How do you know if a function is increasing or decreasing?

Look at the sign of its derivative: increasing where f'(x)>0 and decreasing where f'(x)<0. A sign chart finds those intervals.

What is the difference between the first and second derivative tests?

The first derivative test checks whether f' changes sign at a critical number. The second derivative test checks the sign of f'' there. The first always gives an answer; the second fails when f''(c)=0.

What does the second derivative tell you?

Whether the slope is increasing or decreasing, which is how the graph bends: concave up (like a cup) when f''>0 and concave down (like a cap) when f''<0.

Is an inflection point where the second derivative is zero?

Only if the concavity actually changes there. f''=0 is a candidate, like a critical number; check that f'' changes sign.

What does concave up mean in real life?

That the rate of change is increasing. For a position graph, concave up means positive acceleration: the velocity is going up.

Increasing, Decreasing and Concavity: Derivative Tests

Q: What is the difference between the first and second derivative tests?

The first derivative test checks whether f' changes sign at a critical number. The second derivative test checks the sign of f'' there. The first always gives an answer; the second fails when f''(c)=0.

Q: What does the second derivative tell you?

Whether the slope is increasing or decreasing, which is how the graph bends: concave up (like a cup) when f''>0 and concave down (like a cap) when f''<0.

Q: Is an inflection point where the second derivative is zero?

Only if the concavity actually changes there. f''=0 is a candidate, like a critical number; check that f'' changes sign.

Q: What does concave up mean in real life?

That the rate of change is increasing. For a position graph, concave up means positive acceleration: the velocity is going up.

How do you find where a function is increasing or decreasing?

A function is increasing on an interval if its outputs go up as the inputs go up (the graph rises from left to right), and decreasing if they go down. The derivative is the slope of the graph, so its sign tells you which is happening:

The increasing/decreasing test

f^{'} (x) > 0

for every

x

in an interval,

f

is increasing there. If

f^{'} (x) < 0

for every

x

in an interval,

f

is decreasing there.

Why it works

Take any two inputs $x_{1} < x_{2}$ in the interval.
By the Mean Value Theorem, $\frac{f (x_{2}) - f (x_{1})}{x_{2} - x_{1}} =$ $f^{'} (c)$ for some $c$ between them.
If $f^{'} (c) > 0$ , the top $f (x_{2}) - f (x_{1})$ is positive (the bottom is positive), so $f (x_{2}) > f (x_{1})$ : the function went up.

$f^{'}$ can only change sign where it is $0$ or undefined (for the functions in a first course), so those points cut the number line into intervals on which the sign stays the same. A sign chart records that sign in each interval:

Find $f^{'} (x)$ and factor it if you can.
Find the critical numbers: where $f^{'} (x) = 0$ or $f^{'} (x)$ doesn't exist. Also mark any $x$ where $f$ itself is undefined.
These points split the number line into intervals. Test one number in each interval to find the sign of $f^{'}$ there.
$+$ means increasing; $-$ means decreasing.

Worked example: Increasing and decreasing intervals

Find where $f (x) = x^{3} - 3 x^{2} - 24 x + 2$ is increasing and where it is decreasing.

$f^{'} (x) =$ $3 x^{2} - 6 x - 24 =$ $3 (x^{2} - 2 x - 8) =$ $3 (x - 4) (x + 2)$ .
Critical numbers: $f^{'} (x) = 0$ at $x = - 2$ and $x = 4$ .
Test $x = - 3$ : $3 (- 7) (- 1) = 21 > 0$ . Test $x = 0$ : $3 (- 4) (2) = - 24 < 0$ . Test $x = 5$ : $3 (1) (7) = 21 > 0$ .
So $f^{'} > 0$ on $(- \infty, - 2)$ , $f^{'} < 0$ on $(- 2, 4)$ , and $f^{'} > 0$ on $(4, \infty)$ .

Answer: Increasing on $(- \infty, - 2)$ and $(4, \infty)$ ; decreasing on $(- 2, 4)$ .

What is the first derivative test?

Picture walking along the graph from left to right. If you climb up to a point and then go down after it, that point is a hilltop: a local maximum. If you go down and then climb, it's a valley: a local minimum. If you climb on both sides, it's neither. In terms of the derivative, "climbing" means $f^{'} > 0$ , so the sign chart settles it.

The first derivative test

At a critical number

c

: if

f^{'}

changes from positive to negative,

f

has a local maximum at

c

; from negative to positive, a local minimum; and if

f^{'}

has the same sign on both sides, neither.

Worked example: Local maximum and minimum

Find the local maximum and minimum values of $f (x) = x^{3} - 3 x^{2} - 24 x + 2$ .

From the sign chart above, $f^{'}$ changes from $+$ to $-$ at $x = - 2$ , and from $-$ to $+$ at $x = 4$ .
Local maximum at $x = - 2$ : $f (- 2) = - 8 - 12 + 48 + 2 = 30$ .
Local minimum at $x = 4$ : $f (4) = 64 - 48 - 96 + 2 = - 78$ .

