Concavity and points of inflection

Concave up and concave down, using the second derivative to find concavity and points of inflection, and the second-derivative test for classifying stationary points.

Concavity

A point of inflection is a point where the concavity changes. At such a point $f''(x) = 0$ (or is undefined), and $f''$ changes sign.

Concavity along y = x³ − 6x² + 9x + 1

Drag P. The curve changes from concave down to concave up at x = 2.

Try this

Find the point of inflection. How is it related to the two turning points?

The second-derivative test

Example 1●●

Find the turning points and point of inflection of $y = x^3 - 6x^2 + 9x + 1$.

$y' = 3x^2 - 12x + 9 = 3(x - 1)(x - 3)$ and $y'' = 6x - 12$.

$y''(1) = -6 < 0$, so $(1, 5)$ is a maximum. $\;y''(3) = 6 > 0$, so $(3, 1)$ is a minimum.

$y'' = 0$ at $x = 2$, and it changes sign, so $(2, 3)$ is a point of inflection.

y″ = 6x − 12 is linear, so it changes sign at its zero.

Example 2: finding a coefficient●●

$y = x^3 + bx^2 + 5$ has a point of inflection at $x = 2$. Find $b$.

$y'' = 6x + 2b$, and $y''(2) = 0$ gives $12 + 2b = 0$, so $b = -6$.

Flashcards

$f''(x) > 0$

Concave up.

Point of inflection

$f''$ is zero and changes sign.

$f'(a) = 0$, $f''(a) < 0$

Maximum turning point.

$f'(a) = 0$, $f''(a) = 0$

Test fails: use a table of slopes.

Practice

Find the $x$-coordinate of the point of inflection of $y = x^3 - 6x^2 + 9x + 1$.
Find its $y$-coordinate.
For the same curve, $y''(1) = -6$. So $x = 1$ is:
$y = x^3 + bx^2 + 5$ has an inflection at $x = 2$. Find $b$.
Does $y = x^4$ have a point of inflection at the origin?
Where is $y = x^3 - 6x^2 + 9x + 1$ concave down?
If $f'(a) = 0$ and $f''(a) = 0$, then:
Find the $x$-coordinate of the point of inflection of $y = x^3 - 3x^2$.
On a concave up curve, the gradient is: