Visual walkthrough — Curve sketching — systematic approach

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Step 1 — Where is the machine even allowed to run? (Domain)

WHAT. Our machine is $f(x)=\frac{x^2}{x^2-1}$. The bar means "divide". Division by zero is the one thing arithmetic forbids — it has no answer. So first we hunt for any $x$ that makes the bottom (the denominator) equal zero.

$$x^2-1=0⟹\underset{\text{a square}}{\underbrace{x^2}}=\underset{\text{the number }1}{\underbrace{1}}⟹x=\pm 1.$$

• $x^2$ means $x$ times itself.
• The two dangerous inputs are $x=+1$ and $x=-1$, because $(\pm 1{)}^2=1$ and $1-1=0$.

WHY this step first. If we forgot the domain we might cheerfully draw the curve through $x=1$ — but there is literally no point there. The whole picture would be a lie. (See the parent mistake box: domain always comes first.)

PICTURE. Two dashed vertical "forbidden lines" split the number line into three rooms the curve must live inside.

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Step 2 — Where does the curve touch the axes? (Intercepts)

WHAT. Two easy anchor dots.

• $y$-intercept: feed in $x=0$. $f(0)=\frac{0^2}{0^2-1}=\frac{0}{-1}=0.$ So the curve passes through $(0,0)$.
• $x$-intercepts: where is the output zero? A fraction is zero only when its top is zero: $x^2=0\Rightarrow x=0$. Same point again.

$$f(x)=0⟺\underset{\text{top}}{\underbrace{x^2}}=0⟺x=0.$$

WHY. Intercepts are the cheapest true dots on the whole graph — pure arithmetic, no calculus. They pin down the middle of the picture.

PICTURE. A single pinned dot at the origin, sitting between the two forbidden walls.

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Step 3 — Is the picture a mirror image? (Symmetry)

WHAT. We test what happens if we flip $x$ to $-x$:

$$f(-x)=\frac{(-x{)}^2}{(-x{)}^2-1}=\frac{x^2}{x^2-1}=f(x).$$

• $(-x{)}^2=x^2$ because a negative times a negative is positive. So nothing changed.

Because $f(-x)=f(x)$, the machine is even: whatever it does at $x$, it does the identical thing at $-x$.

WHY. This literally halves our work. Once we know the shape for $x\ge 0$, we flip it across the $y$-axis to get $x\le 0$ for free.

PICTURE. A butterfly: the right half and its mirror-reflected left half.

Recall Even vs odd — one-line reminder

Even test ::: $f(-x)=f(x)$, mirror across the $y$-axis. Odd test ::: $f(-x)=-f(x)$, spin $180^{\circ }$ about the origin.

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Step 4 — The invisible walls and floor (Asymptotes)

WHAT — vertical walls. Near $x=1$ the bottom $x^2-1$ shrinks toward $0$ while the top $x^2$ stays near $1$. A fixed number divided by a tinier and tinier number explodes in size. So the output races off to $\pm \infty$: the line $x=1$ (and its mirror $x=-1$) is a vertical asymptote — a wall the curve hugs but never reaches.

Which sign of infinity? Check the sign of the bottom on each side, because the top $x^2>0$ is always positive:

approaching	$x^2-1$ sign	$f=\frac{+}{\text{that}}$
$x\to 1^{+}$ (just above 1)	$+$ (e.g. $1.1^2-1>0$)	$+\infty$
$x\to 1^{-}$ (just below 1)	$-$ (e.g. $0.9^2-1<0$)	$-\infty$

WHAT — horizontal floor/ceiling. Push $x$ enormously large. Divide top and bottom by $x^2$:

$$f(x)=\frac{x^2}{x^2-1}=\frac{1}{1-\frac{1}{x^2}}\underset{x\to \pm \infty }{\to }\frac{1}{1-0}=1.$$

• The term $\frac{1}{x^2}$ is a big-number-denominator, so it fades to $0$.
• The output settles toward $1$: the line $y=1$ is a horizontal asymptote.

WHY. Walls and floor are the "cage" — they tell each of our three rooms which way the curve must escape. These are all limit statements; see Limits and continuity and Rational functions and asymptotes.

PICTURE. The cage: two vertical dashed walls at $x=\pm 1$, one horizontal dashed line at $y=1$, with little arrows showing which way the curve shoots off near each wall.

