Day 14: Derivative Tests & Concavity

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→

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 Exercises

Example 01Medium

Find the intervals of increase and decrease for

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

.

NEED A HINT?

Factor

f'(x)

completely, then build a sign chart using its three roots.

SHOW DETAILED EXPLANATION

Step 1: Differentiate and Factor

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

.

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)

.

Step 3: State the Intervals

Decreasing on

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

. Increasing on

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

.

Example 02Medium

Prove that

e^x \ge 1+x

for all

x \ge 0

.

NEED A HINT?

Define

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

, show

g(0)=0

, then show

g

is increasing on

[0,\infty)

.

SHOW DETAILED EXPLANATION

Step 1: Define a Difference Function

Let

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

. Then

g(0) = 1-1 = 0

.

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)

.

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

.

Example 03Medium

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 DETAILED EXPLANATION

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}

.

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

.

Step 3: State the Intervals

Increasing on

(-1,1)

. Decreasing on

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

.

Example 04Medium

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 DETAILED EXPLANATION

Step 1: Apply the Product Rule

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

.

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

.

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→

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 Exercises

Example 01Medium

Classify all critical points of

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

using the First Derivative Test.

NEED A HINT?

You already built the sign chart for

f'

— just read off the sign changes at

x=-1,0,2

.

SHOW DETAILED EXPLANATION

Step 1: Recall the Sign Pattern

From the sign chart:

f'

goes

- , +, -, +

across the intervals split by

x=-1,0,2

.

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.

Step 3: Find the Extreme Values

f(-1)=0

(local min),

f(0)=5

(local max),

f(2)=-27

(local min).

Example 02Hard

Use the First Derivative Test to classify the critical points of

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

.

NEED A HINT?

Factor

f'(x)

to find both critical points, then check whether the sign actually changes at each one.

SHOW DETAILED EXPLANATION

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)

.

Step 2: Find the Critical Points

x=0

(where

f'

is undefined, but

f(0)=0

is defined) and

x=1

(where

f'=0

).

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

.

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

.

Example 03Easy

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 DETAILED EXPLANATION

Conclusion

f

has a local maximum at

x=2

.

Example 04Medium

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 DETAILED EXPLANATION

Step 1: Recall the Critical Point

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

is zero only at

x=1

(

e^{-x}

is never zero).

Step 2: Check the Sign Change

f'>0

for

x<1

and

f'<0

for

x>1

— a change from positive to negative.

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→

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 Exercises

Example 01Medium

Find the intervals of concavity and any inflection points of

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

.

NEED A HINT?

Find

f''(x)

, factor it, then build a sign chart just like you would for

f'

.

SHOW DETAILED EXPLANATION

Step 1: Find f''(x)

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

.

Step 3: State Concavity

Concave up on

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

. Concave down on

(0,2)

.

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)

.

Example 02Hard

Discuss the concavity of

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

.

NEED A HINT?

f''

is undefined at

x=0

— check the sign on each side separately, since a real cube root exists for negative numbers too.

SHOW DETAILED EXPLANATION

Step 1: Find f''(x)

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

.

Step 2: Check the Sign for x > 0

For

x>0

,

x^{-5/3}>0

, so

f''(x)<0

— concave down.

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.

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).

Example 03Medium

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 DETAILED EXPLANATION

Step 1: Find y''

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

.

Step 2: Find Where y'' = 0

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

within

[0,2\pi]

.

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).

Step 4: State the Inflection Point

Concavity changes at

x=\pi

(interior point):

y(\pi)=3+0=3

. Inflection point at

(\pi, 3)

.

Example 04AB/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 DETAILED EXPLANATION

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

.

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}

.

Step 3: Use the Pythagorean Identity

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

.

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!→

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 Exercises

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

).

NEED A HINT?

Find

f''(x)

and evaluate its sign at each of the three critical points.

SHOW DETAILED EXPLANATION

Step 1: Find f''(x)

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

.

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

.

Step 3: Classify

Local min at

x=0

. Local max at

x=1

. Local min at

x=3

.

Example 02Hard

Use the Second Derivative Test to classify the critical points of

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

.

NEED A HINT?

Factor

f'(x)

to find the critical points first, then compute

f''(x)

and evaluate.

SHOW DETAILED EXPLANATION

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

.

Step 2: Find f''(x)

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

.

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.

Example 03Medium

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 DETAILED EXPLANATION

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.

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.

Step 3: Conclude

f

has a local (in fact, absolute) minimum at

x=0

.

Example 04Medium

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 DETAILED EXPLANATION

Step 1: Try the Second Derivative Test

f'(x)=3x^2

, critical point at

x=0

.

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

— inconclusive.

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

.

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 $x=0$ ).

Differentiate and Fully Factor

Find All Critical Points

Test One Point Per Interval

Practice Exercises

Step 1: Differentiate and Factor

Step 2: Build a Sign Chart

Step 3: State the Intervals

Step 1: Define a Difference Function

Step 2: Show g Is Increasing on [0,∞)

Step 3: Conclude

Step 1: Apply the Quotient Rule

Step 2: Analyze the Sign

Step 3: State the Intervals

Step 1: Apply the Product Rule

Step 2: Analyze the Sign

Step 3: State the Intervals

Build (or Reuse) the Sign Chart for f'

Read Off the Sign Change at Each Critical Point

Compute the Extreme Value

Practice Exercises

Step 1: Recall the Sign Pattern

Step 2: Classify Each Critical Point

Step 3: Find the Extreme Values

Step 1: Differentiate and Factor

Step 2: Find the Critical Points

Step 3: Analyze the Sign

Step 4: Classify Each Point

Conclusion

Step 1: Recall the Critical Point

Step 2: Check the Sign Change

Step 3: Conclude

Find f''(x) and Fully Factor

Build a Sign Chart for f''

Confirm an Actual Sign Change

Practice Exercises

Step 1: Find f''(x)

Step 2: Build a Sign Chart for f''

Step 3: State Concavity

Step 4: Identify Inflection Points

Step 1: Find f''(x)

Step 2: Check the Sign for x > 0

Step 3: Check the Sign for x < 0

Step 4: Conclude

Step 1: Find y''

Step 2: Find Where y'' = 0

Step 3: Check the Sign

Step 4: State the Inflection Point

Step 1: Find y''

Step 2: Set y'' = 0

Step 3: Use the Pythagorean Identity

Step 4: Substitute sin x = y/x

Step 5: Clear Denominators

Confirm f'(c) = 0

Evaluate f''(c)

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

Practice Exercises

Step 1: Find f''(x)

Step 2: Evaluate f'' at Each Critical Point

Step 3: Classify

Step 1: Find and Factor f'(x)

Step 2: Find f''(x)

Step 3: Evaluate and Classify

Step 1: Try the Second Derivative Test

Step 2: Fall Back to the First Derivative Test

Step 3: Conclude

Step 1: Try the Second Derivative Test

Step 2: Fall Back to the First Derivative Test

Step 3: Conclude

Gary Chang