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Μάθημα: AP®︎ Λογισμός BC > Ενότητα 5

Μάθημα 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

\cap

or vice versa).

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

Practice set 1: Analyzing inflection points graphically

Πρόβλημα 1.1

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 (x) = \frac{1}{2} x^{4} + x^{3} - 6 x^{2}

The second derivative of

f

f^{″} (x) = 6 (x - 1) (x + 2)

f^{″} (x) = 0

for

x = - 2, 1

, and it's defined everywhere.

x = - 2

and

x = 1

divide the number line into three intervals:

Let's evaluate

f^{″}

at each interval to see if it's positive or negative on that interval.

Interval	$x$ ‍-value	$f^{″} (x)$ ‍	Verdict
$x < - 2$ ‍	$x = - 3$ ‍	$f^{″} (- 3) = 24 > 0$ ‍	$f$ ‍ is concave up $\cup$ ‍
$- 2 < x < 1$ ‍	$x = 0$ ‍	$f^{″} (0) = - 12 < 0$ ‍	$f$ ‍ is concave down $\cap$ ‍
$x > 1$ ‍	$x = 2$ ‍	$f^{″} (2) = 24 > 0$ ‍	$f$ ‍ is concave up $\cup$ ‍

We can see that the graph of

f

changes concavity at both

x = - 2

and

x = 1

, so

f

has inflection points at both of those

x

-values.

Πρόβλημα 2.1

g (x) = x^{4} + 4 x^{3} - 18 x^{2}

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.

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