Day 13: Analyzing Functions with Derivatives

Increasing, Decreasing, and Monotonicity

The sign of the derivative is a live readout of which direction a function is heading — positive means climbing, negative means falling. Reading a sign chart off $f'$ is the single most useful skill in this entire unit.

Core Theorem

Suppose

f

is continuous on

[a,b]

and differentiable on

(a,b)

.
If

f'(x) > 0

for all

x \in (a,b)

, then

f

is increasing on

(a,b)

.
If

f'(x) < 0

for all

x \in (a,b)

, then

f

is decreasing on

(a,b)

.

Watch the TikTok ExplanationIncreasing, Decreasing, and Monotonic Functions→

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

Find the intervals of increase and decrease for

f(x) = 3x^4-4x^3-12x^2+5

.

How to think about it

Key idea: Factor $f'(x)$ completely, then build a sign chart using its three roots.

Full solution

1. Step 1: Differentiate and Factor

f'(x) = 12x^3-12x^2-24x = 12x(x^2-x-2) = 12x(x-2)(x+1)

.

2. Step 2: Build a Sign Chart

Roots at

x=-1,0,2

split the number line into 4 intervals. Testing each: negative on

(-\infty,-1)

, positive on

(-1,0)

, negative on

(0,2)

, positive on

(2,\infty)

.

3. Step 3: State the Intervals

Decreasing on

(-\infty,-1) \cup (0,2)

. Increasing on

(-1,0) \cup (2,\infty)

.

Worked Example 02Medium

Prove that

e^x \ge 1+x

for all

x \ge 0

.

How to think about it

Key idea: Define $g(x)=e^x-(1+x)$ , show $g(0)=0$ , then show $g$ is increasing on $[0,\infty)$ .

Full solution

1. Step 1: Define a Difference Function

Let

g(x) = e^x - (1+x)

. Then

g(0) = 1-1 = 0

.

2. Step 2: Show g Is Increasing on [0,∞)

g'(x) = e^x - 1

. For

x \ge 0

,

e^x \ge 1

, so

g'(x) \ge 0

—

g

is increasing on

[0,\infty)

.

3. Step 3: Conclude

Since

g(0)=0

and

g

is increasing from there,

g(x) \ge 0

for all

x \ge 0

, i.e.

e^x \ge 1+x

.

Step-by-Step SOP

1
Differentiate and Fully Factor
Find $f'(x)$ and factor it into simple linear or irreducible pieces.
2
Find All Critical Points
These are the boundaries between your sign-chart intervals.
3
Test One Point Per Interval
Plug in a convenient test value from each interval to determine the sign of $f'$ there, then read off increasing/decreasing directly.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

Find the intervals of increase and decrease for

f(x) = \dfrac{x}{x^2+1}

.

Need a hint?

The denominator of

f'(x)

is always positive after the quotient rule — the sign of

f'

depends only on the numerator.

Show solution

1. Step 1: Apply the Quotient Rule

f'(x) = \frac{(1)(x^2+1)-x(2x)}{(x^2+1)^2} = \frac{1-x^2}{(x^2+1)^2}

.

2. Step 2: Analyze the Sign

The denominator is always positive, so the sign of

f'

matches the sign of

1-x^2

: positive when

-1<x<1

, negative when

|x|>1

.

3. Step 3: State the Intervals

Increasing on

(-1,1)

. Decreasing on

(-\infty,-1) \cup (1,\infty)

.

Practice 02Medium

Find the intervals of increase and decrease for

f(x) = xe^{-x}

.

Need a hint?

Apply the product rule, then factor out

e^{-x}

(which is always positive) before analyzing the sign.

Show solution

1. Step 1: Apply the Product Rule

f'(x) = e^{-x} + x(-e^{-x}) = e^{-x}(1-x)

.

2. Step 2: Analyze the Sign

e^{-x}>0

always, so the sign of

f'

matches the sign of

(1-x)

: positive for

x<1

, negative for

x>1

.

3. Step 3: State the Intervals

Increasing on

(-\infty,1)

. Decreasing on

(1,\infty)

.

Common Pitfalls

⚠
Always Factor Before Building the Sign ChartTrying to determine the sign of an unfactored derivative by plugging in random numbers is slow and error-prone — factor completely first so the roots (and the sign pattern between them) are obvious.
⚠
The Denominator Doesn't Always VanishWhen $f'$ is a fraction, check whether the denominator is always positive (like $(x^2+1)^2$ ) — if so, you only need to analyze the sign of the numerator.

