{\sqrt{2}}, + \infty)

Graph of y = e^(-x^2), a bell curve, concave up outside and concave down between the inflection points — Graph of $y = e^{- x^{2}}$ . Concave down between $- 1 / \sqrt{2}$ and $1 / \sqrt{2}$ , concave up outside this interval.

Concavity and Tangent Lines

If a function is concave up on an interval, its graph lies above all of its tangent lines on that interval. If it is concave down, the graph lies below all of its tangent lines.

Let $f$ be differentiable on an open interval $I$ and $x_{0}$ be an arbitrary point of $I$ . Show that if $f$ is concave up on $I$ , the curve $y = f (x)$ lies above the tangent line at $(x_{0}, f (x_{0}))$ , except at $(x_{0}, f (x_{0}))$ itself.

Solution

The tangent line at

(x_{0}, f (x_{0}))

is L(x)=f(x_0)+f'(x_0)(x-x_0). We need to show

f (x) > L (x)

for all

x \neq x_{0}

I

. Assume

x > x_{0}

. By the Mean Value Theorem applied to

[x_{0}, x]

, there exists

c \in (x_{0}, x)

such that: \frac{f(x)-f(x_0)}{x-x_0}=f'(c). Since

f

is concave up, f' is increasing, and

c > x_{0}

implies f'(c)>f'(x_0). Therefore: f'(x_0)<\frac{f(x)-f(x_0)}{x-x_0}. Multiplying both sides by

x - x_{0} > 0

: f'(x_0)(x-x_0) A similar argument works for

x < x_{0}

Points of Inflection

Most curves are concave up on some intervals and concave down on others. A point where the direction of concavity changes is called an inflection point.

Graph showing a point of inflection where the curve changes from concave up to concave down — An inflection point is where a curve changes from concave upward to concave downward (or vice versa).

We say $(x_{0}, f (x_{0}))$ is an inflection point of the graph of $f$ if:

(a) $f$ is continuous at $x_{0}$ ,

(b) the direction of concavity changes (from upward to downward or from downward to upward) at $x_{0}$ .

Diagram showing four types of inflection points where concavity changes and the tangent line crosses the curve — At an inflection point, the slope changes from increasing to decreasing or vice versa, and the curve crosses the tangent line.

At an inflection point, if f''(x_0) is continuous, the sign of f'' changes there, so f''(x_0) must equal zero by the Intermediate Value Theorem. Even without continuity of f'', if f''(x_0) exists it must be zero. However, f''(x_0) may simply not exist at an inflection point.

If $(x_{0}, f (x_{0}))$ is an inflection point of the graph of $f$ , then either f''(x_0)=0 or f''(x_0) does not exist.

Caution: The converse is false. Even if f''(x_0)=0, the point need not be an inflection point.

Examples of Inflection Points

Find the inflection points of:

(a) $f (x) = 2 x^{2} - 5 x - 7$

(b) $f (x) = x^{3}$

Solution

**(a)** We saw f''(x)=4>0 everywhere, so concavity never changes. The graph has **no inflection points**.

Graph of y = 2x^2 - 5x - 7 with no inflection points — $f (x) = 2 x^{2} - 5 x - 7$ is always concave up; no inflection points.

**(b)**

f

is concave down on

(- \infty, 0)

and concave up on

(0, + \infty)

. Since

f

is continuous at

x = 0

and the concavity changes there,

(0, f (0)) = (0, 0)

is an inflection point.

Graph of y = x^3 with the inflection point at the origin labeled — $(0,0)$ is the inflection point of $y = x^{3}$ .

**(c)** The concavity changes at

x = 1

. Since

f

is continuous everywhere:

(1, f (1)) = (1, 1 - 3 + 1) = (1, - 1)

is an inflection point.

Graph of y = x^3 - 3x^2 + 1 with inflection point labeled at (1, -1) — $(1, - 1)$ is the inflection point of $y = x^{3} - 3 x^{2} + 1$ .

Determine the concavity and find the inflection points of $f (x) = x - \cos x$ on $[- π, π]$ .

Solution

f'(x)=1+\sin x\Rightarrow f''(x)=\cos x.

