Inflection Points
An inflection point is a point on the graph of a function where the concavity changes.
Let $ y=f(x) $ be a function defined and continuous on an interval \(I\) and differentiable throughout the interval, except possibly at a single point \(x_0\). The point \((x_0,f(x_0))\) is an inflection point if the graph changes concavity as it passes through \(x_0\).
In practical terms, the curve bends differently on the two sides of an inflection point.
For example, the graph may be concave down before the inflection point and concave up after it. The opposite change can also occur.
- The tangent line at an inflection point
- How to classify an inflection point
- Inflection points and tangent lines
- From concave down to concave up
- From concave up to concave down
- Stationary Points of Inflection: Conditions and Theorem
- How to Find Inflection Points Using the Second Derivative
- Higher-Order Derivative Test (Inflection Points with Horizontal Tangents)
- Higher-Order Derivative Test for Inflection Points with Nonhorizontal Tangents
The tangent line at an inflection point
If the function is differentiable at an inflection point, the graph has a tangent line at that point.
The tangent line may be horizontal or oblique. An inflection point may also have a vertical tangent, in which case the ordinary derivative at that point is not finite.
For a sufficiently smooth curve with a well-defined tangent at the inflection point, the curve typically crosses its tangent line at the point of inflection.
This geometric behavior is closely related to the change in concavity that occurs at the inflection point.
Why? Consider an inflection point \(F\) and a secant line that intersects the curve at \(A\), \(F\), and \(B\). As \(A\) and \(B\) move closer and closer to \(F\), the secant line approaches the tangent line. In the limiting configuration, the three intersections approach \(F\). For sufficiently smooth algebraic curves, this behavior can be described by saying that the tangent line has an odd order of contact with the curve at an ordinary inflection point.
How to classify an inflection point
There are two useful ways to describe an inflection point.
The first is based on the direction of the tangent line, when a tangent exists.
- If the tangent line is parallel to the \(x\)-axis, the inflection point has a horizontal tangent.
- If the tangent line is parallel to the \(y\)-axis, the inflection point has a vertical tangent.
- If the tangent line is parallel to neither axis, the inflection point has an oblique tangent.
The second is based on the direction of the change in concavity.
- The graph may change from concave down to concave up.
- The graph may change from concave up to concave down.
Note. These two descriptions are independent. For example, an inflection point may have a horizontal tangent while the graph changes from concave down to concave up. Likewise, it may have an oblique tangent while the graph changes from concave up to concave down.
The key idea is simple: what defines an inflection point is always a change in concavity. The direction of the tangent line, when one exists, provides additional information about the geometry of the graph.
Inflection points and tangent lines
When a tangent line exists at an inflection point, we can further describe the point according to the direction of that tangent.
The tangent line may be horizontal, vertical, or oblique.
Inflection point with a horizontal tangent
An inflection point has a horizontal tangent when its tangent line is parallel to the \(x\)-axis.
In this case, the slope of the tangent line is zero. Therefore, if the function is differentiable at \(x_0\), the derivative satisfies $ f'(x_0)=0 $.
Note. The condition $ f'(x_0)=0 $ by itself is not enough to show that \(x_0\) is an inflection point. The graph must also change concavity at \(x_0\). If the concavity does not change, the point may instead be a local maximum, a local minimum, or another type of stationary point.
Inflection point with a vertical tangent
An inflection point has a vertical tangent when its tangent line is parallel to the \(y\)-axis.
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In this case, the ordinary derivative at the point is not finite and may be undefined, while the graph still has a vertical tangent. A standard example is the graph of \(f(x)=\sqrt[3]{x}\) at \(x=0\).
Inflection point with an oblique tangent
An inflection point has an oblique tangent when its tangent line is parallel to neither the \(x\)-axis nor the \(y\)-axis.
In this case, the tangent line has a finite, nonzero slope.
If the function is differentiable at \(x_0\), the derivative at the inflection point is finite and nonzero: $ f'(x_0)\neq 0 $.
From concave down to concave up
One possible type of inflection occurs when the graph changes from concave down to concave up.
More specifically:
- to the left of the inflection point, the graph is concave down;
- to the right of the inflection point, the graph is concave up.
As we move through the inflection point from left to right, the way the graph bends changes.
