Concavity

A function may increase or decrease, but its manner of doing so can differ significantly. Some functions rise in such a way that they bend upwards, while others rise but bend downwards. Similarly, a function that decreases may do so while curving in either direction. To distinguish these behaviors, we introduce the concept of concavity, which describes how a function bends in relation to its secant lines.

A function \( f \) defined on an interval \( I \) is said to be concave upwards (also called convex downwards) if, for any two points on its graph, the function lies below or on the secant line connecting them. This means that if we take any two points \( x_1, x_2 \) in the domain and consider an intermediate point between them, the function value at that point does not exceed the corresponding value on the secant line.

Conversely, \( f \) is concave downwards (also called convex upwards) if the function lies above or on all secant lines. This means that for any two points in the domain, the function value at an intermediate point is at least as large as the corresponding secant line value.

These conditions provide a way to determine whether a function bends upwards or downwards without having to graph it explicitly. To express these ideas precisely, we introduce the secant line condition.

Definition of Concavity Using Secants

Consider two points \( (x_1, f(x_1)) \) and \( (x_2, f(x_2)) \) on the graph of \( f \), where \( x_1 < x_2 \). The secant line connecting these points is the unique linear function given by

\[ S(x) = f(x_1) + \frac{f(x_2) - f(x_1)}{x_2 - x_1} (x - x_1). \]

To analyze how \( f \) behaves relative to this secant line, take an intermediate point \( x \) between \( x_1 \) and \( x_2 \). Expressing \( x \) in terms of a parameter \( \lambda \), we write

\[ x = \lambda x_1 + (1-\lambda)x_2, \quad \text{where } 0 \leq \lambda \leq 1. \]

This representation ensures that \( x \) remains between \( x_1 \) and \( x_2 \), with \( \lambda = 1 \) corresponding to \( x_1 \) and \( \lambda = 0 \) corresponding to \( x_2 \).

Substituting this value into the secant line equation,

\[ S(\lambda x_1 + (1-\lambda)x_2) = f(x_1) + \frac{f(x_2) - f(x_1)}{x_2 - x_1} \big( (\lambda x_1 + (1-\lambda)x_2) - x_1 \big). \]

Since

\[ (\lambda x_1 + (1-\lambda)x_2) - x_1 = (1-\lambda)(x_2 - x_1), \]

it follows that

\[ S(\lambda x_1 + (1-\lambda)x_2) = f(x_1) + \frac{f(x_2) - f(x_1)}{x_2 - x_1} (1-\lambda)(x_2 - x_1). \]

Canceling \( x_2 - x_1 \) in the fraction,

\[ S(\lambda x_1 + (1-\lambda)x_2) = f(x_1) + (1-\lambda)(f(x_2) - f(x_1)). \]

Rewriting terms,

\[ S(\lambda x_1 + (1-\lambda)x_2) = (1-\lambda) f(x_2) + \lambda f(x_1). \]

A function \( f \) is concave upwards (convex downwards) on \( I \) if, for all \( x_1, x_2 \in I \) with \( x_1 < x_2 \) and for all \( \lambda \in [0,1] \),

\[ f(\lambda x_1 + (1-\lambda)x_2) \leq (1-\lambda) f(x_2) + \lambda f(x_1). \]

This condition states that at every intermediate point, the function value is at most the corresponding secant line value. That is, the function lies below or on every secant line, forming an upward-bending shape.

Similarly, \( f \) is concave downwards (convex upwards) on \( I \) if, for all \( x_1, x_2 \in I \) and all \( \lambda \in [0,1] \),

\[ f(\lambda x_1 + (1-\lambda)x_2) \geq (1-\lambda) f(x_2) + \lambda f(x_1). \]

This means that at every intermediate point, the function value is at least the corresponding secant line value. That is, the function lies above or on every secant line, forming a downward-bending shape.

If \( f \) is concave upwards (convex downwards), its graph has a shape resembling a bowl (\(\smile\)). Every secant line lies above or on the function.
If \( f \) is concave downwards (convex upwards), its graph has a shape resembling an inverted bowl (\(\frown\)). Every secant line lies below or on the function.

Example

Consider the function

\[ f(x) = |x|. \]

This function is continuous everywhere but not differentiable at \( x = 0 \).