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Step 5 — Where does it climb, where does it fall? ($f^{\prime }$)

WHAT. The derivative $f^{\prime }(x)$ is the slope of the tangent line — steepness at each point. Positive slope = climbing, negative slope = falling. Using the quotient rule on $\frac{x^2}{x^2-1}$:

$$f^{\prime }(x)=\frac{\stackrel{\text{deriv of top}}{\overbrace{2x}}(x^2-1)-\stackrel{\text{top}}{\overbrace{x^2}}\stackrel{\text{deriv of bottom}}{\overbrace{(2x)}}}{(x^2-1{)}^2}=\frac{-2x}{(x^2-1{)}^2}.$$

Now read the signs, one factor at a time:

• Denominator $(x^2-1{)}^2$ is something squared, so it is $\ge 0$ — always positive (except at the walls where it's undefined). It never flips the sign.
• Numerator $-2x$: positive when $x<0$, negative when $x>0$.

$$f^{\prime }(x)>0\text{ for }x<0(\text{climbing}),f^{\prime }(x)<0\text{ for }x>0(\text{falling}).$$

At $x=0$ the slope is exactly $0$ and it switches $+\to -$: by the First Derivative Test that is a local maximum, a peak at $(0,0)$. (This is First derivative and monotonicity in action; the underlying guarantee is the Mean Value Theorem.)

WHY. Slope tells the direction of travel without plotting extra points. One sign line captures the entire up/down story.

PICTURE. A slope sign-line: green "$+$ climbing" tangents left of $0$, coral "$-$ falling" tangents right of $0$, a flat tangent at the peak.

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Step 6 — How does it bend? ($f^{\prime \prime }$)

WHAT. The second derivative $f^{\prime \prime }$ is the rate of change of the slope — it measures bending. Differentiate $f^{\prime }=\frac{-2x}{(x^2-1{)}^2}$ once more:

$$f^{\prime \prime }(x)=\frac{6x^2+2}{(x^2-1{)}^3}.$$

Sign detective work again:

• Numerator $6x^2+2$: a square times $6$, plus $2$ — always $\ge 2>0$. Never negative.
• Denominator $(x^2-1{)}^3$: an odd power, so it keeps the sign of $x^2-1$.
  • $|x|>1$ (outer rooms): $x^2-1>0\Rightarrow$ cube $>0\Rightarrow f^{\prime \prime }>0$ → concave up $⌣$.
  • $|x|<1$ (middle room): $x^2-1<0\Rightarrow$ cube $<0\Rightarrow f^{\prime \prime }<0$ → concave down $⌢$.

Edge case — is there an inflection point? An inflection needs $f^{\prime \prime }$ to change sign at a real point of the curve. Here $f^{\prime \prime }$ flips sign only across $x=\pm 1$ — but those are holes (the walls), not points on the curve. So there is no inflection point. This is exactly why Step 4 mattered: sign changes at forbidden $x$ don't count.

WHY. Concavity decides whether the climbing/falling happens along a smile or a frown, which fixes the exact curvature between our anchor points. See Second derivative test.

PICTURE. Three rooms shaded by concavity — cup, cap, cup — with a $⌣/⌢/⌣$ label in each.

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Step 7 — Assemble every clue into one curve

WHAT. Put the room facts together. One more sign check tells us which side of the floor $y=1$ each room lives on, using the sign of the bottom (top $x^2\ge 0$ always):

• Outer rooms $|x|>1$: bottom $>0$, and $x^2>x^2-1$, so $f>1$ → curve sits above the floor, concave up, plunging up the walls to $+\infty$.
• Middle room $|x|<1$: bottom $<0$, so $f<0$ (except the peak touches $0$) → curve sits below the floor, concave down, a single hump peaking at $(0,0)$ and diving to $-\infty$ at both walls.

WHY. Every earlier answer — domain, intercept, symmetry, walls, slope, bend — snaps into exactly one consistent picture. No freedom left; the curve is forced.

PICTURE. The finished graph, all three pieces drawn, cage lines dashed.

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The one-picture summary

Everything at once: forbidden walls (Step 1), origin dot + peak (Steps 2, 5), mirror symmetry (Step 3), the floor $y=1$ (Step 4), climb/fall arrows (Step 5), $⌣/⌢/⌣$ shading (Step 6).

Recall Feynman retelling — the whole walkthrough in plain words

I've got a dividing machine, $\frac{x^2}{x^2-1}$. First I ask: can it ever break? Yes — at $x=\pm 1$ the bottom is zero, so I paint two invisible walls there and split the world into three rooms. Then the cheap dots: it goes through the origin. I flip $x$ to $-x$ and nothing changes, so the picture is a mirror butterfly — I only need the right half. Far away, dividing by a huge number, the machine calms down to $1$, so there's an invisible floor at $y=1$. Next I feel the slope: it climbs while $x<0$, tips over at a peak on the origin, and falls while $x>0$. Then I feel the bend: outside the walls it smiles (cup), inside it frowns (cap), and because the smile-to-frown switch only happens at the forbidden walls, there's no real inflection. Finally I check heights: the two outer pieces float above the floor and rocket up the walls; the middle piece is a lonely hump below zero that dives down both inner walls. Snap all six answers together and there's only one curve that fits — done, no calculator needed.

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Active Recall