Note. If the function is twice differentiable on intervals immediately to the left and right of the inflection point \(x_0\), the second derivative satisfies $ f''(x)<0 $ to the left of \(x_0\) (concave down) and $ f''(x)>0 $ to the right of \(x_0\) (concave up).
This type of inflection point may have a horizontal, vertical, or oblique tangent, depending on the behavior of the function at the point.
There is an important distinction to keep in mind: a change from concave down to concave up does not mean that the function itself is necessarily increasing.
Concavity describes how the slope of the graph changes, not whether the function is moving upward or downward.
These are two separate ideas. Increasing and decreasing behavior concerns the monotonicity of the function and is related to the sign of \(f'(x)\). Concavity, on the other hand, concerns how the slope changes and, when the second derivative exists, is related to the sign of \(f''(x)\).
Keeping this distinction in mind helps avoid confusing the monotonicity of a function with the concavity of its graph.
From concave up to concave down
The opposite type of change occurs when the graph goes from concave up to concave down.
In this case:
- to the left of the inflection point, the graph is concave up;
- to the right of the inflection point, the graph is concave down.
Note. If the function is twice differentiable on intervals immediately to the left and right of the inflection point \(x_0\), the second derivative satisfies $ f''(x)>0 $ to the left of \(x_0\) (concave up) and $ f''(x)<0 $ to the right of \(x_0\) (concave down).
This type of inflection point may also have a horizontal, vertical, or oblique tangent, depending on the behavior of the function at the point.
Here again, the defining feature is the change in concavity as the graph passes through the point, in this case from concave up to concave down.
This does not necessarily mean that the function itself is decreasing.
Whether a function is increasing or decreasing and whether its graph is concave up or concave down are separate properties. The first is determined by the behavior of the first derivative, while the second describes how the slope itself changes.
Stationary Points of Inflection: Conditions and Theorem
Let \(y=f(x)\) be a function that is defined, continuous, and differentiable on an open interval \(I_{x_0}\) containing \(x_0\). A stationary point of inflection is a point on the graph where the tangent line is horizontal and the function changes concavity.
To determine whether \(x_0\) could be a stationary point of inflection, we can check two conditions:
- The first derivative is zero at the point: \[ f'(x_0)=0 \]
- The first derivative keeps the same sign on both sides of \(x_0\), for all \(x\neq x_0\) sufficiently close to \(x_0\).
These conditions help us identify possible stationary points of inflection. They tell us that the tangent line is horizontal and that the function continues to increase or decrease as it passes through the stationary point.
However, these conditions alone are not enough to confirm a point of inflection. Even if the first derivative remains positive or negative on both sides of the point, the function may not change concavity.
To establish that \(x_0\) is a stationary point of inflection, we must also verify that the function changes concavity as \(x\) passes through \(x_0\).
Note. A stationary point of inflection has two essential features: a horizontal tangent line and a change in concavity. At the point itself, the first derivative is zero. In the cases considered here, the derivative has the same sign on both sides. This means the function keeps increasing or decreasing, even though the way its slope changes is different. Unlike a local maximum or minimum, the function does not switch from increasing to decreasing or vice versa.
In practice, identifying a stationary point of inflection means looking at both the first derivative and the concavity of the function.
Let's examine the two possible cases.
1] Stationary Point of Inflection on an Increasing Function
Suppose the first derivative is positive on both sides of \(x_0\):
\[ f'(x)>0 \qquad \text{for }x\neq x_0 \]
where \(x\) is sufficiently close to \(x_0\).
Because the derivative is positive, the function is increasing on both sides of the stationary point.
At \(x_0\), the derivative becomes zero, so the tangent line is horizontal. If the function also changes concavity at this point, then \(x_0\) is a stationary point of inflection.
Example
Consider the function:
\[ f(x)=x^3 \]
First, calculate its derivative:
\[ f'(x)=3x^2 \]
At \(x=0\), the derivative is zero:
\[ f'(0)=0 \]
For every \(x\neq 0\), however, the derivative is positive:
\[ 3x^2>0 \]
This tells us that the function is increasing on both sides of the origin, even though its tangent line is horizontal at \(x=0\).
But is the origin actually a point of inflection? To find out, we need to check whether the concavity changes.
