Single-Variable-Calculus · Unit 8 · Video 3 · Interactive Practice

Reading a Graph from the Sign of the Derivative: Critical Points of 3x − x³

IKey Formulas

Formula	Name	What it tells you
$f' > 0 \Rightarrow f$ increasing, $\ f' < 0 \Rightarrow f$ decreasing	First principle	The tangent line points up or down, and $f$ follows it
$f'' > 0 \Rightarrow f'$ increasing $\Rightarrow f$ concave up	Second principle	The slope grows, e.g. from $-4$ to $-3$
$f'(x) = 3 - 3x^2 = 3(1 - x)(1 + x)$	Derivative of $3x - x^3$ , factored	Its sign, one factor at a time
$f'(x_0) = 0$ , $\ y_0 = f(x_0)$	Critical point $x_0$ , critical value $y_0$	An input on the $x$ -axis vs. the height of the graph there

Key Insight: For $f(x) = 3x - x^3$ the sign of $f'$ runs $-,\ +,\ -$ across $x = -1$ and $x = 1$ , so $(-1, -2)$ is a valley and $(1, 2)$ a hilltop, read from the first derivative alone.

IIVisualization 1 — The Sign Chart of $f'(x) = 3(1 - x)(1 + x)$

The signs of the two factors decide the sign of $f'$ , and with it whether the tangent line points up or down.

x 0.5

IIIVisualization 2 — Critical Points, Then Critical Values

Solving $f'(x) = 0$ gives the inputs; plugging them into $f$ gives the heights of the two points to plot.

Step 1

Step 1 — Set the derivative to zero

f'(x) = 3(1 - x)(1 + x) = 0

Step 2 — Critical points

1 - x = 0 \ \text{ or } \ 1 + x = 0 \quad\Longrightarrow\quad x = \pm 1

Step 3 — Critical value at $x = 1$

f(1) = 3 \cdot 1 - 1^3 = 3 - 1 = 2

Step 4 — Critical value at $x = -1$

f(-1) = 3(-1) - (-1)^3

= -3 + 1 = -2

Step 5 — Shapes from the sign of $f'$

Down then up across

x = -1

: a valley at

(-1, -2)

. Up then down across

x = 1

: a hilltop at

(1, 2)

. Both have a horizontal tangent.

IVVisualization 3 — Valley, Hilltop, Level-Off or Corner?

The signs of $f'$ on either side of $x_0$ , and whether $f'(x_0)$ exists, fix the local shape.

Case

VQuiz Questions

Problem 1 · Where Is It Decreasing?

Given: $f(x) = x^3 - 12x$ — find the interval where $f$ is decreasing.

✅ Correct!

f'(x) = 3(x - 2)(x + 2)

has factors of opposite sign exactly when

-2 < x < 2

, so

f' < 0

there.

❌ That is where $f$ increases. For

x > 2

both factors

x - 2

and

x + 2

are positive, so

f' > 0

❌ That interval comes from the zeros of $f$ , not of $f'$ .

x^3 - 12x = 0

gives

x = 0

and

x = \pm 2\sqrt{3}

; where the graph crosses the axis says nothing about rising or falling.

❌ Not quite. Differentiate, factor

f'

, and read its sign interval by interval.

Show solution

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

Sign one factor at a time:

$x < -2$ : $x - 2 < 0$ , $x + 2 < 0$ , so $f' > 0$ (increasing).
$-2 < x < 2$ : $x - 2 < 0$ , $x + 2 > 0$ , so $f' < 0$ (decreasing).
$x > 2$ : both factors positive, so $f' > 0$ (increasing).

So $f$ is decreasing on $-2 < x < 2$ .

Problem 2 · Points vs. Values

Given: $g(x) = x^3 - 3x^2$ — find the critical values of $g$ .

✅ Correct! The critical points are

x = 0

and

x = 2

; the critical values are the heights

g(0) = 0

and

g(2) = -4

❌ Those are the critical points. A critical point is an input with

g'(x_0) = 0

; the critical value is

g(x_0)

, the height of the graph there.