Answer: Local maximum $30$ at $x = - 2$ ; local minimum $- 78$ at $x = 4$ .

f (x) = x^{3} - 3 x^{2} - 24 x + 2

: rising, then falling between the local maximum at

x = - 2

and the local minimum at

x = 4

, then rising again. The bend changes direction at the inflection point

x = 1

What is concavity?

Knowing a graph rises doesn't say how it rises: it could curve upward like a cup, getting steeper, or curve downward like a cap, leveling off. This bending is called concavity, and it is measured by the second derivative $f^{″}$ , which tells you whether the slope $f^{'}$ is itself increasing or decreasing.

Definition: concave up and concave down

The graph of $f$ is concave up on an interval if its slope $f^{'}$ is increasing there (the graph bends upward, like a cup); it is concave down if $f^{'}$ is decreasing (it bends downward, like a cap).

The concavity test

f^{″} (x) > 0

on an interval, the graph is concave up there; if

f^{″} (x) < 0

, it is concave down.

Definition: inflection point

An inflection point is a point on the graph where the concavity changes, from up to down or from down to up.

Worked example: Concavity and inflection points

Find the concavity of $f (x) = x^{3} - 3 x^{2} - 24 x + 2$ and its inflection point.

$f^{'} (x) = 3 x^{2} - 6 x - 24$ , so $f^{″} (x) = 6 x - 6 = 6 (x - 1)$ .
$f^{″} (x) = 0$ at $x = 1$ . Test $x = 0$ : $f^{″} = - 6 < 0$ . Test $x = 2$ : $f^{″} = 6 > 0$ .
So the graph is concave down on $(- \infty, 1)$ and concave up on $(1, \infty)$ .
The concavity changes at $x = 1$ , and $f (1) = 1 - 3 - 24 + 2 = - 24$ .

Answer: Concave down on $(- \infty, 1)$ , concave up on $(1, \infty)$ ; inflection point $(1, - 24)$ .

Common mistake

Calling every point where $f^{″} = 0$ an inflection point. For $f (x) = x^{4}$ , $f^{″} (x) = 12 x^{2}$ is $0$ at $x = 0$ , but it is positive on both sides, so the graph is concave up throughout and $x = 0$ is not an inflection point. The sign of $f^{″}$ must actually change.

What is the second derivative test?

Concavity gives a second way to classify a critical number where $f^{'} (c) = 0$ . The tangent line there is flat. If the graph bends upward like a cup, that flat spot is the bottom of the cup; if it bends downward like a cap, it's the top.

The second derivative test

Suppose

f^{'} (c) = 0

. If

f^{″} (c) > 0

f

has a local minimum at

c

. If

f^{″} (c) < 0

, a local maximum. If

f^{″} (c) = 0

, the test gives no answer; use the first derivative test instead.

Worked example: Both tests on one function

For $f (x) = x e^{- x}$ , find where $f$ increases and decreases, classify the critical number, and find the inflection point.

Product and chain rules: $f^{'} (x) =$ $1 \cdot e^{- x} + x \cdot (- e^{- x}) =$ $(1 - x) e^{- x}$ .
$e^{- x}$ is always positive, so $f^{'}$ has the sign of $1 - x$ : positive for $x < 1$ , negative for $x > 1$ . Increasing on $(- \infty, 1)$ , decreasing on $(1, \infty)$ .
Differentiate again: $f^{″} (x) =$ $- e^{- x} + (1 - x) (- e^{- x}) =$ $e^{- x} (- 1 - 1 + x) =$ $(x - 2) e^{- x}$ .
Second derivative test at $x = 1$ : $f^{″} (1) =$ $(1 - 2) e^{- 1} =$ $- \frac{1}{e} < 0$ , so a local maximum, with value $f (1) =$ $e^{- 1} =$ $\frac{1}{e} \approx 0.368$ .
$f^{″}$ has the sign of $x - 2$ : negative for $x < 2$ , positive for $x > 2$ . Concavity changes at $x = 2$ , where $f (2) =$ $2 e^{- 2} \approx 0.271$ .

Answer: Local maximum $\frac{1}{e}$ at $x = 1$ ; inflection point $(2, \frac{2}{e^{2}})$ .

y = x e^{- x}

rises to its maximum at

x = 1

, then falls toward

0

; it changes from concave down to concave up at

x = 2

How do the signs of f′ and f″ describe a graph?

$f^{'}$	$f^{″}$	The graph is
$+$	$+$	rising and getting steeper (bending up)
$+$	$-$	rising but leveling off (bending down)
$-$	$+$	falling but leveling off (bending up)
$-$	$-$	falling and getting steeper (bending down)

Together with intercepts and asymptotes, these signs are everything you need to sketch a graph by hand. In applications, an inflection point marks where a rate is fastest or slowest, such as the point of diminishing returns in business. To find the absolute highest and lowest values on an interval, see finding maximum and minimum values.

Increasing, decreasing, and concavity: the first and second derivative tests