\cos x > 0

when

- π / 2 < x < π / 2

, so

f

is concave up on

(- π / 2, π / 2)

\cos x < 0

when

- π < x < - π / 2

π / 2 < x < π

, so

f

is concave down on those intervals. The concavity changes at

x = \pm π / 2

(- \frac{π}{2}, f (- \frac{π}{2})) = (- \frac{π}{2}, - \frac{π}{2}), (\frac{π}{2}, f (\frac{π}{2})) = (\frac{π}{2}, \frac{π}{2}) .

Graph of f(x) = x - cos(x) showing inflection points at (±pi/2, ±pi/2) — Graph of $f (x) = x - \cos x$ , with inflection points at $(\pm π / 2, \pm π / 2)$ .

Find the inflection points of $f (x) = \sqrt[3]{x}$ .

Solution

f'(x)=\frac{1}{3}x^{-2/3}\Rightarrow f''(x)=-\frac{2}{9}x^{-5/3}=-\frac{2}{9\sqrt[3]{x^5}}. Although f''(0) does not exist: f''(x)>0 when

x < 0

(concave up on

(- \infty, 0)

), f''(x)<0 when

x > 0

(concave down on

(0, + \infty)

). Since the concavity changes at

x = 0

and

f

is continuous there, $(0,0)$ is an inflection point.

Graph of y = x^(1/3) with inflection point at the origin and a vertical tangent there — Graph of $y = x^{1 / 3}$ . The origin is an inflection point with a vertical tangent.

Find the inflection points of

f(x)=\begin{cases}x^2 & \text{if }x\geq0\\ -x^2 & \text{if }x<0.\end{cases}

Solution

f'(x)=\begin{cases}2x & x\geq0\\ -2x & x<0,\end{cases}\qquad f''(x)=\begin{cases}2 & x>0\\ -2 & x<0.\end{cases}

f

is concave down on

(- \infty, 0)

and concave up on

(0, + \infty)

. The function is continuous at

x = 0

, so $(0,0)$ is an inflection point even though f''(0) does not exist.

Graph of the piecewise function showing the inflection point at the origin where f'' is undefined — $(0,0)$ is an inflection point although f''(0) does not exist.

Determine the inflection points of $f (x) = x^{4}$ .

Solution

f'(x)=4x^3\Rightarrow f''(x)=12x^2. f''(x)=12x^2\geq0 for all

x

, with equality only at

x = 0

. The concavity does not change. So even though f''(0)=0, the point $(0,0)$ is **not** an inflection point.

Graph of y = x^4 showing the graph is always concave up with no inflection point

How to Find Inflection Points

To find inflection points of $f$ :

Find f''(x).
Find all $x_{0}$ where f''(x_0)=0 or f''(x_0) does not exist.
Check whether the sign of f'' actually changes at each candidate $x_{0}$ and whether $f$ is continuous there.
If both conditions hold, $(x_{0}, f (x_{0}))$ is an inflection point.

Note on definitions. Some textbooks require a tangent line to exist at an inflection point. Under the definition used in this book, continuity alone (not differentiability) is required; inflection points with vertical tangents are permitted.

Frequently Asked Questions

Can an inflection point occur where the second derivative does not exist?

Yes. The function

y = x^{1 / 3}

has an inflection point at the origin, where f''(0) does not exist. What matters is that the concavity changes and the function is continuous there.

If f''(x_0) = 0, is (x_0, f(x_0)) necessarily an inflection point?

No. For

f (x) = x^{4}

, f''(0)=0 but $(0,0)$ is not an inflection point, the graph is concave up on both sides of

x = 0

. Always check whether the sign of f'' actually changes.

What is the relationship between concavity and the tangent line?

If a function is concave up on an interval, its graph lies entirely above each tangent line (except at the point of tangency). If it is concave down, the graph lies entirely below each tangent line. At an inflection point, the graph crosses its tangent from one side to the other.

Concavity

Examples

Concavity and Tangent Lines

Points of Inflection

Examples of Inflection Points

How to Find Inflection Points

Frequently Asked Questions

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