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### Course: APĀ®ļø/College Calculus ABĀ >Ā Unit 5

Lesson 7: Determining concavity of intervals and finding points of inflection: algebraic- Analyzing concavity (algebraic)
- Inflection points (algebraic)
- Mistakes when finding inflection points: second derivative undefined
- Mistakes when finding inflection points: not checking candidates
- Analyzing the second derivative to find inflection points
- Analyze concavity
- Find inflection points
- Concavity review
- Inflection points review

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# Inflection points review

Review your knowledge of inflection points and how we use differential calculus to find them.

## What are inflection points?

Inflection points (or points of inflection) are points where the graph of a function changes concavity (from $\u0101\x88\u0156$ to $\u0101\x88\copyright $ or vice versa).

*Want to learn more about inflection points and differential calculus? Check out this video.*

## Practice set 1: Analyzing inflection points graphically

*Want to try more problems like this? Check out this exercise.*

## Practice set 2: Analyzing inflection points algebraically

Inflection points are found in a way similar to how we find extremum points. However, instead of looking for points where the derivative changes its sign, we are looking for points where the

*second derivative*changes its sign.Let's find, for example, the inflection points of $f(x)={\displaystyle \frac{1}{2}}{x}^{4}+{x}^{3}\u0101\x88\x926{x}^{2}$ .

The second derivative of $f$ is ${f}^{\u0101\x80\xb3}(x)=6(x\u0101\x88\x921)(x+2)$ .

Let's evaluate ${f}^{\u0101\x80\xb3}$ at each interval to see if it's positive or negative on that interval.

Interval | Verdict | ||
---|---|---|---|

We can see that the graph of $f$ changes concavity at both $x=\u0101\x88\x922$ and $x=1$ , so $f$ has inflection points at both of those $x$ -values.

*Want to try more problems like this? Check out this exercise.*

## Want to join the conversation?

- Can you use the third derivative to find inflection points? I want to say that for the second derivative to change sign, then it must actually
**cross**the x-axis. So, when the second derivative is zero, the derivative of that can't be zero, it must be either positive or negative depending on whether f''(x) is crossing the x-axis from above or below. Is that right, or are there cases where this wouldn't work?(10 votes)- Your intuition was good, but unfortunately there are exceptions that limit how useful this test is.

If the third derivative does**not**= 0, then you do have an inflection point. Unfortunately, there are cases where the third derivative = 0 that**are**inflection points.

For an example:

f(x) = xā¶ + 6xāµ

f'(x) = 6xā“(5 + x)

f"(x) = 30xĀ³(4 + x)

f'"(x) = 120xĀ²(3 + x)

At x = 0 all of these functions including the third derivative equal zero, however x=0**is**an inflection point, as you can see if you look at the graphs:

https://goo.gl/98QXuV(20 votes)

- Is it possible to have an inflection point at a point where the third derivative is undefined? Thanks!(5 votes)
- Yes, consider the function f(x)=

-xĀ² for xā¤0

xĀ² for xā„0

f''(x) is -2 for negative x and 2 for positive x. Because the second derivative changes between positive and negative, we have an inflection point at x=0.

But obviously, f'' is discontinuous, and therefore nondifferentiable at 0, so the third derivative is undefined there.(6 votes)

- Is it possible to find the concavity of a function that is similar to an absolute value function? I know that the "vertex" of the absolute value is not differentiable at that point, so does that mean that there is no point of inflection for that sort of function?(1 vote)
- Points of inflection are points where the second derivative changes between positive and negative. The second derivative of |x| is undefined at 0 and is 0 everywhere else, so it has no inflection points.(9 votes)

- I think Problem 1.3 has 6 inflection points.(4 votes)
- See that the derivative of the function is graphed. So, inflection points would essentially be the maxima and minima. And there are only five of them(2 votes)

- In practice set 1-(1/3), I don't understand why there is two inflection points, in my view it should have 3 inflection points -(3.5, 0 and 5). i have watched the inflection points couple of time , but doesn't make sense!(1 vote)
- I think you are getting inflection points confused with critical points. Inflection points are where the second derivative changes sign, what you are thinking of is where the first derivative changes sign ...(4 votes)

- so can i make something similar to the mean value theorem, for the second derivative:

If in an interval [a,b], a function(say f) is differentiable, and f'(a)=f'(b), where a and b are distinct values, then there must exist atleast one c in the interval [a,b], such that at c the point of inflection occurs.

I hope i didn't mess it up.(2 votes) - Does a sin inverse graph have inflection point?(1 vote)
- Yes - it has multiple inflection points! A POI is where the second derivative of a function is equal to 0 or where the graph changes concavity. The graph of inverse sign has POIs whenever it crosses the x-axis, I would recommend looking up the graph to see how it changes concavity at these points.(2 votes)

- Does the second derivative need to change signs to be an inflection point? could the second derivative be equal to zero but not change signs and be an inflection point?(1 vote)
- Yes, in order for it to be an inflection point, the second derivative must change signs. Hope this helps!(2 votes)

- I just want to make sure I'm following and this concept is analogous to what we've covered so far.

In terms of the 3rd derivative when dealing with functions involving motion, the Jerk is represented by the 3rd derivative, or the rate of change of acceleration. So would this mean that an extremum point of the 2nd derivative graph mark the x value where the Acceleration is constant (not changing), and thus the particle would be accelerating at a constant rate ?

Similarly, if the slope of the tangent line on the 2nd derivative graph (acceleration) is positive, it means the particle's rate of acceleration is increasing (and negative slope would indicate it's acceleration rate is decreasing (deceleration?).

And does this also relate to the idea of inflection points (although Im sure they may have a different name)? Where the value of f''(x) = 0 and the 3rd derivative is used to test if that x value is an 'inflection' value of the Jerk function.

Am I in the right ballpark here ?(1 vote)- Your second paragraph does not make sense. If you want something to be constant you would need a horizontal line.

Your third paragraph is close however not quite there. For deacceleration to occur the acceleration would need to become negative so you need to go under the x-axis.

f''(x)<0 would lead to deacceleration

You can use the derivatives or higher if you want determine whether point is maximum or minimum or saddle point or just a point of inflection.

Inflexion points are located where f''(x) = 0. In the case f''(x)=0 & f'(x)= 0 then you will need to do further testing

I will let you do your own research(1 vote)

- When you look for critical points, you look for places where the first derivative is either zero or undefined. Do inflection points also exist where the second derivative is undefined in addition to when it is equal to zero?(1 vote)
- Yes, inflection points exist when the second derivative is either zero or undefined.(1 vote)