Calculate the second derivative:
\[ f''(x)=6x \]
Now examine its sign:
- For \(x<0\), the second derivative is negative, so the graph is concave down.
- For \(x>0\), the second derivative is positive, so the graph is concave up.
The concavity changes at the origin. Since the first derivative is also zero there, the origin is a stationary point of inflection.
Therefore, \(f(x)=x^3\) is increasing and has a stationary point of inflection at \(O(0,0)\), rather than a local maximum or minimum.
2] Stationary Point of Inflection on a Decreasing Function
Now consider the opposite situation. Suppose the first derivative is negative on both sides of \(x_0\):
\[ f'(x)<0 \qquad \text{for }x\neq x_0 \]
where \(x\) is sufficiently close to \(x_0\).
In this case, the function is decreasing on both sides of the stationary point.
As before, if the first derivative is zero at \(x_0\) and the function changes concavity there, the point is a stationary point of inflection.
Example
Consider the function:
\[ f(x)=-x^3 \]
Its first derivative is:
\[ f'(x)=-3x^2 \]
At the origin, the derivative is zero:
\[ f'(0)=0 \]
For every \(x\neq 0\), however:
\[ -3x^2<0 \]
The derivative is always negative except at the origin. Therefore, the function is decreasing on both sides of \(x=0\).
Next, we calculate the second derivative to see whether the concavity changes:
\[ f''(x)=-6x \]
Its sign depends on the value of \(x\):
- For \(x<0\), the second derivative is positive, so the graph is concave up.
- For \(x>0\), the second derivative is negative, so the graph is concave down.
The function changes concavity at the origin, while its first derivative is zero there.
We can therefore conclude that \(f(x)=-x^3\) is decreasing and has a stationary point of inflection at \(O(0,0)\).
Stationary Point of Inflection vs. Local Maximum and Minimum
How can we tell a stationary point of inflection apart from a local maximum or minimum?
The key is to examine how the sign of the first derivative behaves on either side of the stationary point.
- Local maximum: The first derivative changes from positive to negative. The function switches from increasing to decreasing.
- Local minimum: The first derivative changes from negative to positive. The function switches from decreasing to increasing.
- Stationary point of inflection on an increasing function: The first derivative remains positive on both sides of the point.
- Stationary point of inflection on a decreasing function: The first derivative remains negative on both sides of the point.
At a local maximum or minimum of the types described above, the function changes its direction of monotonicity. At the stationary points of inflection considered here, the function continues to increase or decrease, but its concavity changes.
How to Find a Stationary Point of Inflection
When working with a differentiable function, you can identify stationary points of inflection by following four steps:
- 1. Find the first derivative \(f'(x)\).
- 2. Locate the stationary points by solving \(f'(x)=0\).
- 3. Check the sign of the first derivative immediately to the left and right of each stationary point.
- 4. Test for a change in concavity, for example, by examining the sign of the second derivative.
If the first derivative retains its sign on both sides of a stationary point and the function changes concavity there, the point is a stationary point of inflection.
Alternative Method: The Third Derivative Test
There is another useful way to identify stationary points of inflection when the function is sufficiently differentiable: the third derivative test.
Suppose both the first and second derivatives are zero at \(x_0\):
\[ f'(x_0)=0,\qquad f''(x_0)=0 \]
Now check the third derivative. If it exists and is nonzero at that point:
\[ f'''(x_0)\neq 0 \]
then \(x_0\) is a stationary point of inflection, assuming the function satisfies the required differentiability conditions in a neighborhood of \(x_0\).
These conditions are sufficient but not necessary. This means they guarantee a stationary point of inflection, but a function may still have one even if its third derivative is zero.
For example, consider:
\[ f(x)=x^5 \]
This function has a stationary point of inflection at the origin, even though its third derivative also vanishes there.
Therefore, a zero third derivative does not rule out a stationary point of inflection. It simply means that this test is inconclusive, and we need to examine the function further.
In such cases, we can investigate the behavior of the second derivative or use higher-order derivatives to determine whether the function changes concavity.
Note. The previous section focused on inflection points with horizontal tangents. However, an inflection point can also have a nonhorizontal or vertical tangent. In this section, we'll look at how to identify inflection points in general and determine the type of tangent at each one.