The First Derivative Test

Once you have a sign chart for $f'$ , classifying every critical point as a local max, local min, or neither is just a matter of reading how the sign changes as you pass through it.

Core Theorem

Let

c

be a critical point of a continuous function

f

.
If

f'

changes from negative to positive at

c

,

f

has a local minimum at

c

.
If

f'

changes from positive to negative at

c

,

f

has a local maximum at

c

.
If

f'

does not change sign at

c

,

f

has neither a local max nor a local min there.

Watch the TikTok ExplanationFirst Derivative Test→

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

Classify all critical points of

f(x) = 3x^4-4x^3-12x^2+5

using the First Derivative Test.

How to think about it

Key idea: You already built the sign chart for $f'$ — just read off the sign changes at $x=-1,0,2$ .

Full solution

1. Step 1: Recall the Sign Pattern

From the sign chart:

f'

goes

- , +, -, +

across the intervals split by

x=-1,0,2

.

2. Step 2: Classify Each Critical Point

At

x=-1

:

-

to

+

\implies

local min. At

x=0

:

+

to

-

\implies

local max. At

x=2

:

-

to

+

\implies

local min.

3. Step 3: Find the Extreme Values

f(-1)=0

(local min),

f(0)=5

(local max),

f(2)=-27

(local min).

Worked Example 02Hard

Use the First Derivative Test to classify the critical points of

f(x) = x^{4/3}-4x^{1/3}

.

How to think about it

Key idea: Factor $f'(x)$ to find both critical points, then check whether the sign actually changes at each one.

Full solution

1. Step 1: Differentiate and Factor

f'(x) = \frac{4}{3}x^{1/3} - \frac{4}{3}x^{-2/3} = \frac{4}{3}x^{-2/3}(x-1)

.

2. Step 2: Find the Critical Points

x=0

(where

f'

is undefined, but

f(0)=0

is defined) and

x=1

(where

f'=0

).

3. Step 3: Analyze the Sign

x^{-2/3}

is always positive, so the sign of

f'

matches

(x-1)

: negative for

x<1

, positive for

x>1

.

4. Step 4: Classify Each Point

At

x=0

: the sign is negative on BOTH sides (no change) — NOT an extremum. At

x=1

: sign changes from

-

to

+

\implies

local min, with

f(1) = 1-4 = -3

.

Step-by-Step SOP

1
Build (or Reuse) the Sign Chart for f'
Identify every critical point and the sign of $f'$ immediately to its left and right.
2
Read Off the Sign Change at Each Critical Point
Negative-to-positive is a local min; positive-to-negative is a local max; no change means neither.
3
Compute the Extreme Value
Plug the classified critical point back into $f$ (not $f'$ ) to state the actual local max/min value.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Easy

If

f'(x)

changes from positive to negative at

x=2

, and

f

is continuous there, classify

x=2

.

Need a hint?

Positive-to-negative means the function stops climbing and starts falling.

Show solution

1. Conclusion

f

has a local maximum at

x=2

.

Practice 02Medium

Use the First Derivative Test to classify the critical point of

f(x) = xe^{-x}

.

Need a hint?

You already found

f'(x) = e^{-x}(1-x)

has exactly one critical point.

Show solution

1. Step 1: Recall the Critical Point

f'(x) = e^{-x}(1-x)

is zero only at

x=1

(

e^{-x}

is never zero).

2. Step 2: Check the Sign Change

f'>0

for

x<1

and

f'<0

for

x>1

— a change from positive to negative.

3. Step 3: Conclude

f

has a local maximum at

x=1

, with

f(1) = \frac{1}{e}

.

Common Pitfalls

⚠
No Sign Change Means No ExtremumA critical point where $f'$ has the SAME sign on both sides (like $x=0$ for $x^{4/3}-4x^{1/3}$ , or $x=0$ for $x^3$ ) is not a local max or min — don't assume every critical point is automatically an extremum.
⚠
Reuse Your Sign ChartThe First Derivative Test doesn't require new work if you've already built a sign chart for increasing/decreasing — the classifications fall right out of it.

Concavity and Inflection Points

Concavity describes how a curve bends — whether it cups upward like a bowl or curves downward like a dome. That bend is controlled entirely by the sign of the second derivative, and a point where the bend flips is called an inflection point.

Core Theorem

Let

f

be twice differentiable on an interval

I

.
If

f''(x) > 0

on

I

, the graph of

f

is concave up (ccu) on

I

.
If

f''(x) < 0

on

I

, the graph of

f

is concave down (ccd) on

I

.
A point

(c, f(c))

is an inflection point if

f

is continuous there and the concavity of

f

actually changes at

c

(which requires

f''(c)=0

or undefined there).