Observing the graph of \( f(x) = |x| \), we see that for any two points chosen on the function, the secant line connecting them either lies on the graph itself or above the graph. The function never rises above any of its secant lines.

This confirms that \( |x| \) satisfies the secant definition of concave upwards on \( \mathbb{R} \). The graph bends upwards like a bowl, ensuring that every secant line stays above or touches the curve.

Defining Concavity using Tangents

A function’s concavity describes how its graph bends—whether it curves upwards like a bowl or downwards like an arch. Previously, concavity was defined using secant lines, which compare the function’s values at two different points. We can also define concavity using tangent lines. While secant lines examine the function’s behavior over an interval, tangent lines describe how the function bends at every individual point.

The tangent line to a function \( f \) at a point \( x \) is the unique straight line that just touches the curve at \( x \) without crossing it locally. The equation of the tangent line at \( x \) is given by:

\[ T_x(y) = f(x) + m(y - x), \]

where \( m \) is the slope of the tangent at \( x \). If \( f \) is differentiable, the slope is given by \( m = f'(x) \), so the equation simplifies to:

\[ T_x(y) = f(x) + f'(x)(y - x). \]

This equation tells us how the function behaves in a small neighborhood around \( x \). To determine concavity, we compare the function \( f(y) \) with its tangent line \( T_x(y) \) for all points \( y \) in the interval.

A function \( f \) is concave upwards on an interval \( I \) if, for every \( x \in I \) and for all \( y \in I \),

\[ f(y) \geq T_x(y). \]

This means that at every point \( x \), the function lies above or on its tangent line. The graph never dips below any of its tangent lines.

Similarly, \( f \) is concave downwards on \( I \) if, for every \( x \in I \) and for all \( y \in I \),

\[ f(y) \leq T_x(y). \]

This means that at every point \( x \), the function lies below or on its tangent line. The graph never rises above any of its tangent lines.

Concavity and the Behavior of the First Derivative

We can also characterize concavity by examining how the slope of the function changes across an interval. The function’s derivative provides a direct way to determine whether the function bends upwards or downwards.

A function \( f \) defined on an interval \( I \) is said to be concave upwards if its slope increases as \( x \) moves from left to right. That is, at any two points in the interval, the slope at the later point is greater than or equal to the slope at the earlier point. To ensure a well-defined notion of concavity at every point, we require that \( f \) be continuous on \( I \), though it may not necessarily be differentiable everywhere.

Formally, a function \( f \) that is continuous on \( I \) is concave upwards on \( I \) if \( f' \), wherever it exists, is an increasing function on \( I \). That is, for any \( x_1, x_2 \in I \) with \( x_1 < x_2 \),

\[ f'(x_1) \leq f'(x_2), \]

where \( f'(x) \) exists.

Similarly, \( f \) is concave downwards on \( I \) if its slope decreases as \( x \) moves from left to right. That is, the derivative, wherever it exists, is decreasing across the interval. Formally, a function \( f \) that is continuous on \( I \) is concave downwards on \( I \) if \( f' \), wherever it exists, satisfies

\[ f'(x_1) \geq f'(x_2), \]

for all \( x_1, x_2 \in I \) with \( x_1 < x_2 \).

Consider the function \( f(x) = x^2 \). To determine whether it is concave upwards or downwards, we analyze how its slope behaves. The derivative of \( f \) is

\[ f'(x) = 2x. \]

Since \( f'(x) \) increases as \( x \) increases, this suggests that the function bends upwards. More formally, concavity is established by checking whether \( f' \) is increasing. Computing the second derivative,

\[ f''(x) = 2, \]

we see that it is positive for all \( x \). This confirms that \( f' \) is strictly increasing, meaning that \( f \) satisfies the definition of concave upwards.

Visually, the function \( x^2 \) curves upwards at every point, never flattening out or bending downwards. The slopes of the tangent lines become steeper as \( x \) moves from left to right. This example illustrates how the second derivative provides a direct method to determine concavity, without needing to compare secants or examine specific values of \( f' \).

