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Inequality Visual Explorer


Curve, sign chart, and explanation all sync to the marble's position. Click any factor, column, or row to navigate.
nudge · shift+drag snap · [ ] step · space play · r reset
(x + 2)(x − 1)(x − 5) < 0
f(x)-1e+201e+2x-4-3-2-101234567zero — drag to change-2zero — drag to change1zero — drag to change5
excludedincluded / zeropoledomain edgesolution/trail
f(-4) = -90satisfies
speed0.6×
interval(−∞, -2) ∪ (1, 5)
set-builder{ x ∈ ℝ : f(x) < 0 }
Templates & parameters
f(x) = (x − r₁)(x − r₂)(x − r₃)
root r₁
−2
-8-208
root r₂
+1
-8018
root r₃
+5
-8058
Sign chart
x < -2-2-2 – 111 – 55x > 5
(x + 2)+++
(x − 1)++
(x − 5)+
f(x)++
Explanation
at x = -4
(x + 2)= ((-4) + 2)= -2
(x − 1)= ((-4) − 1)= -5
(x − 5)= ((-4) − 5)= -9
product of signs: · ·  = 
f(-4) = -90 < 0 satisfies f(x) < 0
Each factor's sign at x = -4: (x + 2) is negative; (x − 1) is negative; (x − 5) is negative. Combining them gives f(-4) = -90, which is indeed negative. Since the inequality asks for f(x) < 0, this x is in the solution set.
Three simple roots alternate the sign through four intervals, and the solution is every second one: (,2)(1,5)(-\infty, -2) \cup (1, 5). Learn more about three distinct roots · All polynomial states
Strict less-than: the negative intervals with every boundary excluded — zero is not less than zero. Learn more about this comparison · Direction and strictness







Key Terms


  • Inequality f(x)>0f(x) > 0 (or <,,<, \geq, \leq) — a statement asking which xx make f(x)f(x) positive, negative, non-negative, or non-positive
  • Solution set — the set of all xx satisfying the inequality, typically a union of intervals on the real line
  • Direction — which comparison the inequality uses: >>, <<, \geq, or \leq
  • Strictness — whether the inequality is strict (>>, <<) or non-strict (\geq, \leq); affects whether boundary points belong to the solution
  • Sign chart — a table tracking the sign of f(x)f(x) across intervals separated by its zeros and undefined points
  • Critical point — a zero of f(x)f(x) or a point where ff is undefined; the only places the sign of f(x)f(x) can change
  • Factor — one of the building blocks of f(x)f(x) when written as a product; each gets its own row in the sign chart
  • Interval — a maximal piece of the real line on which f(x)f(x) has constant sign
  • Marble — the draggable probe positioned at some xx in the visualizer; lets you read off the sign of f(x)f(x) at that point
  • Pole — a point where ff is undefined (typically a denominator zero); excluded from the solution set regardless of strictness


Getting Started with the Visualizer

Open the explorer and you see a curve y=f(x)y = f(x) with a draggable marble sitting on it. The inequality currently being solved is displayed symbolically above the graph (e.g. f(x)>0f(x) > 0), with each factor of ff rendered as a clickable element.

The layout has two columns. On the left, the Hero panel shows the inequality and the curve with the marble; the Controls panel below it holds the type tabs, parameter sliders, templates, direction and strictness toggles, and interaction modes. On the right, the Sign chart panel displays signs of every factor across every interval, and the Explanation panel narrates the current step or live reading.

The whole interface is wired together: hovering or clicking a factor in the inequality highlights its row in the sign chart; clicking a column in the sign chart moves the marble; clicking a row in the explanation panel does the same. Everything stays in sync.

Selecting an Inequality Type

The type bar at the top of the page is a row of tabs, one per inequality family the visualizer supports. Each tab carries a tooltip describing its structure. Click a tab to switch families.

Switching the type does three things at once:

• The graph updates to show the new function f(x)f(x)
• The parameter sliders below reconfigure to match the new family's parameters
• The sign chart rebuilds with a new set of factors and intervals

The currently active type is highlighted in blue. The five families are polynomial, quadratic, absolute value, rational, and radical — each leading to a different sign-chart structure. Higher-degree types produce more factors and more critical points, but the solution-set logic is identical across all of them.