Watch the TikTok ExplanationConcavity of Function Graphs→

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

Find the intervals of concavity and any inflection points of

f(x) = x^4-4x^3

.

How to think about it

Key idea: Find $f''(x)$ , factor it, then build a sign chart just like you would for $f'$ .

Full solution

1. Step 1: Find f''(x)

f'(x)=4x^3-12x^2 \implies f''(x) = 12x^2-24x = 12x(x-2)

.

2. Step 2: Build a Sign Chart for f''

Roots at

x=0,2

. Testing: positive on

(-\infty,0)

, negative on

(0,2)

, positive on

(2,\infty)

.

3. Step 3: State Concavity

Concave up on

(-\infty,0) \cup (2,\infty)

. Concave down on

(0,2)

.

4. Step 4: Identify Inflection Points

Concavity changes at both

x=0

and

x=2

:

f(0)=0

and

f(2)=16-32=-16

. Inflection points at

(0,0)

and

(2,-16)

.

Worked Example 02Hard

Discuss the concavity of

f(x) = x^{1/3}

.

How to think about it

Key idea: $f''$ is undefined at $x=0$ — check the sign on each side separately, since a real cube root exists for negative numbers too.

Full solution

1. Step 1: Find f''(x)

f'(x) = \frac{1}{3}x^{-2/3} \implies f''(x) = -\frac{2}{9}x^{-5/3}

.

2. Step 2: Check the Sign for x > 0

For

x>0

,

x^{-5/3}>0

, so

f''(x)<0

— concave down.

3. Step 3: Check the Sign for x < 0

For

x<0

,

x^{1/3}<0

, so

x^{5/3}=(x^{1/3})^5<0

, making

x^{-5/3}<0

too. So

f''(x) = -\frac{2}{9}(\text{negative}) > 0

— concave up.

4. Step 4: Conclude

Concave up on

(-\infty,0)

, concave down on

(0,\infty)

. Since the concavity flips and

f

is continuous at

x=0

,

(0,0)

is an inflection point — even though

f''(0)

doesn't exist (there's a vertical tangent there).

Step-by-Step SOP

1
Find f''(x) and Fully Factor
Differentiate twice, then factor completely to expose the roots clearly.
2
Build a Sign Chart for f''
Mark every zero (and every point where $f''$ is undefined) and test the sign of $f''$ on each resulting interval.
3
Confirm an Actual Sign Change
Only report an inflection point where the concavity genuinely flips from one side to the other.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

Find the concavity and inflection points of

y = 3+\sin x

on

[0, 2\pi]

.

Need a hint?

y'' = -\sin x

— find where this is zero within

[0,2\pi]

and check the sign on each side.

Show solution

1. Step 1: Find y''

y' = \cos x \implies y'' = -\sin x

.

2. Step 2: Find Where y'' = 0

-\sin x = 0 \implies x = 0, \pi, 2\pi

within

[0,2\pi]

.

3. Step 3: Check the Sign

On

(0,\pi)

:

\sin x>0 \implies y''<0

(concave down). On

(\pi,2\pi)

:

\sin x<0 \implies y''>0

(concave up).

4. Step 4: State the Inflection Point

Concavity changes at

x=\pi

(interior point):

y(\pi)=3+0=3

. Inflection point at

(\pi, 3)

.

Practice 02AB/BC Standard

Show that the inflection point of

y = x\sin x

lies on the curve

y^2(x^2+4) = 4x^2

.

Need a hint?

Find where

y''=0

to get an equation relating

\cos x

and

y

, then combine it with the Pythagorean identity and

y=x\sin x

itself to eliminate

x

and

\theta

.

Show solution

1. Step 1: Find y''

y' = \sin x + x\cos x \implies y'' = \cos x + \cos x - x\sin x = 2\cos x - x\sin x

.

2. Step 2: Set y'' = 0

2\cos x = x \sin x

. Since

y = x\sin x

, this says

2\cos x = y

, i.e.

\cos x = \frac{y}{2}

.

3. Step 3: Use the Pythagorean Identity

\sin^2 x = 1-\cos^2 x = 1 - \frac{y^2}{4}

.

4. Step 4: Substitute sin x = y/x

From

y=x\sin x

:

\sin x = \frac{y}{x} \implies \sin^2 x = \frac{y^2}{x^2}

. Setting the two expressions for

\sin^2 x

equal:

\frac{y^2}{x^2} = 1-\frac{y^2}{4}

.