How to Find Inflection Points Using the Second Derivative
Suppose a function \(f(x)\) is continuous in a neighborhood of \(x_0\) and twice differentiable throughout that neighborhood, except possibly at \(x_0\) itself. To find an inflection point, we examine the sign of the second derivative \(f''(x)\). This tells us whether the graph is concave up or concave down and helps us identify where its concavity changes.
Here's how to find inflection points step by step:
- Find the second derivative \(f''(x)\).
- Identify possible inflection points by finding all values of \(x\) in the function's domain where \(f''(x)=0\) or where the second derivative does not exist.
- Check for a change in concavity. Make sure the function is continuous at each candidate point. Then examine the sign of \(f''(x)\) on either side. If the second derivative changes from positive to negative or from negative to positive, the graph changes concavity, confirming an inflection point.
- Determine the type of tangent. Once you've identified an inflection point at \(x_0\), examine the first derivative:
- If \(f'(x_0)=0\), the graph has an inflection point with a horizontal tangent.
- If \(f'(x_0)\ne 0\) and the derivative is finite, the graph has an inflection point with a nonhorizontal tangent.
- If the function is continuous at \(x_0\) but not differentiable there, it may have a vertical tangent. If the graph also changes concavity across the point, it is an inflection point with a vertical tangent. One way to verify the existence of a vertical tangent is to examine the limit of the first derivative: \[ \lim_{x\to x_0}f'(x)=+\infty \] or \[ \lim_{x\to x_0}f'(x)=-\infty \] If either limit holds and the graph changes concavity at \(x_0\), the point is an inflection point with a vertical tangent. The equation of the tangent line is \(x=x_0\).
Important: If a function has an inflection point at \(x_0\) and its second derivative exists there, then \(f''(x_0)=0\).
However, this is a necessary but not sufficient condition. A second derivative equal to zero does not automatically mean there's an inflection point. The graph must actually change concavity.
Also, keep in mind that an inflection point can occur even when the second derivative is undefined.
Example 1: An Inflection Point with a Horizontal Tangent
Let's start with the function:
\[ f(x)=x^3 \]
First, calculate the first derivative:
\[ f'(x)=3x^2 \]
Then find the second derivative:
\[ f''(x)=6x \]
Set the second derivative equal to zero:
\[ f''(x)=0 \]
\[ 6x=0 \]
\[ x=0 \]
We've found a possible inflection point at \(x=0\). Now we need to check whether the graph changes concavity there.
For \(x<0\), the second derivative is negative, so the graph is concave down.
For \(x>0\), the second derivative is positive, so the graph is concave up.
Because the concavity changes at \(x=0\), the origin is an inflection point.
Next, let's determine the slope of the tangent line at this point.
\[ f'(0)=3\cdot0^2=0 \]
Since \(f'(0)=0\), the tangent line is horizontal.
Therefore, the origin \(O(0,0)\) is an inflection point with a horizontal tangent.
Example 2: An Inflection Point with a Nonhorizontal Tangent
Now consider a slightly different function:
\[ f(x)=x^3+x \]
Its first derivative is:
\[ f'(x)=3x^2+1 \]
Its second derivative is:
\[ f''(x)=6x \]
Set the second derivative equal to zero:
\[ f''(x)=0 \]
\[ 6x=0 \]
\[ x=0 \]
Once again, \(x=0\) is a possible inflection point.
To confirm it, check the sign of the second derivative on either side of the origin.
For \(x<0\), \(f''(x)<0\), so the graph is concave down.
For \(x>0\), \(f''(x)>0\), so the graph is concave up.
Since the concavity changes at \(x=0\), the origin is an inflection point.
Now evaluate the first derivative at \(x=0\):
\[ f'(0)=3\cdot0^2+1=1 \]
The derivative is not zero, so the tangent line is not horizontal. Instead, it has a slope of 1.
Therefore, the origin \(O(0,0)\) is an inflection point with a nonhorizontal tangent.
Notice how the curve crosses its tangent line at the origin. As it passes through the inflection point, the graph changes from concave down to concave up.
How do you find the equation of a nonhorizontal tangent line?
Once you've found an inflection point, you may also want to determine the equation of its tangent line.