Direction and Strictness

Two controls determine *which* inequality you are solving for the current f(x)f(x):

Direction — choose between >>, <<, \geq, or \leq. Selecting >> asks for xx where f(x)f(x) is positive; selecting \leq asks for xx where f(x)f(x) is non-positive
Strictness — toggles between strict (>>, <<) and non-strict (\geq, \leq). The strict and non-strict versions of an inequality differ only at the boundary points (zeros of ff): strict excludes them, non-strict includes them

Strictness has a visible effect on the solution set: boundary points render as open circles for strict comparisons and filled circles for non-strict ones. Poles — points where ff is undefined — are always excluded, regardless of strictness, because f(x)f(x) has no value there to compare against zero.
Each of the four comparisons gets a dedicated frozen frame below, all on the same polynomial: strictly less than zero, at most zero, strictly greater than zero, and at least zero.

Strictly Less Than Zero

The default comparison, frozen on the default polynomial: (x+2)(x1)(x5)<0(x + 2)(x - 1)(x - 5) < 0. Two open-ended bars — (,2)(-\infty, -2) and (1,5)(1, 5) — with every boundary dot drawn open.
f(x)-1000100x-4-3-2-101234567-215(x + 2)(x − 1)(x − 5) < 0
Operator <, frozen

The reference frame: negative intervals shaded, every boundary dot open — zero is not less than zero.

Strict less-than is the reference state the other three operators perturb. Its signature is the open circles: the roots themselves give f(x)=0f(x) = 0, and zero is not less than zero, so all three boundary points are excluded.

The four operator states share one curve and one set of critical points; only membership at and around those points changes. Clicking through them in the live tool with the sign chart in view is the fastest way to see that direction picks *which intervals* and strictness picks *the boundary dots* — two independent switches. Compare at most zero for the first switch flipped.

At Most Zero

The same polynomial under \leq: identical bars, but the three boundary dots at 2-2, 11, 55 now render filled — the roots joined the solution set.
f(x)-1000100x-4-3-2-101234567-215(x + 2)(x − 1)(x − 5) ≤ 0
Operator ≤, frozen

Same bars, filled dots: non-strict comparison admits the roots themselves, and nothing else changes.

Non-strict comparison admits equality, and equality happens exactly at the zeros of ff. The solution grows by precisely three points: (,2][1,5](-\infty, -2] \cup [1, 5]. Nothing else moves — the interiors of the intervals were already decided by the sign chart.

The open-versus-filled dot convention is the entire visual difference between this frame and the strict version, which is the point: strictness is a boundary-only phenomenon. (For rational types the pole dot would stay open even here — poles never join; see the rational family.)

Strictly Greater Than Zero

The same polynomial under >>: the shading jumps to the complementary intervals (2,1)(-2, 1) and (5,)(5, \infty), boundary dots open again.
f(x)-1000100x-4-3-2-101234567-215(x + 2)(x − 1)(x − 5) > 0
Operator >, frozen

Direction flipped: the complementary intervals light up, boundaries open again.

Flipping direction selects the intervals where the sign chart reads ++ instead of - — the exact complement of the less-than state, minus the boundary points, which belong to neither strict solution. Between the two strict frames, every point of the line is claimed exactly once, except the three roots, claimed by neither.

That near-partition is a useful sanity check when solving by hand: if your << answer and your >> answer overlap, or jointly miss an interval, a sign was charted wrong.

At Least Zero

The final operator: \geq shades [2,1][-2, 1] and [5,)[5, \infty) with filled boundary dots — the positive intervals plus their root endpoints.
f(x)-1000100x-4-3-2-101234567-215(x + 2)(x − 1)(x − 5) ≥ 0
Operator ≥, frozen

The fourth corner of the two-by-two: positive intervals with their endpoints included.

The fourth frame completes a tidy two-by-two: direction chooses the interval family, strictness chooses the dots. Together the \leq and \geq frames cover the whole line with the three roots shared — every real number satisfies at least one of the two non-strict comparisons, since every f(x)f(x) is 0\leq 0 or 0\geq 0.

Frozen side by side, the four frames are a truth table for the operator pair — and a compact answer to the perennial question of when to use round versus square brackets in interval notation: round follows open dots, square follows filled ones, always.