5. Step 5: Clear Denominators

Multiply both sides by

4x^2

:

4y^2 = 4x^2-x^2y^2 \implies y^2(4+x^2) = 4x^2

, exactly as required.

Common Pitfalls

⚠
f''(c) = 0 Alone Doesn't Guarantee an Inflection PointJust like a critical point needs a sign CHANGE in $f'$ to be an extremum, a candidate point needs concavity to actually flip on either side — otherwise it's not a true inflection point.
⚠
Inflection Points Can Occur Where f'' Is Undefined TooDon't only check zeros of $f''$ — also check points where $f''$ fails to exist (like $x=0$ for $x^{1/3}$ ), as long as $f$ itself is still continuous there.

The Second Derivative Test

At a critical point where the tangent is flat, checking concavity there tells you immediately whether you're sitting in a valley or on a hilltop — no sign chart needed, as long as the second derivative isn't also zero.

Core Theorem

Let

c

be a critical point of

f

with

f'(c)=0

.
If

f''(c) > 0

,

f

has a local minimum at

c

.
If

f''(c) < 0

,

f

has a local maximum at

c

.
If

f''(c) = 0

, the test is inconclusive — fall back to the First Derivative Test.

Watch the TikTok ExplanationSecond Derivative Test!→

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

Use the Second Derivative Test to classify the critical points of

f(x) = 3x^4-16x^3+18x^2

(critical points

x=0,1,3

).

How to think about it

Key idea: Find $f''(x)$ and evaluate its sign at each of the three critical points.

Full solution

1. Step 1: Find f''(x)

f'(x)=12x^3-48x^2+36x \implies f''(x) = 36x^2-96x+36 = 12(3x^2-8x+3)

.

2. Step 2: Evaluate f'' at Each Critical Point

f''(0)=12(3)=36>0

.

f''(1)=12(3-8+3)=-24<0

.

f''(3)=12(27-24+3)=72>0

.

3. Step 3: Classify

Local min at

x=0

. Local max at

x=1

. Local min at

x=3

.

Worked Example 02Hard

Use the Second Derivative Test to classify the critical points of

f(x) = (x^2-3)e^x

.

How to think about it

Key idea: Factor $f'(x)$ to find the critical points first, then compute $f''(x)$ and evaluate.

Full solution

1. Step 1: Find and Factor f'(x)

f'(x) = 2xe^x+(x^2-3)e^x = e^x(x^2+2x-3) = e^x(x+3)(x-1)

. Critical points:

x=-3, 1

.

2. Step 2: Find f''(x)

f''(x) = e^x(x^2+2x-3) + e^x(2x+2) = e^x(x^2+4x-1)

.

3. Step 3: Evaluate and Classify

f''(-3) = e^{-3}(9-12-1) = -4e^{-3} < 0 \implies

local max.

f''(1) = e(1+4-1) = 4e > 0 \implies

local min.

Step-by-Step SOP

1
Confirm f'(c) = 0
The Second Derivative Test only applies where the critical point comes from a zero derivative, not an undefined one.
2
Evaluate f''(c)
Positive means local min, negative means local max.
3
If f''(c) = 0, Switch to the First Derivative Test
Build a sign chart for $f'$ around $c$ instead — that always works, even when the Second Derivative Test can't decide.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

Show that the Second Derivative Test fails for

f(x)=x^4

at

x=0

, and classify it using the First Derivative Test instead.

Need a hint?

Compute

f''(0)

first — if it's zero, you'll need a different approach.

Show solution

1. Step 1: Try the Second Derivative Test

f'(x)=4x^3

, critical point at

x=0

.

f''(x)=12x^2 \implies f''(0)=0

— inconclusive.

2. Step 2: Fall Back to the First Derivative Test

f'(x)=4x^3

is negative for

x<0

and positive for

x>0

— a sign change from negative to positive.

3. Step 3: Conclude

f

has a local (in fact, absolute) minimum at

x=0

.

Practice 02Medium

Show that the Second Derivative Test also fails for

f(x)=x^3

at

x=0

, and explain what the First Derivative Test reveals instead.

Need a hint?

Again compute

f''(0)

, and if inconclusive, check whether

f'(x)

actually changes sign around

x=0

.

Show solution

1. Step 1: Try the Second Derivative Test

f'(x)=3x^2

, critical point at

x=0

.

f''(x)=6x \implies f''(0)=0

— inconclusive.

