Single-Variable-Calculus · Unit 8 · Video 4 · Interactive Practice

Every Piece Must Agree: Finishing the Sketch of 3x − x³

IKey Formulas

Formula	Name	What it tells the sketch
$f(-x) = -f(x)$	Odd symmetry	$(x, y)$ on the graph $\Rightarrow$ $(-x, -y)$ on the graph: the left half is the right half turned half a turn about the origin
$3x - x^3 \approx -x^3$ for large $\|x\|$	The ends	$f(x) \to -\infty$ as $x \to +\infty$ and $f(x) \to +\infty$ as $x \to -\infty$
$f''(x) = -6x$	Concavity	$f'' > 0$ : concave up (here $x < 0$ ); $f'' < 0$ : concave down (here $x > 0$ )
$f''(x_0) = 0$ and $f''$ changes sign	Inflection point	Here at the origin: concave up on the left, concave down on the right

Key Insight: The sign of $f'$ , the critical values, $f(0) = 0$ , the symmetry, the ends and the concavity all describe one curve, so each piece must agree with every other; a clash means a mistake somewhere.

IIVisualization 1 — The Left Half for Free

Because $f(-x) = -f(x)$ , a half turn about the origin carries the right half of the graph onto the left half.

Turn 0° Tracked point

IIIVisualization 2 — Off the Screen: The Ends

In a wide enough window, the $3x$ term of $3x - x^3$ is negligible and the graph looks like $y = -x^3$ .

Window half-width W 1.5

💡 Each window is scaled so that $y = -x^3$ always fills it the same way; as $W$ grows, $3x$ shrinks to $3/W^2$ of $x^3$ at the window's edge.

IVVisualization 3 — Concavity and the Inflection Point

The sign of $f''(x) = -6x$ decides whether the tangent line lies below or above the curve.

Point x −1.5

VQuiz Questions

Problem 1 · The Ends of a Quartic

Given: $h(x) = 2x + 5x^2 - x^4$ — find what $h(x)$ does at each end, as $x \to -\infty$ and as $x \to +\infty$ .

✅ Correct! For large

|x|

the term

-x^4

dominates, and

x^4

is positive for negative

x

too, so

h \to -\infty

at both ends.

❌ Not quite. The ends of

3x - x^3

point opposite ways because

x^3

is odd. Here the dominant term is

-x^4

, and an even power is positive for negative

x

, so

-x^4

is negative at both ends.

❌ Not quite. Opposite ends come from an odd highest power. Here the highest power is

x^4

, which is even, so both ends go the same way, and the minus sign in

-x^4

sends both down.

❌ Close, but… the two ends do match, but they point down. The dominant term is

-x^4

, not the positive

5x^2

: at

x = \pm 10

5x^2 = 500

while

-x^4 = -10000

❌ Not quite. A polynomial follows its highest-power term at both ends, here

-x^4

Show solution

For large $|x|$ the highest power dominates: $h(x) \approx -x^4$ . Since $x^4 > 0$ for every $x \ne 0$ , $-x^4 \to -\infty$ as $x \to -\infty$ .

Check with numbers:

h(-10) = -20 + 500 - 10000 = -9520, \qquad h(10) = 20 + 500 - 10000 = -9480

So $h(x) \to -\infty$ as $x \to -\infty$ and as $x \to +\infty$ : unlike $3x - x^3$ , both ends of this graph point down.

Problem 2 · Is It Odd?

Given: $g(x) = x^3 - 4x + 2$ , with $g(1) = -1$ — find $g(-1)$ .

✅ Correct! The constant

2

is an even power (

2 = 2x^0

), so

g

is not odd and

g(-1) \ne -g(1)

❌ Not quite.

g(-1) = -g(1)

needs every power of

x

to be odd, and the constant term

2

breaks that. Substitute directly instead.

❌ Not quite.

g

is not even either:

x^3

and

-4x

change sign when

x

does. Substitute

x = -1

directly.

❌ Check your signs.

(-1)^3 = -1

and

-4(-1) = +4

, so

g(-1) = -1 + 4 + 2

❌ Not quite. Substitute

x = -1

into each term.