Three Interaction Modes and Keyboard Shortcuts

The Controls panel offers three modes for moving the marble, each grouped behind its own button. Only one is active at a time.

Drag — grab the marble with the mouse and slide it along the xx-axis. Hold shift to snap to integer values
Step — Previous and Next buttons jump the marble between named stops: critical points, midpoints of intervals, and other landmarks
Auto — the marble plays back the sequence of stops automatically, with a speed slider for playback rate

Keyboard shortcuts work whenever the page has focus and you are not in an input:

Arrow Left / Arrow Right — nudge the marble by 0.10.1; Shift nudges by 11
[ and ] — step the marble to the previous or next named stop
Space — toggle play/pause in auto mode, or switch to auto mode
R — reset all parameters, marble position, and mode to defaults

Adjusting Parameters and Using Templates

Each inequality type has its own parameter sliders, laid out in a three-column grid below the graph. Drag a slider, click a tick to snap to a notable value, or type directly into the numeric input. Sliders that hit invalid combinations (such as a denominator forced to zero) display a red error chip with the reason.

Each parameter has a value chip showing whether it is positive (blue), negative (amber), or zero (dashed). The mode toggle next to the chip switches between slider and numeric input for finer control.

Above the sliders, the Templates strip offers a few preset parameter combinations for the current type — useful starting points for common shapes like "no solution", "solution is a single interval", "solution is two disjoint intervals", and similar. Click a template to load it.

Reading the Sign Chart

The Sign chart panel on the right is a compact table tracking the sign of f(x)f(x) across the real line. Reading it top to bottom:

Header row — the critical xx-values (zeros of ff and any poles), in increasing order, dividing the real line into intervals
Factor rows — for each factor of ff, a row of ++, -, or 00 entries showing the sign of that factor in each interval
Product row — highlighted, gives the sign of f(x)f(x) itself in each interval, computed by multiplying the factor signs
Pole columns — points where ff is undefined, marked in red

The chart is interactive. Hover or click any factor in the inequality above to highlight its row. Click an interval cell to send the marble there. Click a critical-point column to land the marble exactly on the boundary. The same highlighting flows from the explanation panel and the curve, so every part of the interface points at the same intervals.

Reading the Explanation Panel

The Explanation panel below the sign chart has two tabs.

Steps — a numbered list reconstructing the standard solving procedure: identify the factors of ff, locate their zeros and any poles, build the sign chart, pick the intervals matching the chosen direction, and assemble the solution set with the right boundary inclusion. Each step is tied directly to what is on screen.
Live — a compact table that recomputes whenever the marble moves. It shows the marble's xx, the sign of each factor at that xx, the combined sign of f(x)f(x), and ends with a verdict: does the inequality hold at this xx?

If the marble sits at a pole, the Live tab flags f(x)f(x) as undefined and notes that this xx is excluded from the solution set. A short verbal summary below the table phrases the conclusion in plain language.

Polynomial Inequalities in the Explorer

The Polynomial tab solves (xr1)(xr2)(xr3)<0(x - r_1)(x - r_2)(x - r_3) < 0 — a cubic handed over in factored form, so every root is visible in the formula before the graph draws it. The three templates cover the shapes that matter: three distinct roots, a double root, and a tight cluster of adjacent roots.

Factored form is the pedagogical gift of this tab: the sign chart's factor rows correspond one-to-one with the parentheses, and the product's sign is literally the product of the rows. Dragging any root dot on the axis rewrites the factorization live.

Three Distinct Roots

The default state of the whole tool: (x+2)(x1)(x5)<0(x + 2)(x - 1)(x - 5) < 0, with roots at 2-2, 11, and 55 splitting the line into four intervals. The blue bars shade where the product is negative: left of 2-2, and between 11 and 55.
f(x)-1000100x-4-3-2-101234567-215(x + 2)(x − 1)(x − 5) < 0
(x+2)(x−1)(x−5) < 0, frozen

Four intervals, alternating signs, two shaded: the default state of the tool, with the marble certifying the leftmost interval from x = −4.

Three simple roots means the sign alternates through all four intervals — negative, positive, negative, positive — because each crossing flips exactly one factor. The solution (,2)(1,5)(-\infty, -2) \cup (1, 5) reads straight off the alternation: every second interval.