Using the Second Derivative to Decide Concavity

First, let \( f \) be a function that is continuous on an interval \( I \) but possibly non-differentiable at some points. If \( f \) is concave upwards on \( I \), then the second derivative, wherever it exists, must be nonnegative:

\[ f''(x) \geq 0 \quad \text{for all } x \text{ where } f''(x) \text{ exists}. \]

Similarly, if \( f \) is concave downwards on \( I \), then

\[ f''(x) \leq 0 \quad \text{for all } x \text{ where } f''(x) \text{ exists}. \]

This condition provides a necessary test for concavity, meaning that if \( f \) is known to be concave upwards, then \( f''(x) \geq 0 \) must hold at all points where \( f''(x) \) exists.

Examples

Consider \( f(x) = e^x \). Its derivatives are

\[ f'(x) = e^x, \quad f''(x) = e^x. \]

Since \( f''(x) > 0 \) for all \( x \in \mathbb{R} \), this confirms that \( f \) is concave upwards everywhere.
Consider \( f(x) = |x| \). The first derivative is

\[ f'(x) = \begin{cases} 1, & x > 0, \\ -1, & x < 0. \end{cases} \]

Since \( f' \) is not continuous at \( x = 0 \), the second derivative does not exist at \( x = 0 \). However, for \( x \neq 0 \),

\[ f''(x) = 0 \quad \text{for all } x \in \mathbb{R} \setminus \{0\}. \]

This suggests that \( f''(x) = 0 \) almost everywhere.

The Converse is Not True

The previous conditions only provide a necessary test, but not a sufficient one. Consider the function \( f(x) = |x| \). As observed, \( f''(x) = 0 \) for all \( x \neq 0 \), but the same would be true for another function, such as \( g(x) = -|x| \), which has

\[ g''(x) = 0 \quad \text{for all } x \in \mathbb{R} \setminus \{0\}. \]

However, \( f(x) = |x| \) is concave upwards, while \( g(x) = -|x| \) is concave downwards, as we can understand from secant analysis. This shows that knowing only the second derivative is not enough to determine concavity: two functions with different concavity can have the same second derivative at all points where it exists.

Take another example:

Consider the function

\[ f(x) = e^{-|x|} \]

for \( x \in \mathbb{R} \). This function is continuous on \( \mathbb{R} \) because both its left-hand and right-hand expressions match at \( x = 0 \), and exponentials are continuous everywhere.

Now, we examine its differentiability. The function can be rewritten as a piecewise function:

\[ f(x) = \begin{cases} e^{-x}, & x \geq 0 \\ e^{x}, & x < 0 \end{cases} \]

Since exponentials are differentiable everywhere, we only need to check at \( x = 0 \). Computing the derivatives on either side,

\[ f'(x) = \begin{cases} - e^{-x}, & x > 0 \\ e^x, & x < 0 \end{cases} \]

Taking the limits,

\[ \lim_{x \to 0^+} f'(x) = -1, \quad \lim_{x \to 0^-} f'(x) = 1. \]

Since these limits are not equal, \( f'(x) \) is discontinuous at \( x = 0 \), implying that \( f(x) \) is not differentiable at \( x = 0 \).

Next, consider the second derivative:

\[ f''(x) = \begin{cases} e^{-x}, & x > 0 \\ e^x, & x < 0. \end{cases} \]

Since both expressions are positive for \( x \neq 0 \), we conclude that \( f''(x) > 0 \) for all \( x \neq 0 \).

To analyze concavity, we examine the behavior of secants. If both endpoints of a secant are within the same branch (i.e., either both in \( x > 0 \) or both in \( x < 0 \)), the function appears concave upward. But if the endpoints belong to different branches (one in \( x > 0 \) and the other in \( x < 0 \)), then the secant lies below the function, contradicting concavity.

Thus, while \( f(x) \) is concave upward separately for \( x > 0 \) and \( x < 0 \), that is it is piecewise concave upwards, it fails to be globally concave upwards or downwards on \( \mathbb{R} \).

A Stricter Condition: The Second Derivative as a Sufficient Test

To make a stronger claim, we require a stricter assumption. Suppose \( f \) is differentiable on \( I \) (not just continuous). Then we can indeed use the second derivative to decide concavity.

If \( f''(x) \geq 0 \) wherever it exists for all \( x \in I \), then \( f \) is concave upwards on \( I \).
If \( f''(x) \leq 0 \) wherever it exists for all \( x \in I \), then \( f \) is concave downwards on \( I \).

If \( f \) is only continuous but not differentiable, the second derivative test cannot always be applied, as we saw in examples.

