Finding the point of inflection for a function is a key concept in calculus that often trips up students. I remember struggling with this when I first learned it in high school! But with some practice and a clear understanding of what an inflection point is, finding inflection points can become second nature.
In this comprehensive guide I’ll walk you through the step-by-step process I use to identify inflection points. I’ll explain inflection points in simple terms, show you how to take derivatives and provide tips for double checking your work. My goal is to break this down into easy-to-follow steps so you can master finding inflection points!
What is a Point of Inflection?
Let’s start with the basics – what exactly is an inflection point?
An inflection point is a point on a curve where the curve changes concavity. This means it switches from being concave up to concave down, or vice versa. Visually, it’s the point where the curve stops curving one way and starts curving the other way.
Here’s a quick example to illustrate:
![Graph showing an inflection point where curve changes concavity]
See how the curve switches from concave up to concave down at the inflection point? That’s the key thing to look for.
Inflection points also have an important property – the second derivative of the function equals zero at the inflection point. This is a key fact we’ll use later when finding inflection points algebraically.
- Is where the curve changes concavity
- Makes the second derivative equal to zero
Easy enough, right? Now let’s get into the step-by-step process for actually finding inflection points.
Step 1: Take the First Derivative
To find inflection points algebraically, we’ll need to take the first and second derivatives of the function. Let’s start with the first derivative.
The first step is always to take the derivative of the original function. You can use any method you’re comfortable with – the power rule, quotient rule, chain rule, etc.
For example, say we have the function:
f(x) = x^3 – 3x^2 + 4
To find the first derivative, we take the derivative of each term using the power rule:
f'(x) = 3x^2 – 6x
Make sure you simplify the derivative properly. Taking derivatives is likely review for you if you’re learning about inflection points. So this first step should be straightforward.
Step 2: Take the Second Derivative
Now that we have the first derivative f'(x), we simply take the derivative again to get the second derivative f”(x).
Revisiting our example:
f'(x) = 3x^2 – 6x
f”(x) = 6x – 6
Again, apply the power rule to each term to find the second derivative. Simplify if needed.
Finding the second derivative is the key step that will allow us to locate inflection points algebraically. This is because inflection points make the second derivative equal to 0.
- Take derivative to get f'(x)
- Take derivative again to get f”(x)
Step 3: Set f”(x) = 0 and Solve
Here’s where the magic happens! To find candidate inflection points, we set the second derivative equal to 0 and solve:
f”(x) = 0
Solving this will reveal the x-values where the second derivative could be zero, which are our candidate inflection points.
Using our example:
f”(x) = 6x – 6
Set f”(x) equal to 0:
6x – 6 = 0
6x = 6
x = 1
Therefore, x = 1 is a candidate for being an inflection point of the original function.
But we’re not done yet…
Step 4: Verify with the First Derivative
Finding where f”(x) = 0 gives us potential inflection points. However, we need to verify that these points are actual inflection points by checking that the first derivative f'(x) changes sign at these points.
Here’s what to do:
-
Plug values less than and greater than the candidate x-value into f'(x)
-
If f'(x) is positive on one side of x and negative on the other side, then x is an inflection point
Let’s see this in action with our example:
Candidate inflection point: x = 1
Plug numbers into f'(x) = 3x^2 – 6x
When x = 0, f'(x) = -6 (negative value)
When x = 2, f'(x) = 12 (positive value)
Since f'(x) went from negative to positive as x went through 1, we confirm x = 1 is an inflection point!
Verifying with the first derivative is an essential step – it ensuresInflection points make the second derivative 0, but the converse isn’t necessarily true.
Tips for Finding Inflection Points
Here are some handy tips as you practice finding inflection points:
- Double check your algebra when taking derivatives
- Clearly label f'(x) and f”(x) to avoid mix-ups
- Be methodical about checking signs of f'(x) on both sides
- Graphing the function can provide visual intuition
- Remember the second derivative = 0 is necessary but not sufficient for inflection points
With some algebraic manipulation and methodical verification, finding inflection points doesn’t have to be hard. I hope breaking down the step-by-step process makes this concept more approachable.
The key is taking it slow – work through some examples methodically, and the process will become second nature before you know it. Understanding inflection points unlocks the ability to fully analyze and describe functions, so it’s worth investing the time to master it.
4 Answers 4 Sorted by:
The best you can do in general is use finite differences to approximate the 2nd derivative and find where that changes sign.
Given two columns X and Y estimate the 2nd derivative Ypp away from the ends with the following formula. Assumes the x values increment with a fixed value h
| X | Y | ≈Ypp |
|---|---|---|
x_1 |
y_1 |
|
x_2 |
y_2 |
=(y_3-2*y_2+y_1)/(h*h) |
x_3 |
y_3 |
The inflection point is where the 2nd derivative switches signs. You can simply find where two consecutive values multiply to a negative value ypp_2*ypp_1 <= 0.
If you want more precision then you need to fit a model to the data, or go with cubic splines.
Note that if the X data isnt with equal intervals you have to use the more complex formula below
You can - as an approximation - calculate DeltaY / DeltaX for each subsequent pair of lines and check for change of the sign of this.
Example (starting in [A1] - copy all formulas down from their starting cell)
change of slope occurs at relative maxima or minima (1st differential quotient equal 0)
If there is any noise in the data, computing differences will amplify that noise, so there is a greater chance of finding spurious inflection points. A way to reduce the noise is to fit a curve to the data, and then compute the inflection points for that curve. E.g. fit a cubic polynomial to the data, and find the inflection point of that.
May I suggest doing this would by using regression. Not a linear, but a typical multiple order regression, AKA polynomial regression (y = a_0 + a_1*x + a_2*x^2 + ... + a_n*x^n). See this thread for more details on how to do it. This can be done directly in Excel, no need to code anything in VBA. However, youll probably need to deal with array formulas (AKA CTRL+ Enter formulas).
Then, once, youve find a regression that fits your distribution (r² > 0.9 or what suits you), you could simply do a derivative of this equation. Since this is a polynomial equation, the equation is quite easy : y = a_1 + 2*a_2*x + ... + n*a_n+1.
The fun part now starts! We need to find what values of x makes y = 0. If your regression is below the 4th order, there is an analytical solution possible (i.e. theres an equation that can gives you the x value, because your derivative will be of order 3). If you are over the 4th order, then, you need to use a numerical method. Yes, you can use VBA to get a bisection algorithm going, but do you know that Excel has a numerical solver integrated? Use it to get the values you are looking for (assuming at least one value is real).
As you didnt supplied exemple of dataset, this is though to figure, but if we use MikeD exemple, we would get this !
With the values, instead of the formulas:
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