MCV4U Concavity, Points of Inflection and the Second Derivative Test: Notes and Practice

Two news headlines: "housing prices are still rising" and "housing prices are rising, but more slowly than last year". Both describe an increasing function, so the first derivative is positive in both. The difference is in how the SLOPE is changing. In the second headline the slope is getting smaller, so the graph is still climbing but bending over, like the top of a hill. Describing that bend needs a new tool: the derivative of the derivative.

The SECOND DERIVATIVE $f^{''} (x)$ is the derivative of $f^{'} (x)$ . Since $f^{'} (x)$ is the rate of change of $f$ , the second derivative is the rate of change of the rate of change: it measures how fast the slope of the tangent is changing. In motion, if $s (t)$ is position, $s^{'} (t)$ is velocity and $s^{''} (t)$ is acceleration, the rate at which velocity changes.

f^{''} (x) = \frac{d}{d x} [f^{'} (x)] = \frac{d ^{2} y}{d x ^{2}}

Two common notations for the second derivative. Differentiate once to get f′, then differentiate f′.

For a polynomial, just apply the power rule twice. If $f (x) = x^{4} - 6 x^{2}$ , then $f^{'} (x) = 4 x^{3} - 12 x$ and $f^{''} (x) = 12 x^{2} - 12$ . For a simple rational function, rewrite with negative exponents or use the quotient rule twice. If $f (x) = \frac{1}{x} = x^{- 1}$ , then $f^{'} (x) = - x^{- 2}$ and $f^{''} (x) = 2 x^{- 3} = \frac{2}{x ^{3}}$ . If $f (x) = x + \frac{4}{x}$ , then $f^{'} (x) = 1 - \frac{4}{x ^{2}}$ and $f^{''} (x) = \frac{8}{x ^{3}}$ .

Concavity

	Concave up	Concave down
Sign of $f^{''} (x)$	$f^{''} (x) > 0$	$f^{''} (x) < 0$
Tangent slopes $f^{'} (x)$	increasing as $x$ increases	decreasing as $x$ increases
Graph compared with its tangents	lies above its tangent lines	lies below its tangent lines
Shape	bends upward, like a cup	bends downward, like a cap
Simplest example	$y = x^{2}$	$y = - x^{2}$

Concavity describes which way a graph bends. It is separate from whether the graph is rising or falling.

Concavity and direction are independent. A graph can be increasing and concave up (rising faster and faster), increasing and concave down (rising more and more slowly, like the housing headline), decreasing and concave up (falling, but levelling off) or decreasing and concave down (falling faster and faster). The first derivative tells you the direction; the second derivative tells you the bend.

Fig 1The graph of f(x) = x³ − 3x² + 1, concave down to the left of x = 1 and concave up to the right, with the point of inflection (1, −1), the sign chart of f″(x) = 6x − 6 underneath, and the second derivative test classifying the local maximum (0, 1) and local minimum (2, −3).

Points of inflection

Point of inflection

A POINT OF INFLECTION is a point on the graph where the concavity changes, from up to down or from down to up. Candidates are the

x

-values in the domain where

f^{''} (x) = 0

f^{''} (x)

does not exist; a candidate is an inflection point only if

f^{''}

actually changes sign there.

At an inflection point the tangent slopes switch from increasing to decreasing, or the reverse, so the first derivative reaches a local maximum or minimum there. Geometrically, an inflection point is where the graph is steepest (or least steep) compared with nearby points. For the curriculum sample $f (x) = x^{3} + 2 x^{2} - 3 x$ : $f^{''} (x) = 6 x + 4$ is zero at $x = - \frac{2}{3}$ and changes from negative to positive, so $(- \frac{2}{3}, \frac{70}{27})$ is a point of inflection. Right there, $f^{'} (x) = 3 x^{2} + 4 x - 3$ has its minimum value $f^{'} (- \frac{2}{3}) = - \frac{13}{3}$ : the steepest downhill slope on the graph.

A zero second derivative is not enough

For

f (x) = x^{4}

f^{''} (x) = 12 x^{2}

, so

f^{''} (0) = 0

. But

12 x^{2} > 0

on both sides of

0

: the graph is concave up on both sides, and

(0, 0)

is a minimum, not an inflection point. Always check for a sign change. Also, a concavity change at a vertical asymptote (such as

x = 0

for

y = \frac{1}{x}

) is not an inflection point, because there is no point on the graph there.

VideoFinding intervals of concavity and points of inflection from a sign chart of the second derivative, the same method as this lesson.Watch on YouTube

The second derivative test

Find a critical number $c$ with $f^{'} (c) = 0$ , then evaluate $f^{''} (c)$ .
If $f^{''} (c) > 0$ , the graph is concave up at $c$ (a cup), so $f$ has a LOCAL MINIMUM at $c$ .
If $f^{''} (c) < 0$ , the graph is concave down at $c$ (a cap), so $f$ has a LOCAL MAXIMUM at $c$ .
If $f^{''} (c) = 0$ , the test is INCONCLUSIVE: $x^{4}$ , $- x^{4}$ and $x^{3}$ all have $f^{'} (0) = f^{''} (0) = 0$ , yet they have a minimum, a maximum and neither at $0$ . Go back to the first derivative test.

The other two sample functions are the simplest cases. $f (x) = 4 x + 1$ has $f^{''} (x) = 0$ everywhere: a straight line does not bend, so it is neither concave up nor concave down and has no inflection points. $f (x) = x^{2} + 3 x - 10$ has $f^{''} (x) = 2 > 0$ everywhere, so the parabola is concave up on all of $R$ , and the second derivative test confirms that its critical number $x = - \frac{3}{2}$ is a minimum.

Loading interactive…

Turn on the second derivative and drag a point along the curve: the graph is concave up where f″ is above the axis, concave down where it is below, and the inflection points are marked where f″ changes sign.

Concavity, Points of Inflection and the Second Derivative Test

Key things to remember

Concavity

Points of inflection

The second derivative test

Worked examples

Where marks get lost

Flashcards 8

Practice