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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB>Unit 5

Lesson 7: Determining concavity of intervals and finding points of inflection: algebraic

# 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 \cup to \cap or vice versa).

## Practice set 1: Analyzing inflection points graphically

Problem 1.1
• Current
How many inflection points does the graph of f have?

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, left parenthesis, x, right parenthesis, equals, start fraction, 1, divided by, 2, end fraction, x, start superscript, 4, end superscript, plus, x, cubed, minus, 6, x, squared.
The second derivative of f is f, start superscript, prime, prime, end superscript, left parenthesis, x, right parenthesis, equals, 6, left parenthesis, x, minus, 1, right parenthesis, left parenthesis, x, plus, 2, right parenthesis.
f, start superscript, prime, prime, end superscript, left parenthesis, x, right parenthesis, equals, 0 for x, equals, minus, 2, comma, 1, and it's defined everywhere. x, equals, minus, 2 and x, equals, 1 divide the number line into three intervals:
Let's evaluate f, start superscript, prime, prime, end superscript at each interval to see if it's positive or negative on that interval.
Intervalx-valuef, start superscript, prime, prime, end superscript, left parenthesis, x, right parenthesisVerdict
x, is less than, minus, 2x, equals, minus, 3f, start superscript, prime, prime, end superscript, left parenthesis, minus, 3, right parenthesis, equals, 24, is greater than, 0f is concave up \cup
minus, 2, is less than, x, is less than, 1x, equals, 0f, start superscript, prime, prime, end superscript, left parenthesis, 0, right parenthesis, equals, minus, 12, is less than, 0f is concave down \cap
x, is greater than, 1x, equals, 2f, start superscript, prime, prime, end superscript, left parenthesis, 2, right parenthesis, equals, 24, is greater than, 0f is concave up \cup
We can see that the graph of f changes concavity at both x, equals, minus, 2 and x, equals, 1, so f has inflection points at both of those x-values.
Problem 2.1
• Current
g, left parenthesis, x, right parenthesis, equals, x, start superscript, 4, end superscript, plus, 4, x, cubed, minus, 18, x, squared
For what values of x does the graph of g have a point of inflection?

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? • 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) = x⁶ + 6x⁵f'(x) = 6x⁴(5 + x)f"(x) = 30x³(4 + x)f'"(x) = 120x²(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
• Is it possible to have an inflection point at a point where the third derivative is undefined? Thanks! • 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) • 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) • 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. • I think Problem 1.3 has 6 inflection points. • Does a sin inverse graph have inflection point?
(1 vote) • 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) • 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. 