❌ Check the arithmetic.

g(2) = 8 - 3 \cdot 4 = 8 - 12

, which is negative.

❌ Check the factoring.

3x^2 - 6x = 3x(x - 2)

, so the second critical point is

x = 2

, not

x = -2

❌ Not quite. Solve

g'(x) = 0

, then evaluate

g

at each solution.

Show solution

g'(x) = 3x^2 - 6x = 3x(x - 2) = 0 \quad\Longrightarrow\quad x = 0 \ \text{ or } \ x = 2

These are the critical points. The critical values are the outputs:

g(0) = 0, \qquad g(2) = 2^3 - 3 \cdot 2^2 = 8 - 12 = -4

So the critical values are $0$ and $-4$ , and the points to plot are $(0, 0)$ and $(2, -4)$ .

Problem 3 · Three Factors, Three Critical Points

Given: $f(x) = x^4 - 2x^2$ — find all critical points, then use the sign of $f'$ to identify the hilltop.

What are the critical points?

Which point is a hilltop?

✅ Correct!

f'

runs

-,\ +,\ -,\ +

, so

(0, 0)

is a hilltop between the valleys

(-1, -1)

and

(1, -1)

❌ A factor went missing.

4x^3 - 4x = 4x(x^2 - 1)

: the factor

x

gives a third critical point,

x = 0

❌ Those come from $f(x) = 0$ . Critical points solve

f'(x) = 0

;

x = \pm\sqrt{2}

are where the graph crosses the axis.

❌ Check the critical points. Factor

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

completely and set each factor to zero.

❌ Check the shape. A hilltop needs

f' > 0

on its left and

f' < 0

on its right.

Show solution

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

Critical points: $x = -1$ , $x = 0$ , $x = 1$ . Sign of $f'$ on each interval:

$x < -1$ : three negative factors, $f' < 0$ (down).
$-1 < x < 0$ : two negative factors, $f' > 0$ (up).
$0 < x < 1$ : one negative factor, $f' < 0$ (down).
$x > 1$ : all positive, $f' > 0$ (up).

Down–up at $x = \pm 1$ gives valleys at $f(\pm 1) = 1 - 2 = -1$ ; up–down at $x = 0$ gives the hilltop $(0, f(0)) = (0, 0)$ .

Problem 4 · A Critical Point That Does Not Turn

Given: $f(x) = x^3 - 3x^2 + 3x$ — find what happens at $x = 1$ .

✅ Correct!

f'(x) = 3(x - 1)^2

is zero at

x = 1

but positive on both sides, so

f

flattens at

(1, 1)

and keeps going up.

❌ $f'(1) = 0$ alone does not make a turn. A valley or hilltop needs

f'

to change sign, and

3(x - 1)^2

is never negative.

❌ Check the definition. A critical point only requires

f'(x_0) = 0

; turning around is not part of it.

❌ Not quite. Factor

f'

and look at its sign on each side of

x = 1

Show solution

f'(x) = 3x^2 - 6x + 3 = 3(x^2 - 2x + 1) = 3(x - 1)^2

$f'(1) = 0$ , so $x = 1$ is a critical point, with critical value $f(1) = 1 - 3 + 3 = 1$ .

For every $x \ne 1$ , $(x - 1)^2 > 0$ , so $f' > 0$ on both sides: $f$ is increasing to the left and to the right of $1$ . The graph levels off to a horizontal tangent at $(1, 1)$ and keeps rising, which is why critical points are not called turning points.

Solved: 0 / 4

Reading a Graph from the Sign of the Derivative: Critical Points of 3x − x³

IKey Formulas

IIVisualization 1 — The Sign Chart of f′(x)=3(1−x)(1+x)f'(x) = 3(1 - x)(1 + x)f′(x)=3(1−x)(1+x)

IIIVisualization 2 — Critical Points, Then Critical Values

IVVisualization 3 — Valley, Hilltop, Level-Off or Corner?

VQuiz Questions

IIVisualization 1 — The Sign Chart of $f'(x) = 3(1 - x)(1 + x)$