Show solution

g(-1) = (-1)^3 - 4(-1) + 2 = -1 + 4 + 2 = 5

In general $g(-x) = -x^3 + 4x + 2$ , while $-g(x) = -x^3 + 4x - 2$ . They differ by $4$ , so $g$ is not odd: its graph is the odd curve $x^3 - 4x$ shifted up by $2$ , symmetric about the point $(0, 2)$ rather than the origin.

Problem 3 · Locate the Inflection Point

Given: $k(x) = x^3 - 6x^2 + 9x + 1$ — find its inflection point and its concavity to the left of it.

Where is the inflection point?

For $x$ to the left of the inflection point, the graph is…

✅ Correct!

k''(x) = 6x - 12

is zero at

x = 2

, negative to its left and positive to its right;

k(2) = 3

❌ Not quite. The inflection point of

3x - x^3

happened to be the origin; here

k''(0) = -12 \ne 0

. Solve

k''(x) = 0

❌ Close, but… that is a critical point, where

k'(x) = 0

. An inflection point comes from

k''(x) = 0

❌ Not quite. Differentiate twice and solve

k''(x) = 0

❌ Check the sign of $k''$ . Test one point left of the inflection point, for instance

k''(0) = -12

Show solution

k'(x) = 3x^2 - 12x + 9 = 3(x - 1)(x - 3), \qquad k''(x) = 6x - 12 = 6(x - 2)

$k''(x) = 0$ at $x = 2$ , and $k(2) = 8 - 24 + 18 + 1 = 3$ . For $x < 2$ , $k''(x) < 0$ (e.g. $k''(0) = -12$ ): concave down. For $x > 2$ , $k''(x) > 0$ : concave up. The sign changes, so $(2, 3)$ is an inflection point.

Cross-check: the critical points are the hilltop $(1, 5)$ , where $k''(1) = -6 < 0$ , and the valley $(3, 1)$ , where $k''(3) = 6 > 0$ . The inflection point sits between them, exactly where the bending switches.

Problem 4 · Sketching an Even Quartic

Given: $q(x) = x^4 - 6x^2$ , an even function: $q(-x) = q(x)$ — find its inflection points and its left end.

Where are the inflection points?

As $x \to -\infty$ , $q(x) \to$ …

✅ Correct!

q''(x) = 12(x - 1)(x + 1)

changes sign at

x = \pm 1

, and an even function's ends match: both go to

+\infty

❌ Close, but…

(\pm\sqrt{3}, -9)

are the two valleys, where

q'(x) = 4x(x^2 - 3) = 0

. Inflection points come from

q''(x) = 0

❌ Not quite.

x = 0

is the hilltop, where

q''(0) = -12 < 0

, so the graph is concave down there, not switching.

❌ Not quite.

\pm\sqrt{6}

are where the graph crosses the axis (

q(x) = x^2(x^2 - 6) = 0

), not where the bending changes.

❌ Not quite. Solve

q''(x) = 12x^2 - 12 = 0

❌ Not quite. Opposite ends belong to odd functions like

3x - x^3

. Here

q(-x) = q(x)

, so the left end mirrors the right end, and

x^4

dominates.

❌ Not quite. For large

|x|

q(x) \approx x^4

, which is positive and unbounded.

Show solution

q'(x) = 4x^3 - 12x = 4x(x^2 - 3), \qquad q''(x) = 12x^2 - 12 = 12(x - 1)(x + 1)

$q''$ is positive for $|x| > 1$ and negative for $|x| < 1$ , so the graph is concave up, then concave down, then concave up, with inflection points at $x = \pm 1$ , where $q(\pm 1) = 1 - 6 = -5$ .

For large $|x|$ , $q(x) \approx x^4 \to +\infty$ at both ends, as the even symmetry requires (the graph is its own mirror image in the $y$ -axis).

Everything agrees: the valleys $(\pm\sqrt{3}, -9)$ lie in the concave-up regions ( $q''(\sqrt{3}) = 24 > 0$ ), the hilltop $(0, 0)$ lies in the concave-down region ( $q''(0) = -12 < 0$ ), and the axis crossings at $\pm\sqrt{6} \approx \pm 2.45$ sit beyond the valleys on the way up to $+\infty$ .

Solved: 0 / 4