Consider the function

\[ f(x) = x^2 |x|, \quad x \in \mathbb{R}. \]

Since \( |x| \) is continuous and polynomial functions are continuous, \( f(x) \) is continuous on \( \mathbb{R} \). To examine its differentiability, rewrite it in piecewise form:

\[ f(x) = \begin{cases} x^3, & x \geq 0 \\ - x^3, & x < 0. \end{cases} \]

This function is differentiable everywhere except possibly at \( x = 0 \). Compute its derivative separately for \( x > 0 \) and \( x < 0 \):

\[ f'(x) = \begin{cases} 3x^2, & x > 0 \\ -3x^2, & x < 0. \end{cases} \]

Taking the limit as \( x \to 0 \),

\[ \lim_{x \to 0^+} f'(x) = 0, \quad \lim_{x \to 0^-} f'(x) = 0. \]

Since these limits agree, \( f'(x) \) exists at \( x = 0 \), and defining \( f'(0) = 0 \) ensures \( f'(x) \) is continuous everywhere. Thus,

\[ f'(x) = \begin{cases} 3x^2, & x \geq 0 \\ -3x^2, & x < 0. \end{cases} \]

Proceeding to the second derivative,

\[ f''(x) = \begin{cases} 6x, & x > 0 \\ -6x, & x < 0. \end{cases} \]

Evaluating at \( x = 0 \),

\[ \lim_{x \to 0^+} f''(x) = 0, \quad \lim_{x \to 0^-} f''(x) = 0. \]

Since these limits agree, define \( f''(0) = 0 \), ensuring \( f''(x) \) exists for all \( x \in \mathbb{R} \). Thus,

\[ f''(x) = \begin{cases} 6x, & x \geq 0 \\ -6x, & x < 0. \end{cases} \]

Since \( f''(x) \geq 0 \) for all \( x \in \mathbb{R} \), the function is concave upwards everywhere. Here is the graph:

Take another example:

Consider the function

\[ f(x) = x^{\frac{4}{3}}, \quad x \in \mathbb{R}. \]

Since power functions are continuous, \( f(x) \) is continuous for all \( x \). Compute its derivative:

\[ f'(x) = \frac{4}{3} x^{\frac{1}{3}}. \]

This function is differentiable everywhere since \( x^{\frac{1}{3}} \) is defined for all \( x \in \mathbb{R} \). Proceeding to the second derivative,

\[ f''(x) = \frac{4}{9} x^{-\frac{2}{3}}. \]

For all \( x \neq 0 \), \( f''(x) \) exists and satisfies

\[ f''(x) > 0. \]

Even though \( f''(x) \) does not exist at \( x = 0 \), it is positive wherever defined, ensuring \( f(x) \) is concave upwards everywhere. Here is its graph:

Critical Points of second kind and finding intervals of Concavity

In the study of monotonicity, we identify critical points where a function might change from increasing to decreasing or vice versa. Analogously, to determine the concavity of a function, we seek points where the function might transition between concave upward and concave downward behavior. These points serve as candidates for changes in concavity and allow us to partition the domain into intervals where the function exhibits uniform concavity.

Let \( f \) be a function that is differentiable on an open interval \( (a, b) \). A point \( c \in (a, b) \) is said to be a critical point of second kind if the second derivative \( f''(c) \) either vanishes, i.e.,

\[ f''(c) = 0, \]

or fails to exist. Such points are potential locations where the concavity of \( f \) may change. To determine the precise nature of concavity in different regions, we analyze the sign of \( f''(x) \) on the intervals defined by these points.

Consider the function

\[ f(x) = x^3 - 3x + 1. \]

To determine where the function is concave upward or concave downward, we analyze its second derivative.

Differentiating,

\[ f'(x) = 3x^2 - 3. \]

Differentiating again,

\[ f''(x) = 6x. \]

A critical point of second kind occurs where \( f''(x) = 0 \) or where \( f''(x) \) is undefined. Since \( f''(x) = 6x \) is defined everywhere, we solve

\[ 6x = 0 \implies x = 0. \]

This is a candidate for a change in concavity.