2. Step 2: Fall Back to the First Derivative Test

f'(x)=3x^2 \ge 0

for ALL

x

— it never changes sign around

x=0

.

3. Step 3: Conclude

Since

f'

doesn't change sign,

x=0

is NOT a local extremum for

x^3

— it's a saddle-like inflection point instead.

Common Pitfalls

⚠
The Second Derivative Test Only Applies at f'(c) = 0If $f'(c)$ is undefined rather than zero (like a corner), the Second Derivative Test doesn't apply at all — go straight to the First Derivative Test.
⚠
f''(c) = 0 Means Inconclusive, NOT 'Neither'A common mistake is assuming $f''(c)=0$ means $c$ is not an extremum. It actually means the test simply can't tell you — the point could still be a min, a max, or neither (see $x^4$ vs. $x^3$ at $x=0$ ).

Connecting f, f′, and f″: Graphical Analysis

The AP exam constantly gives you the graph of $f'$ (or a table, or $f''$ ) and asks about $f$ — where it increases, has extrema, changes concavity, or attains an absolute max. The whole skill is a translation table between one graph and the next.

Core Theorem

Reading up (from

f'

to

f

):
•

f' > 0 \Rightarrow f

increasing;

f' < 0 \Rightarrow f

decreasing
•

f'

changes

+\to- \Rightarrow f

has a local max;

-\to+ \Rightarrow

local min
•

f'

increasing

\Rightarrow f

concave up;

f'

decreasing

\Rightarrow f

concave down
•

f'

has a local extremum

\Rightarrow f

has an inflection point there
Reading down (from

f

to

f'

): slopes of

f

become heights of

f'

; where

f

has a horizontal tangent,

f'

crosses zero.

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Easy

The graph of

f'

is a line through

(0,-2)

,

(2,0)

, and

(4,2)

(i.e.

f'(x) = x - 2

). On what interval is

f

decreasing, and where does

f

have a local minimum?

How to think about it

Key idea: $f$ decreases where $f' < 0$ ; a local min is where $f'$ changes from $-$ to $+$ .

Full solution

1. Step 1: Find Where f' < 0

f'(x) = x - 2 < 0

for

x < 2

, so

f

is decreasing on

(-\infty, 2)

.

2. Step 2: Locate the Local Minimum

f'

changes from negative to positive at

x = 2

, so

f

has a local minimum at

x = 2

.

Worked Example 02Medium

The graph of

f'

on

[0,6]

is piecewise linear through

(0,2)

,

(2,-2)

,

(4,-2)

,

(6,2)

. On what interval(s) is the graph of

f

concave up? Justify.

How to think about it

Key idea: $f$ is concave up where $f'$ is increasing — i.e. where the graph of $f'$ has positive slope.

Full solution

1. Step 1: Find Where f' Is Increasing

The graph of

f'

has slope

-2

on

(0,2)

, slope

0

on

(2,4)

, and slope

+2

on

(4,6)

. So

f'

is increasing only on

(4,6)

.

2. Step 2: Conclude

The graph of

f

is concave up on

(4,6)

because

f'

is increasing (equivalently

f'' > 0

) there.

Step-by-Step SOP

1
Identify Which Graph You Were Given
Is it $f$ , $f'$ , or $f''$ ? Every conclusion depends on knowing which curve you are reading.
2
Translate One Level at a Time
From $f'$ : sign → increasing/decreasing of $f$ ; sign change → extremum of $f$ ; increasing/decreasing of $f'$ → concavity of $f$ ; extremum of $f'$ → inflection of $f$ .
3
For a Value of f, Accumulate Area
$f(b) = f(a) + \int_a^b f'(x)\,dx$ — read the definite integral as signed geometric area under the graph of $f'$ .
4
Justify with the Named Fact
" $f$ is concave up on $(4,6)$ because $f'$ is increasing there" — always tie the claim to the sign or behavior of a specific derivative.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Hard

f

is differentiable on

[0,8]

with

f(0) = 3

. The graph of

f'

consists of a semicircle of radius 2 above the

x

-axis on

[0,4]

and the segment from

(4,0)

to

(8,-4)

. Find

f(8)

.

Need a hint?

f(8) = f(0) + \int_0^8 f'(x)\,dx

. Read the integral as signed area: a half-disk minus a triangle.

Show solution

1. Step 1: Area Under the Semicircle (0 to 4)

A semicircle of radius

2

has area

\tfrac{1}{2}\pi(2)^2 = 2\pi

. It is above the axis, so

\int_0^4 f'\,dx = 2\pi

.

