Find the intervals of concavity and the point of inflection of $f(x) = x^3 - 6x^2 + 4$.

Concave down on $(-\infty, 2)$, concave up on $(2, \infty)$; inflection point $(2, -12)$.

Use the second derivative test to classify the critical points of $f(x) = x^3 - 3x^2 - 9x + 1$.

Local max $(-1, 6)$, local min $(3, -26)$.

Find the intervals of concavity and the points of inflection of $f(x) = x^4 - 6x^2$.

Concave up on $(-\infty, -1)$ and $(1, \infty)$, down on $(-1, 1)$; inflection points $(-1, -5)$ and $(1, -5)$.

For $f(x) = x^4$, show that $f''(0) = 0$, then decide whether $(0, 0)$ is a point of inflection and classify it.

Not a point of inflection; $(0, 0)$ is a local (and absolute) minimum.

Find $a$ and $b$ so that $f(x) = x^3 + ax^2 + bx$ has a point of inflection at $(1, 2)$. Then show that this $f$ has no local maximum or minimum.

$a = -3$, $b = 4$; $f'(x) = 3x^2 - 6x + 4 > 0$ for all $x$, so there is no local max or min.

Grade 12 · MCV4U · Calculus & Vectors

MCV4U: Concavity and the Second Derivative Test

The first derivative says whether a graph goes up or down. The second derivative says which way it bends, and that tells you whether a flat point is the top of a hill or the bottom of a valley.

How It Works

Concavity. If $f''(x) > 0$ on an interval, the graph is concave up there: it bends upward like a cup, and its slopes are increasing. If $f''(x) < 0$ , it is concave down, bending like a cap, with slopes decreasing.

Points of inflection. A point of inflection is a point on the graph where the concavity changes. To find them:

1. Find $f''(x)$ and the values where it is zero or undefined.

2. Make a sign chart for $f''$ .

3. Keep only the values where the sign really changes and where $f$ is defined. Substitute into $f$ to get each point.

Step 3 matters. $f''(c) = 0$ does not guarantee an inflection point. For $f(x) = x^4$ , $f''(0) = 0$ , but $f''$ is positive on both sides, so the graph is concave up everywhere.

Second derivative test. Suppose $f'(c) = 0$ , so the tangent at $c$ is horizontal.

If $f''(c) > 0$ , the graph is concave up there, so $c$ gives a local minimum.

If $f''(c) < 0$ , the graph is concave down there, so $c$ gives a local maximum.

If $f''(c) = 0$ , the test says nothing. Go back to the first derivative test and check the sign of $f'$ on each side.

The second derivative test is quicker than a full sign chart when $f''$ is easy to evaluate, because you only substitute the critical numbers. The first derivative test always works, and it is the one to fall back on.

Concavity and direction are separate. A graph can be increasing and concave down, rising but levelling off, or decreasing and concave up, falling but levelling off. A full curve sketch puts both rows on one chart.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
Find the intervals of concavity and the point of inflection of $f(x) = x^3 - 6x^2 + 4$ .
Show solution
1. $f'(x) = 3x^2 - 12x$ and $f''(x) = 6x - 12$ , zero at $x = 2$ .
2. $f''(0) = -12 < 0$ and $f''(3) = 6 > 0$ : the sign changes.
3. $f(2) = 8 - 24 + 4 = -12$ .
Answer: Concave down on $(-\infty, 2)$ , concave up on $(2, \infty)$ ; inflection point $(2, -12)$ .
Easy
Use the second derivative test to classify the critical points of $f(x) = x^3 - 3x^2 - 9x + 1$ .
Show solution
1. $f'(x) = 3x^2 - 6x - 9 = 3(x - 3)(x + 1)$ , zero at $x = 3$ and $x = -1$ .
2. $f''(x) = 6x - 6$ . $f''(3) = 12 > 0$ , so a local min. $f''(-1) = -12 < 0$ , so a local max.
3. $f(3) = 27 - 27 - 27 + 1 = -26$ and $f(-1) = -1 - 3 + 9 + 1 = 6$ .
Answer: Local max $(-1, 6)$ , local min $(3, -26)$ .
Medium
Find the intervals of concavity and the points of inflection of $f(x) = x^4 - 6x^2$ .
Show solution
1. $f''(x) = 12x^2 - 12 = 12(x - 1)(x + 1)$ , zero at $x = \pm 1$ .
2. Signs of $f''$ : $+$ for $x < -1$ , $-$ on $(-1, 1)$ , $+$ for $x > 1$ . It changes at both.
3. $f(\pm 1) = 1 - 6 = -5$ .
Answer: Concave up on $(-\infty, -1)$ and $(1, \infty)$ , down on $(-1, 1)$ ; inflection points $(-1, -5)$ and $(1, -5)$ .
Medium
For $f(x) = x^4$ , show that $f''(0) = 0$ , then decide whether $(0, 0)$ is a point of inflection and classify it.
Show solution
1. $f'(x) = 4x^3$ and $f''(x) = 12x^2$ , so $f'(0) = 0$ and $f''(0) = 0$ .
2. $f''(x) = 12x^2 > 0$ on both sides of $0$ , so the concavity does not change: not a point of inflection.
3. The second derivative test is inconclusive. First derivative test: $f' < 0$ for $x < 0$ and $f' > 0$ for $x > 0$ .
Answer: Not a point of inflection; $(0, 0)$ is a local (and absolute) minimum.
Test level
Find $a$ and $b$ so that $f(x) = x^3 + ax^2 + bx$ has a point of inflection at $(1, 2)$ . Then show that this $f$ has no local maximum or minimum.
Show solution
1. $f''(x) = 6x + 2a$ . An inflection at $x = 1$ needs $6 + 2a = 0$ , so $a = -3$ . $f''$ is linear and changes sign there.
2. The point is on the curve: $f(1) = 1 - 3 + b = 2$ , so $b = 4$ .
3. $f'(x) = 3x^2 - 6x + 4$ . Its discriminant is $36 - 48 = -12 < 0$ , so $f'$ has no zeros.
4. $f'(0) = 4 > 0$ , so $f'(x) > 0$ everywhere and $f$ is always increasing, with no turning points.
Answer: $a = -3$ , $b = 4$ ; $f'(x) = 3x^2 - 6x + 4 > 0$ for all $x$ , so there is no local max or min.

Learn It on Martian Lab

Martian Lab teaches this in the Concavity and Second Derivative Test lesson of its MCV4U Curve Sketching unit. Watch it, then take the Curve Sketching practice tests to see if it stuck.

Watch the Concavity and Second Derivative Test lesson free

Start Practicing

No credit card required to start.

Get started

MCV4U: Concavity and the Second Derivative Test

How It Works

5 Worked Examples

Learn It on Martian Lab

Keep Going

Start Practicing