To determine the concavity on either side of \( x = 0 \), we check the sign of \( f''(x) \) in the intervals \( (-\infty,0) \) and \( (0,\infty) \). For \( x < 0 \), choosing \( x = -1 \) gives

\[ f''(-1) = 6(-1) = -6 < 0, \]

which implies \( f(x) \) is concave downward for \( x < 0 \).

For \( x > 0 \), choosing \( x = 1 \) gives

\[ f''(1) = 6(1) = 6 > 0, \]

which implies \( f(x) \) is concave upward for \( x > 0 \).

Hence, the function is concave downward on \( (-\infty,0) \) and concave upward on \( (0,\infty) \).

General Procedure to find intervals of concavity

To determine the intervals where a function is concave upward or concave downward, follow these general steps:

Ensure Differentiability:

The function \( f(x) \) must be at least differentiable on the given interval (or on subintervals where differentiability holds). If \( f(x) \) is not differentiable at some points, then we apply general secant method.
Compute the Second Derivative:

Differentiate \( f(x) \) twice to obtain \( f''(x) \). That is,

\[ f''(x) = \frac{d^2}{dx^2} f(x). \]
Find Critical Points of second kind:

Solve for \( x \) where \( f''(x) = 0 \) or where \( f''(x) \) does not exist. These points are candidates where concavity might change.
Determine Concavity on Intervals:

The points obtained in the previous step partition the domain into subintervals. Choose a test point in each interval and evaluate \( f''(x) \) at that point:
- If \( f''(x) > 0 \) on an interval, the function is concave upward there.
- If \( f''(x) < 0 \) on an interval, the function is concave downward there.

A sign graph of \( f''(x) \) can be useful for visualizing how the second derivative behaves across the domain, aiding in determining concavity changes systematically.

Inflection Points

Definition

An inflection point of a function \( f \) is a point where the concavity of \( f \) changes, meaning \( f \) transitions from being concave upward to concave downward or vice versa. However, a mere change in concavity is not sufficient to conclude that a point is an inflection point. A change in concavity can occur at points of discontinuity. Such points do not qualify as inflection points.

Points of discontinuities cannot be inflection points, even though there may be a change in concavity there.

An inflection point of a function \( f \) is a point where the function changes concavity. That is, the function transitions from being concave up to concave down, or vice versa.

Let \( f \) be a continuous function on an open interval \( (a, b) \), and suppose \( c \in (a, b) \) is a point where \( f \) is twice differentiable in a neighborhood of \( c \) (except possibly at \( c \) itself). Then, \( c \) is an inflection point if the second derivative changes sign around \( c \), meaning:

\[ f''(x) > 0 \quad \text{for } x < c, \quad \text{and} \quad f''(x) < 0 \quad \text{for } x > c, \]

or vice versa:

\[ f''(x) < 0 \quad \text{for } x < c, \quad \text{and} \quad f''(x) > 0 \quad \text{for } x > c. \]

Thus, the sign of \( f''(x) \) must change as \( x \) moves through \( c \).

The Role of \( f''(c) \)

If \( f''(c) \) exists, then necessarily \( f''(c) = 0 \), since a change in sign would require passing through zero.
However, \( f''(c) \) does not need to exist at \( c \) for \( c \) to be an inflection point. The function can still change concavity if \( f''(x) \) is positive on one side of \( c \) and negative on the other, even if \( f''(c) \) is undefined.

Example of an Inflection Point with \( f''(c) = 0 \)

Consider \( f(x) = x^3 \), which is twice differentiable everywhere:

\[ f''(x) = 6x. \]

Here, \( f''(x) \) changes sign at \( x = 0 \), and since \( f''(0) = 0 \), \( x = 0 \) is an inflection point.

Example of an Inflection Point Where \( f''(c) \) Does Not Exist

Consider the function

\[ f(x) = x^{\frac{1}{3}}, \quad x \in \mathbb{R}. \]

Differentiating,

\[ f'(x) = \frac{1}{3} x^{-\frac{2}{3}}. \]

For \( x > 0 \), \( f'(x) \) is positive and finite. For \( x < 0 \), \( f'(x) \) is also positive and finite. However, at \( x = 0 \),

\[ \lim_{x \to 0^+} f'(x) = +\infty, \quad \lim_{x \to 0^-} f'(x) = +\infty. \]

Since these limits are infinite, \( f'(0) \) does not exist, meaning the function has a vertical tangent at \( x = 0 \).

