Inflection Point

Inflection Point describes a point where the curvature of a curve changes direction. It represents the transition from a concave to a convex shape or vice versa.

Let's learn about Inflection Points in detail, including Concavity of Function and solved examples.

Inflection Point Definition

Inflection Point of a function f(x), is defined by the following two conditions:

Necessary Condition: At inflection point f''(x) = 0 or f''(x) does not exist. 

Sufficient Condition: At inflection point concavity of graph changes from concave up to concave down or vice versa i.e., f''(x)<0 before inflection point and f''(x) > 0 after inflection point or vice versa.

Point of inflection is the point where the function goes from being concave upwards to concave downwards and vice-versa. Let's say f(x) is the function that has a point x = c as its inflection point. Then, 

• f''(c) = 0
• f''(c) does not exist. 

Let's learn about concavity for a better understanding of the point of inflection.

Concavity of Function

Consider the two functions given in the figure below, while both of these functions look similar.  What differentiates them is their bend, notice that one of them looks opening upwards and the other looks closing downwards. This trend is called concavity. 

Concavity is related to the rate of the increase as we know that derivatives are nothing but the rate of change. This means they tell us if the function is increasing (rate of change is positive) or decreasing (rate of change is negative).

Thus, Concavity is related to the rate of change of the rate of change, as it is defined in terms of second-order derivative which we will learn about in this article further.

Types of Concavity

There are two types of concavity that exist in the function,

• Concave Up
• Concave Down

Concave Up: A function is said to be concave up on an interval if the graph of that function lies above any tangent that is drawn on any point of the interval.

Concave Down: Similarly, the function is said to be concave downwards if the graph of the function lies below any tangent that is drawn on any point of the interval.

Test for Concavity

Consider a function f(x) which is differentiable and has derivatives and double derivatives as f' and f'' on the interval I. 

• f(x) is considered concave up if f''(x) > 0 for all the points in the interval I.
  • f(x) is considered concave down if f''(x) < 0 for all the points in the interval I.

Learn More:

• Concave Function

How to Find Inflection Point

We can find the Inflection Point by two methods i.e., Using the Second Derivative (most common method) and Analyzing Changes in the Concavity of the graph (intuitive method). Let's understand these two methods in detail as follows:

Using the Second Derivative

To find the Inflection Point using the second derivative, let's consider a function f(x) and use the following steps for f(x) further:

Step 1: Find the second-order derivative of the given function f(x) i.e., f''(x).

Step 2: Equate the second-order derivative of the given function to 0 and then solve the equation.

Step 3: Check the sufficient condition i.e., f''(x) changes sign at the point where f''(x) = 0. 

Step 4: If f''(x) changes the sign at the point of consideration then it is required Inflection Point.

Let's consider an example to understand this better.

Example: Find the inflection point for f(x) = x³ - 3x² + 4x if exist.

Solution:

Given function if f(x) = x³ - 3x² + 4x.

Step 1: Find f''(x).

⇒ f'(x) = 3x² - 6x + 4

⇒ f''(x) = 6x - 6

Step 2: Solve the equation f''(x) = 0.

f''(x) = 6x - 6 = 0

⇒ x = 1

Step 3: Check whether f''(x) changes sign at the point of consideration i.e., x = 1.

Let's choose x = 0.5, then f''(0.5) = -3 (Negative)

Let's choose x = 2, then f''(2) = 6 (Positive)

Step 4: Since f''(x) does change sign at x = 1, it is an inflection point.

Therefore, x = 1 is inflection point in the function f(x) = x³ - 3x² + 4x.

Note: It is not necessary that if the double derivative is zero at a point, it is an inflection point. For that to be an inflection point, f''(x) needs to change the sign from positive to negative or vice versa.

Related :

• Implicit Differentiation
• Advanced Differentiation
• Absolute Maxima and Minima

Summary - Inflection Point

The articles discuss the concept of inflection points in mathematical functions, emphasizing the necessary conditions for identifying them and their importance in understanding function concavity. An inflection point is defined as a point on a curve where the curvature changes from concave up to concave down, or vice versa. This transition is identified by either the second derivative of the function being zero or non-existent at that point. The articles further explore how concavity affects the shape of a curve, with concave up sections appearing as upward bends and concave down sections as downward bends. Various methods for determining inflection points and concavity are detailed, primarily focusing on the use of the second derivative test. Practical examples and solved problems are provided to illustrate how to apply these methods to specific functions, helping to clarify the theoretical concepts with real-world applications. Concavity tests and the impact of curvature changes on function behavior are emphasized as vital tools in calculus, offering insights into the properties and extremities of curves.

Sample Problems on Inflection Point

Problem 1: Find out the concavity of the given function between the interval [0,1]. 

f(x) = x² + 4

Solution: 

To check for the concavity, let's look at the second derivative of the function. 

f(x) = x² + 4

Differentiating w.r.t x, 

f'(x) = 2x 

Differentiating w.r.t x again, 

f''(x) = 2 > 0 

This is greater than zero, the function is concave upwards in the given interval. 

Problem 2: Find out the concavity of the given function between the interval [2,4]. 

f(x) = x³ + 4

Solution: 

To check for the concavity, let's look at the second derivative of the function.

f(x) = x³

Differentiating w.r.t x, 

f'(x) = 3x² 

Differentiating w.r.t x again, 

f''(x) = 6x > 0

This is greater than zero, the function is concave upwards in the given interval. 

Problem 3: Find inflection point for the given function. 

f(x) = x³

Solution: 

To check for the point of inflection., let's look at the second derivative of the function.

f(x) = x³ + 4

Differentiating w.r.t x, 

f'(x) = 3x² 

Differentiating w.r.t x again, 

f''(x) = 6x = 0 

Solving f''(x) = 0, it shows that x = 0 is the critical point of f'(x). 

Now let's check the value of f''(x) before and after the critical point. 

for x < 0, f''(x) < 0.  

for x > 0, f''(x) > 0. 

Thus, the function changes from concave downward to concave upward. Thus, x = 0 is the inflection point. 

Problem 4: Find inflection point for the given function. 

f(x) = x³ + x² + 3

Solution: 

To check for the point of inflection., let's look at the second derivative of the function.

f(x) = x³ + x² + 3

Differentiating w.r.t x, 

f'(x) = 3x² + 2x 

Differentiating w.r.t x again, 

f''(x) = 6x + 2 = 0 

Solving f''(x) = 0, it shows that x = -1/3 is the critical point of f'(x). 

Now let's check the value of f''(x) before and after the critical point. 

for x < -1, f''(x) < 0.  

for x > 0, f''(x) > 0. 

Thus, the function changes from concave downward to concave upward. Thus, x = -1/3 is the inflection point. 

Practice Problems on Inflection Point

Q1: Calculate inflection point for 4x³ - 20x² + 3x

Q2: Calculate Inflection point for sin²x + 3sin x

Q3: Calculate Inflection Point for cos³x + 2sin x

Q4: Find the inflection point for 2x³- 9x² + 6x