Concavity and points of inflection

Concavity and points of Inflection
Calculus I
Calculus II
Chemistry
Physics
Oceanography

By Rick Stoll

While the tangent line is a very useful tool, when it comes to investigate the graph of a function, the tangent line fails to say anything about how the graph of a function "bends" at a point. This is where the second derivative comes into play.

Example. Consider the function f(x) = ax². The tangent line at 0 is the x-axis regardless of the value of a. But if we change a, the graph of f(x) bends more or less sharply depending on the size of the parameter a:

Note that the value a is directly related to the second derivative, since f ''(x) = 2a.

Definition. Let f(x) be a differentiable function on an interval I.

(i): We will say that the graph of f(x) is concave up on I iff f '(x) is increasing on I.
(ii): We will say that the graph of f(x) is concave down on I iff f '(x) is decreasing on I.

Some authors use concave for concave down and convex for concave up instead. Usually graphs have regions which are concave up and others which are concave down. Thus there are often points at which the graph changes from being concave up to concave down, or vice versa. These points are called inflection points. Since the monotonicity behavior of a function is related to the sign of its derivative we get the following result:

Let f(x) be a differentiable function on an interval I. Assume that f '(x) is also differentiable on I.

(i): f(x) is concave up on I iff $f''(x) \geq 0$ on I.
(ii): f(x) is concave down on I iff $f''(x) \leq 0$ on I.

It is clear from this result that if c is an inflection point then we must have

$\begin{displaymath}f''(c) = 0\;\;\mbox{or $f''(c)$ does not exist}.\end{displaymath}$

Example. Consider the function f(x) = x^9/5 - x. This function is continuous and differentiable for all x. We have

$\begin{displaymath}f'(x) = \frac{9}{5} x^{4/5} - 1.\end{displaymath}$

Clearly f ''(0) does not exist. In fact, f '(x) has a vertical tangent at 0. More precisely we have for $x\not=0$

$\begin{displaymath}f''(x) = \frac{36}{25} x^{-1/5} \end{displaymath}$

which implies

(1): f ''(x) > 0 for x > 0;
(2): f ''(x) < 0 for x < 0.

In other words, the point 0 is a point of inflection even though f ''(0) does not exist.

Remark. Note that if the graph is concave up (resp. concave down), then the tangent line at any point is below (resp. above) the graph. Therefore, at an inflection point the graph "cuts" through the tangent line.

Exercise 1. Describe the concavity of the graph of

$\begin{displaymath}f(x) = 2 \sin^2(x) - x^2\end{displaymath}$

for $x \in [0, \pi]$ .

Exercise 2. Find $\alpha$ and $\beta$ so that the function

$\begin{displaymath}f(x) = \alpha x^3 + \beta x^2+1\end{displaymath}$

has a point of inflection at $\displaystyle \left(-1,2\right)$ .

Send mail to m_engineer2002@netzero.com with questions or comments about this web site. Copyright © 2001 Math Depot Last modified: April 30, 2001

Send mail to m_engineer2002@netzero.com with questions or comments about this web site.
Copyright © 2001 Math Depot
Last modified: April 30, 2001