Differentiating again,

\[ f''(x) = -\frac{2}{9} x^{-\frac{5}{3}}, \quad x \neq 0. \]

For \( x > 0 \), \( f''(x) < 0 \), indicating concave down behavior. For \( x < 0 \), \( f''(x) > 0 \), indicating concave up behavior. Since \( f''(x) \) changes sign at \( x = 0 \), the function has an inflection point at \( x = 0 \), even though \( f''(0) \) does not exist.

How to find inflection points

Given a continuous function \( f \) on an open interval \( (a, b) \), to determine its inflection points:

Differentiate \( f(x) \) twice to obtain \( f''(x) \). Solve for points where \( f''(x) = 0 \) or where \( f''(x) \) does not exist. These points are called critical points of the second kind.

To confirm an inflection point at \( x = c \), check the sign of \( f''(x) \) on either side of \( c \). If \( f''(x) \) changes sign at \( c \), then \( c \) is an inflection point.

Example: \( f(x) = x^{5/3} \)

Compute the first derivative:

\[ f'(x) = \frac{5}{3} x^{2/3}. \]

Compute the second derivative:

\[ f''(x) = \frac{10}{9} x^{-1/3}. \]

The second derivative does not exist at \( x = 0 \). For \( x > 0 \), \( f''(x) > 0 \) (concave up). For \( x < 0 \), \( f''(x) <> 0 \) (concave down). Since \( f''(x) \) changes sign at \( x = 0 \), there is a inflection point at \( x = 0 \).

Example

Find the intervals in which \( f(x) = \frac{1}{1 + x^2} \) is concave upwards or concave downwards.

Solution:

The function \( f(x) \) is continuous and differentiable for all \( x \in \mathbb{R} \), as the denominator \( 1 + x^2 \) is never zero.

Differentiating,

\[ f'(x) = \frac{-2x}{(1 + x^2)^2}. \]

Differentiating again,

\[ f''(x) = -2 \cdot \frac{(1 + x^2)^2 - 2(1 + x^2) \cdot 2x \cdot x}{(1 + x^2)^4} \]

\[ = -2 (1 + x^2) \frac{(1 + x^2 - 4x^2)}{(1 + x^2)^4} \]

\[ = \frac{2(3x^2 - 1)}{(1 + x^2)^3}. \]

Setting \( f''(x) = 0 \), we solve

\[ 2(3x^2 - 1) = 0 \implies 3x^2 - 1 = 0 \implies x^2 = \frac{1}{3} \implies x = \pm \frac{1}{\sqrt{3}}. \]

Thus, \( x = \pm \frac{1}{\sqrt{3}} \) are critical points of second kind.

Examining the sign of \( f''(x) \):

For \( x \in (-\infty, -\frac{1}{\sqrt{3}}) \), we pick \( x = -1 \). Substituting in \( f''(x) \), we get \( f''(x) > 0 \), so \( f \) is concave upward.
For \( x \in (-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}) \), we pick \( x = 0 \). Substituting, we get \( f''(x) < 0 \), so \( f \) is concave downward.
For \( x \in (\frac{1}{\sqrt{3}}, \infty) \), we pick \( x = 1 \). Substituting, we get \( f''(x) > 0 \), so \( f \) is concave upward.

Thus, \( f(x) \) is concave downward on \( \left(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \) and concave upward on \( \left(-\infty, -\frac{1}{\sqrt{3}}\right) \cup \left(\frac{1}{\sqrt{3}}, \infty\right) \).

Since \( f''(x) \) changes sign at \( x = \pm \frac{1}{\sqrt{3}} \), these points are inflection points.

Example

Find the inflection points and concavity of the function \( f(x) = \sin x \).

Solution:

The function \( f(x) = \sin x \) is continuous and differentiable for all \( x \in \mathbb{R} \).

First, we compute the derivatives:

\[ f'(x) = \cos x. \]

\[ f''(x) = -\sin x. \]

Setting \( f''(x) = 0 \), we solve

\[ -\sin x = 0 \implies \sin x = 0. \]

Since \( \sin x = 0 \) at \( x = n\pi \), where \( n \in \mathbb{Z} \), all such points are critical points of second kind.

From the graph of \( -\sin x \), we observe that for all \( x = n\pi \), \( f''(x) \) changes sign.