For a differentiable function \(f(x)\), the equation of the tangent line at a point with x-coordinate \(x_0\) is given by the point-slope formula:
\[ y-f(x_0)=f'(x_0)(x-x_0) \]
In this formula:
- \(x_0\) is the x-coordinate of the point of tangency.
- \(f(x_0)\) is the y-coordinate of the point of tangency.
- \(f'(x_0)\) is the slope of the tangent line.
Let's use this formula to find the tangent line for the function in Example 2:
\[ f(x)=x^3+x \]
We've already shown that the function has an inflection point at the origin, where \(x_0=0\), and that its tangent line is not horizontal.
Step 1: Find the point of tangency.
Substitute \(x_0=0\) into the function:
\[ y=f(x_0)=f(0)=0^3+0=0 \]
So the point of tangency is \(P(0,0)\).
Step 2: Find the slope of the tangent line.
The first derivative is:
\[ f'(x)=3x^2+1 \]
Evaluate it at \(x_0=0\):
\[ f'(0)=3\cdot0^2+1=1 \]
Therefore, the slope is \(m=1\).
Step 3: Write the equation of the tangent line.
Start with the point-slope formula:
\[ y-f(x_0)=f'(x_0)(x-x_0) \]
Substitute \(x_0=0\), \(f(x_0)=0\), and \(f'(x_0)=1\):
\[ y-0=1\cdot(x-0) \]
Simplify:
\[ y=x \]
Therefore, \(y=x\) is the equation of the tangent line at the inflection point \(P(0,0)\).
This method works for any point where a function is differentiable, whether or not the point is an inflection point.
Higher-Order Derivative Test (Inflection Points with Horizontal Tangents)
Let \(y=f(x)\) be a function with continuous derivatives up to order \(n\) on an open interval. Suppose that, at an interior point \(x_0\), all derivatives from the first through the \((n-1)\)th are zero: \[ f'(x_0)=f''(x_0)=\cdots=f^{(n-1)}(x_0)=0 \] However, the \(n\)th derivative, where \(n\geq 2\), is nonzero: \[ f^{(n)}(x_0)\neq 0 \] In this case, I can classify the stationary point by looking at the order and sign of the first nonzero derivative.
- If \(n\) is even, the function has a strict local maximum at \(x_0\) if \(f^{(n)}(x_0)<0\), or a strict local minimum if \(f^{(n)}(x_0)>0\).
- If \(n\) is odd, the graph has an inflection point with a horizontal tangent at \(x_0\). The function is decreasing through the point if \(f^{(n)}(x_0)<0\), or increasing through the point if \(f^{(n)}(x_0)>0\).
The higher-order derivative test is particularly useful for identifying inflection points with horizontal tangents when both the first and second derivatives are zero.
In this situation, the second derivative test is inconclusive, so I need to examine higher-order derivatives to determine the nature of the stationary point.
Note. The order of a derivative tells me how many times the original function has been differentiated. The first derivative is of order 1, the second derivative is of order 2, the third derivative is of order 3, and so on. Even-order derivatives include the second, fourth, sixth, and so on, while odd-order derivatives include the first, third, fifth, and so on.
Example
I start with the function
\[ f(x)=x^3 \]
First, I calculate the first derivative.
\[ f'(x)=3x^2 \]
To find the stationary points, I set the first derivative equal to zero.
\[ 3x^2=0 \]
Solving the equation gives
\[ x_0=0 \]
Next, I calculate the second derivative.
\[ f''(x)=6x \]
At \(x_0=0\), the second derivative is also zero.
\[ f''(0)=0 \]
Since the second derivative test is inconclusive, I cannot use it to determine whether the stationary point is a local maximum, a local minimum, or an inflection point.
To investigate further, I calculate the third derivative.
\[ f'''(x)=6 \]
The third derivative is positive and nonzero at \(x_0=0\).
\[ f'''(0)=6>0 \]
The first nonzero derivative is therefore of odd order and positive. According to the higher-order derivative test, \(x_0=0\) is an inflection point with a horizontal tangent, where the function is increasing.
Finally, I calculate the function value at this point.
\[ f(0)=0 \]
Therefore, the inflection point has coordinates \((0,0)\).
The same method can be applied to other functions when the first and second derivatives vanish at a stationary point.