The frozen marble sits at the left test point x=4x = -4, where all three factors are negative and the product is negative — one sample point certifying the whole leftmost interval. That "test one point per interval" logic is the entire sign-chart method in miniature. See the double root for what happens when the alternation breaks.

The Double Root

(x+3)(x2)2<0(x + 3)(x - 2)^2 < 0: the squared factor turns the root at 22 into a touch-point. Only the interval left of 3-3 is shaded — the curve bounces off zero at x=2x = 2 without going negative.
f(x)-1000100x-5-4-3-2-101234-32(x + 3)(x − 2)² < 0
(x+3)(x−2)² < 0, frozen

The squared factor cancels one sign change: the curve touches zero at 2 and bounces. Only the far-left interval survives.

A squared factor never changes sign, so the crossing at 22 is cancelled: the curve comes down, touches, and returns. The sign pattern is negative, positive, positive — the alternation of three distinct roots with one flip removed. The solution collapses to (,3)(-\infty, -3).

The subtlety worth dwelling on: x=2x = 2 *is* a zero of ff, so for the non-strict version \leq it belongs to the solution set as an isolated point — a single closed dot disconnected from the interval. Even multiplicity is precisely the case where strict and non-strict answers differ by an isolated point rather than an endpoint.

The Tight Cluster

(x+1)(x)(x1)<0(x + 1)(x)(x - 1) < 0: three roots packed at 1-1, 00, 11. The shading alternates rapidly — left of 1-1, then between 00 and 11.
f(x)-20020x-3-2-10123-101(x + 1)(x)(x − 1) < 0
(x+1)(x)(x−1) < 0, frozen

Three roots in three units: rapid alternation, narrow intervals — x³ − x asked a different question.

Root spacing changes nothing logically and everything visually. The intervals (1,0)(-1, 0) and (0,1)(0, 1) are only one unit wide, and the curve oscillates through them in quick succession — a picture of why densely packed roots demand care with test points: there is little room to sample in.

The cluster is also x3xx^3 - x in factored clothing, the same S-curve the equation explorer freezes as its three-root cubic. Same polynomial, different question: there the interest was *where* it crosses zero; here it is *which side* of zero it spends each interval on.

Quadratic Inequalities in the Explorer

The Quadratic tab works in standard form ax2+bx+cax^2 + bx + c, where the roots must be *earned* — the tool computes them from the discriminant before it can chart signs. The templates hit the three signature configurations: two roots, no real roots, and an opens-down parabola.

The family adds one twist the factored cubic cannot show: an irreducible quadratic, whose sign never changes. When the discriminant goes negative, the whole parabola sits on one side of zero and the inequality's answer is everything or nothing.

A Quadratic with Two Roots

x2x6<0x^2 - x - 6 < 0: the parabola dips below zero between its roots 2-2 and 33, and the single blue bar spans exactly that dip.
f(x)-10010x-4-3-2-1012345-23(x + 2)(x − 3) < 0
x² − x − 6 < 0, frozen

An upward parabola is negative between its roots: one connected bar from −2 to 3.

The tool factors the standard form for you — the sign chart shows (x+2)(x3)(x + 2)(x - 3) — and the answer is the single interval (2,3)(-2, 3): an upward parabola is negative *between* its roots, always. That "between" is worth internalizing as a reflex; its mirror ("outside") appears when the parabola opens down or the direction flips to >>.

Note the contrast with the factored-cubic tab: two roots make three intervals and one sign change fewer, so the solution is one connected piece instead of two. Root count controls solution topology.

A Quadratic with No Real Roots

x2+4<0x^2 + 4 < 0: no roots, no critical points, no shading. The parabola floats entirely above zero, and the solution set is empty.
f(x)-20-1001020x-5-4-3-2-1012345(x² + 4) < 0
x² + 4 < 0, frozen

No roots, one sign, empty answer: the irreducible quadratic as an all-or-nothing gate.

With the discriminant negative, the factor row shows a single irreducible chunk (x2+4)(x^2 + 4) whose sign never changes — the sign chart has one column and it reads ++. The strict inequality <0< 0 therefore has no solutions at all, and the tool's interval notation reports \emptyset.