Thus, at every \( x = n\pi \), there is a transition between concave up and concave down, confirming that all \( x = 2n\pi \) are inflection points.

Hence, \( f(x) \) is concave downward on \( (n\pi, (n+1)\pi) \) for even \( n \) and concave upward on \( (n\pi, (n+1)\pi) \) for odd \( n \).

Example

Find the intervals in which \( f(x) = x^4 \) is concave upwards or concave downwards and determine its inflection points.

Solution:

The function \( f(x) = x^4 \) is continuous and differentiable everywhere.

First, we compute the derivatives:

\[ f'(x) = 4x^3. \]

\[ f''(x) = 12x^2. \]

Setting \( f''(x) = 0 \), we solve

\[ 12x^2 = 0 \implies x^2 = 0 \implies x = 0. \]

Thus, \( x = 0 \) is a critical point of second kind.

Now, analyzing the sign of \( f''(x) = 12x^2 \):

For \( x < 0 \), we observe that \( f''(x) = 12x^2 > 0 \).
For \( x > 0 \), we also observe that \( f''(x) = 12x^2 > 0 \).

Since \( f''(x) \) does not change sign at \( x = 0 \), it follows that \( x = 0 \) is not an inflection point.

Since \( f''(x) >= 0 \) for all \( x \in \mathbb{R} \), the function is concave upward for all \( x \in \mathbb{R} \) and has no points of inflection.

Example

Find the intervals in which \( f(x) = x e^{-x} \) is concave upwards or concave downwards and determine its inflection points.

Solution:
The function \( f(x) = x e^{-x} \) is continuous and differentiable everywhere.

First derivative:

Using the product rule,

\[ f'(x) = e^{-x} + x(-e^{-x}) = e^{-x} (1 - x). \]

Second derivative:

Differentiating again,

\[ f''(x) = e^{-x}(-1) (1 - x) + e^{-x}(-1) = e^{-x} (x - 2). \]

Finding critical points of second kind:

Setting \( f''(x) = 0 \),

\[ e^{-x} (x - 2) = 0. \]

Since \( e^{-x} \) is never zero, we get

\[ x - 2 = 0 \implies x = 2. \]

Thus, \( x = 2 \) is a critical point of second kind.

Analyzing the sign of \( f''(x) \):

For \( x < 2 \), say \( x = 0 \), we have \( f''(0) = e^0(0 - 2) = -2 < 0 \), so \( f(x) \) is concave downward.
For \( x > 2 \), say \( x = 3 \), we have \( f''(3) = e^{-3} (3 - 2) > 0 \), so \( f(x) \) is concave upward.

Since \( f''(x) \) changes sign at \( x = 2 \), we conclude that

\[ x = 2 \quad \text{is an **inflection point**}. \]

Hence,

\( f(x) \) is concave downward on \( (-\infty, 2) \).
\( f(x) \) is concave upward on \( (2, \infty) \).
The inflection point is \( x = 2 \).

Example

Find the inflection points of

\[ f(x) = \begin{cases} x^2, & x < 0, \\ \sqrt{x}, & x \geq 0. \end{cases} \]

The function is continuous on \( \mathbb{R} \), as both \( x^2 \) and \( \sqrt{x} \) match at \( x = 0 \):

\[ \lim_{x \to 0^-} f(x) = 0, \quad \lim_{x \to 0^+} f(x) = 0, \quad f(0) = 0. \]

Differentiate separately for \( x < 0 \) and \( x > 0 \):

\[ f'(x) = \begin{cases} 2x, & x < 0, \\ \frac{1}{2\sqrt{x}}, & x > 0. \end{cases} \]

At \( x = 0 \), \( f'(x) \) does not exist, since

\[ \lim_{x \to 0^-} f'(x) = 0, \quad \lim_{x \to 0^+} f'(x) = +\infty. \]

Differentiate again:

\[ f''(x) = \begin{cases} 2, & x < 0, \\ -\frac{1}{4} x^{-3/2}, & x > 0. \end{cases} \]

For \( x < 0 \), \( f''(x) = 2 \) (concave up). For \( x > 0 \), \( f''(x) < 0 \) (concave down). Since \( f''(x) \) does not exist at \( x = 0 \), but there is a sign change, \( x = 0 \) is an inflection point.