Higher-Order Derivative Test for Inflection Points with Nonhorizontal Tangents
Consider a function \(y=f(x)\) whose derivatives through order \(n\) are continuous on an open interval containing \(x_0\). I assume that the first derivative at \(x_0\) is nonzero:
\[ f'(x_0)\neq 0 \]
Suppose that all derivatives from the second through the \((n-1)\)th are zero at \(x_0\):
\[ f''(x_0)=f'''(x_0)=\ldots=f^{(n-1)}(x_0)=0 \]
However, the \(n\)th derivative is nonzero:
\[ f^{(n)}(x_0)\neq 0,\qquad n\geq 3 \]
Under these conditions, I can determine how the function curves near \(x_0\) by checking whether \(n\) is even or odd and whether \(f^{(n)}(x_0)\) is positive or negative.
If \(n\) is odd, the point \((x_0,f(x_0))\) is an inflection point with a nonhorizontal tangent. As \(x\) increases through \(x_0\):
- If \(f^{(n)}(x_0)>0\), the function changes from concave down to concave up.
- If \(f^{(n)}(x_0)<0\), the function changes from concave up to concave down.
If \(n\) is even, there is no inflection point because the concavity stays the same on both sides of \(x_0\). In a sufficiently small neighborhood of \(x_0\):
- If \(f^{(n)}(x_0)>0\), the function is concave up.
- If \(f^{(n)}(x_0)<0\), the function is concave down.
This test provides a way to identify inflection points with nonhorizontal tangents by examining higher-order derivatives. It also helps determine the direction of concavity near a point.
One important difference from the preceding test is that the first derivative must be nonzero at the point being examined.
A Worked Example
Consider the function:
\[ f(x)=x^3-2x^2+x \]
I want to determine whether the function has an inflection point and, if so, find its coordinates.
I start by calculating the first and second derivatives:
\[ f'(x)=3x^2-4x+1 \]
\[ f''(x)=6x-4 \]
To find a possible inflection point, I set the second derivative equal to zero:
\[ 6x-4=0 \quad \Rightarrow \quad x_0=\frac{2}{3} \]
Next, I check whether the first derivative is nonzero at this point:
\[ f'\left(\frac{2}{3}\right) =3\cdot\frac{4}{9}-4\cdot\frac{2}{3}+1 =-\frac{1}{3}\neq 0 \]
Since the slope is nonzero, the tangent line is not horizontal. I can therefore apply the higher-order derivative test.
I calculate the third derivative:
\[ f'''(x)=6 \]
At \(x_0=\frac{2}{3}\), the second derivative is zero, while the third derivative is positive:
\[ f''\left(\frac{2}{3}\right)=0,\qquad f'''\left(\frac{2}{3}\right)=6>0 \]
The third derivative is the first nonzero derivative of order greater than one. Since 3 is odd, I can conclude that the function has an inflection point with a nonhorizontal tangent.
Because the third derivative is positive, the function changes from concave down to concave up as \(x\) increases through the inflection point.
To find the coordinates of this point, I substitute \(x_0=\frac{2}{3}\) into the original function:
\[ f\left(\frac{2}{3}\right) =\frac{8}{27}-\frac{8}{9}+\frac{2}{3} =\frac{2}{27} \]
The inflection point is therefore:
\[ F\left(\frac{2}{3},\frac{2}{27}\right) \]
The graph below shows the function and its change in concavity.
I can also verify the result directly by examining the sign of the second derivative \(f''(x)=6x-4\):
- For \(x<\frac{2}{3}\), the second derivative is negative, so the function is concave down.
- For \(x>\frac{2}{3}\), the second derivative is positive, so the function is concave up.
This confirms that the concavity changes at \(x_0=\frac{2}{3}\).
Finally, I can find the equation of the tangent line at the inflection point. Its slope is \(f'\left(\frac{2}{3}\right)=-\frac{1}{3}\), so I use the point-slope form:
\[ y-\frac{2}{27}=-\frac{1}{3}\left(x-\frac{2}{3}\right) \]
Simplifying, I obtain:
\[ y=-\frac{1}{3}x+\frac{8}{27} \]
This is the equation of the nonhorizontal tangent line at the inflection point.