The flip side is immediate: the same curve makes x2+4>0x^2 + 4 > 0 true *everywhere*, solution (,)(-\infty, \infty). Irreducible quadratics are all-or-nothing gates — a fact used constantly when factoring higher polynomials, where such factors can be crossed off the sign chart entirely.

A Downward Parabola

x2+2x+3<0-x^2 + 2x + 3 < 0: the parabola opens downward with roots at 1-1 and 33, and the shading covers the two outer rays — the parabola is below zero *outside* its roots.
f(x)-10010x-3-2-1012345-13(x + 1)(x − 3) < 0
−x² + 2x + 3 < 0, frozen

Opens down, so “below zero” means outside the roots: the two outer rays are shaded, not the middle.

Flipping the leading coefficient's sign swaps "between" and "outside": a downward parabola is positive between its roots and negative beyond them, so the solution is (,1)(3,)(-\infty, -1) \cup (3, \infty) — the exact complement (up to endpoints) of what the upward two-root state produces.

This state is the standard trap in textbook problems: students memorize "less than zero means between the roots" from upward parabolas and apply it blindly. The frozen frame is the antidote — the rule is not about the inequality sign, it is about which way the parabola opens.

Absolute-Value Inequalities in the Explorer

The Absolute value tab solves xhk<0|x - h| - k < 0 — distance inequalities in disguise. The templates: the centered V, a shifted V, and the V at zero level, where the vertex itself touches the axis.

The distance reading turns each solution into a sentence: xh<k|x - h| < k means "within kk of hh" — an interval centered at hh — while the >> version means "farther than kk from hh", the two outer rays. The V's two arms are the two linear cases of the definition, drawn simultaneously.

The Centered V

x3<0|x| - 3 < 0: the V-shape with vertex dipping to 3-3, crossing zero at ±3\pm 3. The blue bar spans the valley between them.
f(x)-202x-5-4-3-2-1012345-33|x| − 3 < 0
|x| − 3 < 0, frozen

Distance under 3 from the origin: the valley of the V, shaded from −3 to 3.

Read as distance, x<3|x| < 3 says "within 3 of the origin" — the interval (3,3)(-3, 3), an answer you can state before any chart is drawn. The two crossing points are the two linear arms each solving their own equation, x=3x = 3 and x=3-x = 3.

The V's signature on the sign chart is a single expression row (the tool treats x3|x| - 3 as one factor) with two critical points — unlike polynomial factors, one row can own several sign changes. That is the chart's way of saying the expression is not a polynomial.

The Shifted V

x24<0|x - 2| - 4 < 0: the vertex moves to x=2x = 2, the crossings to 2-2 and 66, and the shaded valley follows — an interval of radius 44 centered at 22.
f(x)-4-2024x-4-3-2-1012345678-26|x − 2| − 4 < 0
|x − 2| − 4 < 0, frozen

Center 2, radius 4, straight from the formula: the shaded interval runs −2 to 6.

The distance sentence updates without friction: "within 4 of 2" is (2,6)(-2, 6). Center and radius are exactly the tool's two sliders hh and kk — the template's whole point is that the interval's midpoint and half-width can be read directly from the formula, no solving required.

Dragging either endpoint dot on the axis adjusts kk live — the explorer inverts the geometry back into the parameter, a small demonstration that the endpoint positions and the formula are the same information.

The V at Zero Level

x+1<0|x + 1| < 0: with k=0k = 0 the vertex touches the axis at 1-1, and the strict inequality asks for negative distance. Nothing is shaded — the solution set is empty.
f(x)-2-1012x-3-2-101-1|x + 1| < 0
|x + 1| < 0, frozen

A distance asked to be negative: the vertex touches zero at −1 and nothing is shaded anywhere.

An absolute value is never negative, so x+1<0|x + 1| < 0 has no solutions regardless of any chart — the frozen frame shows the honest picture: a curve that touches zero once and is positive everywhere else, with an empty axis bar.

The state rewards operator experiments: switch to \leq and exactly one point qualifies, the vertex x=1x = -1, appearing as a single closed dot; switch to >> and everything *except* the vertex qualifies. Three radically different solution sets from one frozen curve — the strongest argument on the page that the operator is a genuine state dimension, treated in Direction and Strictness.