2. Step 2: Area Under the Triangle (4 to 8)

The segment from

(4,0)

to

(8,-4)

forms a triangle below the axis with base

4

and height

4

: signed area

= -\tfrac{1}{2}(4)(4) = -8

.

3. Step 3: Accumulate

f(8) = f(0) + \int_0^8 f'\,dx = 3 + (2\pi - 8) = 2\pi - 5 \approx 1.283

.

Practice 02Medium

The graph of

f'

is positive and increasing on all of

[0,5]

. What can you conclude about

f

on

[0,5]

?

Need a hint?

Use the sign of

f'

for direction and the behavior of

f'

for concavity.

Show solution

1. Step 1: Direction

f' > 0

everywhere on

[0,5]

, so

f

is increasing on

[0,5]

— the absolute maximum of

f

is at

x = 5

and the absolute minimum at

x = 0

.

2. Step 2: Concavity

f'

is increasing, so

f'' > 0

and the graph of

f

is concave up on

[0,5]

— no inflection points.

Common Pitfalls

⚠
Confusing "f' Is Negative" with "f' Is Decreasing" $f' < 0$ tells you $f$ is going down. $f'$ decreasing tells you $f$ is concave down. These are independent — $f'$ can be negative while increasing.
⚠
Reading the Given Graph as f When It Is f'If the problem hands you the graph of $f'$ , a peak on that graph is NOT a max of $f$ — it is an inflection point of $f$ . Label the axes in your head before answering.

Worked Examples

1. Step 1: Differentiate and Factor

2. Step 2: Build a Sign Chart

3. Step 3: State the Intervals

1. Step 1: Define a Difference Function

2. Step 2: Show g Is Increasing on [0,∞)

3. Step 3: Conclude

Differentiate and Fully Factor

Find All Critical Points

Test One Point Per Interval

Practice — Your Turn

1. Step 1: Apply the Quotient Rule

2. Step 2: Analyze the Sign

3. Step 3: State the Intervals

1. Step 1: Apply the Product Rule

2. Step 2: Analyze the Sign

3. Step 3: State the Intervals

Worked Examples

1. Step 1: Recall the Sign Pattern

2. Step 2: Classify Each Critical Point

3. Step 3: Find the Extreme Values

1. Step 1: Differentiate and Factor

2. Step 2: Find the Critical Points

3. Step 3: Analyze the Sign

4. Step 4: Classify Each Point

Build (or Reuse) the Sign Chart for f'

Read Off the Sign Change at Each Critical Point

Compute the Extreme Value

Practice — Your Turn

1. Conclusion

1. Step 1: Recall the Critical Point

2. Step 2: Check the Sign Change

3. Step 3: Conclude

Worked Examples

1. Step 1: Find f''(x)

2. Step 2: Build a Sign Chart for f''

3. Step 3: State Concavity

4. Step 4: Identify Inflection Points

1. Step 1: Find f''(x)

2. Step 2: Check the Sign for x > 0

3. Step 3: Check the Sign for x < 0

4. Step 4: Conclude

Find f''(x) and Fully Factor

Build a Sign Chart for f''

Confirm an Actual Sign Change

Practice — Your Turn

1. Step 1: Find y''

2. Step 2: Find Where y'' = 0

3. Step 3: Check the Sign

4. Step 4: State the Inflection Point

1. Step 1: Find y''

2. Step 2: Set y'' = 0

3. Step 3: Use the Pythagorean Identity

4. Step 4: Substitute sin x = y/x

5. Step 5: Clear Denominators

Worked Examples

1. Step 1: Find f''(x)

2. Step 2: Evaluate f'' at Each Critical Point

3. Step 3: Classify

1. Step 1: Find and Factor f'(x)

2. Step 2: Find f''(x)

3. Step 3: Evaluate and Classify

Confirm f'(c) = 0

Evaluate f''(c)

If f''(c) = 0, Switch to the First Derivative Test

Practice — Your Turn

1. Step 1: Try the Second Derivative Test

2. Step 2: Fall Back to the First Derivative Test

3. Step 3: Conclude

1. Step 1: Try the Second Derivative Test

2. Step 2: Fall Back to the First Derivative Test

3. Step 3: Conclude

Worked Examples

1. Step 1: Find Where f' < 0

2. Step 2: Locate the Local Minimum

1. Step 1: Find Where f' Is Increasing

2. Step 2: Conclude

Identify Which Graph You Were Given

Translate One Level at a Time

For a Value of f, Accumulate Area