Rational Inequalities in the Explorer

The Rational tab solves (xa)/(xb)<0(x - a)/(x - b) < 0 and introduces the sign chart's most dangerous feature: the pole. The zero at aa behaves like any root, but the pole at bb is excluded from the domain no matter what — a red, always-open point. The templates: a simple configuration, zero and pole crossed, and zero and pole adjacent.

The family exists to break a bad habit: multiplying both sides by (xb)(x - b) flips the inequality on half the line and silently erases the domain hole. The sign-chart method needs no multiplication — which is exactly why it is the method of record here.

A Simple Rational Inequality

(x1)/(x+2)<0(x - 1)/(x + 2) < 0: a zero at 11 (blue dot) and a pole at 2-2 (red dot, red stripe). The fraction is negative exactly between them — the bar spans (2,1)(-2, 1).
f(x)-505x-4-3-2-101231-2(x − 1) / (x + 2) < 0
(x−1)/(x+2) < 0, frozen

Blue zero, red pole, and the negative stretch between them. The red stripe marks the domain hole at −2.

A fraction is negative when numerator and denominator disagree in sign, and that happens precisely between the zero and the pole. The sign chart shows the two factor rows disagreeing on exactly that interval — the product row's negative stretch.

The pole's red styling carries the family's core rule: x=2x = -2 is not in the domain, so it can never enter a solution set. Even switching to \leq closes only the zero endpoint at 11; the pole end stays open forever. Domain first, comparison second.

Zero and Pole Crossed

(x+3)/(x4)<0(x + 3)/(x - 4) < 0: this time the zero (3-3) sits left of the pole (44), and the negative stretch — the shaded bar — runs between them across seven units.
f(x)-10-50510x-5-4-3-2-10123456-34(x + 3) / (x − 4) < 0
(x+3)/(x−4) < 0, frozen

Zero left, pole right, seven units of solution between — and the curve diving toward the asymptote at 4.

Swapping which critical point comes first does not change the logic — the fraction is still negative exactly between its two sign-relevant points — but it changes which *kind* of endpoint each end of the solution has: here the left end is a zero (closable under \leq) and the right end is a pole (never closable), the mirror of the simple configuration.

The wide gap also makes the graph's asymptotic behavior legible: the curve dives toward -\infty approaching the pole from the left, having crossed zero calmly at 3-3 far earlier. Zeros are gentle events; poles are violent ones. The chart treats them with one symbol each — 00 versus undefined — and the geometry explains the difference.

Zero and Pole Adjacent

(x2)/(x3)<0(x - 2)/(x - 3) < 0: zero at 22, pole at 33, one unit apart. The shaded solution is the narrow interval (2,3)(2, 3) squeezed between them.
f(x)-4-2024x01234523(x − 2) / (x − 3) < 0
(x−2)/(x−3) < 0, frozen

One unit between zero and pole: the entire solution squeezed into (2, 3).

Proximity stress-tests reading skills: the dashed drop-lines, the differently colored dots, and the thin blue bar all crowd into one unit of axis. The curve behavior is dramatic — it leaves its zero at 22 and almost immediately plunges toward the pole at 33.

The adjacent case is also where the multiply-both-sides error hurts most: multiplying by (x3)(x - 3), negative throughout the solution interval, silently reverses the inequality exactly where the answer lives. The sign chart sidesteps the whole hazard — which is the family's closing argument for it.

Radical Inequalities in the Explorer

The Radical tab solves xak<0\sqrt{x - a} - k < 0 and brings the last complication: a restricted domain. Nothing exists left of x=ax = a; the region is shaded out in red before any sign question can even be asked. The templates: the basic radical, a shifted start, and a high level pushing the zero far to the right.

The standing lesson: a solution set lives *inside* the domain. Every radical answer is the intersection of an interval with [a,)[a, \infty), and forgetting that intersection is the classic radical-inequality error the red shading makes impossible to commit.

The Basic Radical

x2<0\sqrt{x} - 2 < 0: nothing exists left of 00 (red shading), the curve rises from the domain edge, and crosses zero at 44. The bar shades the run from the edge to the crossing.
f(x)-101x-2-1012345604√x − 2 < 0
√x − 2 < 0, frozen

Red nothing left of 0, then a slow climb to the zero at 4. The solution lives between domain edge and crossing.

Two boundary points with two different characters: x=0x = 0 is a domain edge — the amber-flavored "the function starts here" marker — while x=4x = 4 is an honest zero where x\sqrt{x} reaches 22. The solution occupies the stretch between them, where the square root is still small.

Squaring both sides (x<4x < 4) happens to work here, but only because the domain was carried along: the correct answer intersects x<4x < 4 with x0x \geq 0. The red shading does that intersection visually before the algebra can forget it.

The Shifted Radical

x+31<0\sqrt{x + 3} - 1 < 0: the domain now starts at 3-3, and the zero sits just one unit to the right, at 2-2. A short bar spans the gap.
f(x)-1-0.500.51x-5-4-3-2-10-3-2√(x + 3) − 1 < 0
√(x+3) − 1 < 0, frozen

Domain from −3, zero at −2: level 1 keeps the crossing just k² = 1 unit from the edge.

Shifting the radicand moves the entire configuration left: domain edge at 3-3, zero at 3+k2=2-3 + k^2 = -2. The solution (3,2)(-3, -2) is only one unit wide because k=1k = 1 — the zero sits at distance k2k^2 from the edge, so small levels keep the crossing close to the start.

That k2k^2 spacing is the family's quiet parabola-in-reverse: the square root grows slowly, so reaching height kk takes k2k^2 of horizontal distance. The next state stretches this to its extreme.

The High-Level Radical

x14<0\sqrt{x - 1} - 4 < 0: the level is high, so the zero lands far out at x=1+16=17x = 1 + 16 = 17. The shaded run from the domain edge is sixteen units long.
f(x)-202x-1012345678910111213141516171819117√(x − 1) − 4 < 0
√(x−1) − 4 < 0, frozen

Level 4 pushes the zero to 1 + 16 = 17: sixteen units of runway for a square root to climb four.

The frozen frame is the square root's slow growth made spatial: to climb to height 44, the curve needs 42=164^2 = 16 units of runway. The view window stretches to accommodate, and the curve looks nearly flat — an honest picture of how x\sqrt{x} compares with any linear ruler.

Sixteen units of solution from one small parameter change (compare the basic radical) is the family's parting lesson: for radicals, the interesting parameter sensitivity is quadratic, and eyeballing where the zero "should" be will mislead exactly when the level is large.

What an Inequality Means Geometrically

The inequality f(x)>0f(x) > 0 asks for every xx at which the graph y=f(x)y = f(x) sits *above* the xx-axis. The inequality f(x)<0f(x) < 0 asks for every xx where the graph sits *below*. The non-strict versions \geq and \leq include the boundary points where the graph touches the xx-axis.

Unlike equations — whose solutions are typically isolated points where the curve crosses a level — inequalities have solution sets that are *regions* of the real line, almost always unions of intervals. A linear inequality has one half-line as its solution; a quadratic produces either a bounded interval, two unbounded intervals, an empty set, or the whole real line; rational inequalities can have arbitrarily many disjoint pieces.

The solution set changes whenever the sign of f(x)f(x) changes, which happens only at zeros or poles. That is why the sign chart, which catalogs those exact points, is the natural tool for solving any inequality.

For comprehensive theory on inequalities, see inequalities theory.

How the Sign Chart Builds the Solution Set

Given the sign chart, building the solution set is mechanical:

1. Pick the rows matching the direction. For f(x)>0f(x) > 0 or f(x)0f(x) \geq 0, look at intervals where the product row is ++. For f(x)<0f(x) < 0 or f(x)0f(x) \leq 0, look where it is -.
2. Include or exclude boundary zeros based on strictness. Strict comparisons exclude zeros (open intervals); non-strict comparisons include them (closed intervals).
3. Always exclude poles. Even with non-strict comparisons, points where ff is undefined cannot be in the solution.
4. Take the union of all qualifying intervals. The result is the solution set, written in interval notation.

Every inequality of this kind reduces to this procedure once the sign chart is built. The visualizer carries out each step on screen — colored intervals on the curve, highlighted columns in the chart, and a final interval-notation summary in the explanation panel.

For the companion equation case, see equations visualizer.