Curve Sketching

Curve sketching is the synthesis topic of differential calculus: from a formula alone you recover domain, symmetry, end behaviour, asymptotes, and the shape information carried by $f'$ and $f''$, then draw a graph that is correct in every qualitative feature. No new theory appears here — it is limits, l'Hopital's Rule, the monotonicity test (a corollary of the Mean Value Theorem) and the concavity test used together. Work the checklist in order: each step rules out a class of drawing errors, and plotting random points is never a substitute.

The checklist

1. Domain. 2. Intercepts. 3. Symmetry / periodicity. 4. Asymptotes (vertical, horizontal, slant). 5. $f'$: sign chart, monotonicity, critical points, local extrema. 6. $f''$: sign chart, concavity, inflection points. 7. Assemble the sketch.

1. Domain

Exclude zeros of denominators, negative arguments of even roots, and non-positive arguments of $\ln$. Write the domain as a union of intervals; the graph can only break at excluded points. Cube roots are fine for negative inputs: $x^{1/3}$ and $x^{2/3}$ have domain $(-\infty,\infty)$.

2. Intercepts

The $y$-intercept is $f(0)$; it exists only when $0$ belongs to the domain (for instance $\dfrac{e^{x}}{x}$ has none), and there is at most one, because a function assigns one output to each input. The $x$-intercepts solve $f(x)=0$; for a quotient this means numerator zero and denominator nonzero. If the zeros are not findable exactly, locate them approximately with the Intermediate Value Theorem — a sign change of a continuous $f$ on $[a,b]$ guarantees a root inside.

3. Symmetry

• $f$ is even if $f(-x)=f(x)$ for every $x$ in the domain (which must itself be symmetric about $0$): graph symmetric in the $y$-axis.
• $f$ is odd if $f(-x)=-f(x)$: graph symmetric about the origin.
• $f$ is periodic with period $p$ if $f(x+p)=f(x)$: sketch one period and repeat.
• More generally the graph is symmetric about the point $(a,b)$ exactly when $f(a+t)+f(a-t)=2b$ for all admissible $t$. Every cubic has this symmetry about its inflection point.

Symmetry halves the work: analyse $x\ge 0$ and reflect.

4. Asymptotes

Vertical. The line $x=a$ is a vertical asymptote if at least one of $\lim_{x\to a^-}f(x)$, $\lim_{x\to a^+}f(x)$ equals $+\infty$ or $-\infty$. Always compute both one-sided limits with their signs — that is what tells you which way each branch runs. For a rational function written in lowest terms the vertical asymptotes are exactly the zeros of the denominator; a factor that cancels completely gives a hole (removable discontinuity), not an asymptote.

Horizontal. The line $y=L$ is a horizontal asymptote if $\lim_{x\to\infty}f(x)=L$ or $\lim_{x\to-\infty}f(x)=L$. The two ends may give different values, so a function has at most two horizontal asymptotes. A graph is allowed to cross a horizontal asymptote: the statement is only about the tails.

Slant (oblique). The line $y=mx+b$ with $m\ne 0$ is a slant asymptote if $\lim_{x\to\infty}\left[f(x)-(mx+b)\right]=0$ (or the same at $-\infty$). In general $m=\lim_{x\to\infty}\dfrac{f(x)}{x}$ and $b=\lim_{x\to\infty}\left[f(x)-mx\right]$; for a rational function just do polynomial long division. The sign of the remainder tells you whether the graph sits above or below the line on each branch.

With square roots, remember $\sqrt{x^{2}}=|x|$, so for $x\to-\infty$ you must factor out $-x$, not $x$.

l'Hopital's Rule (the tool for indeterminate tails). Suppose $f$ and $g$ are differentiable on an open interval containing $a$, except possibly at $a$ itself, with $g'(x)\ne0$ there, and suppose $\lim_{x\to a}\dfrac{f(x)}{g(x)}$ has the indeterminate form $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$. If $\lim_{x\to a}\dfrac{f'(x)}{g'(x)}$ exists as a finite number or as $\pm\infty$, then $\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$ The same statement holds for one-sided limits and for $a=\pm\infty$. All three hypotheses matter: verify the indeterminate form first, because on a quotient that is not indeterminate the rule usually returns a wrong value — $\lim_{x\to1}\dfrac{x^{2}}{x}=1$, while differentiating top and bottom gives $\lim_{x\to1}\dfrac{2x}{1}=2$. And if the limit of the ratio of derivatives fails to exist (and is not $\pm\infty$) the rule yields no conclusion — return to the original expression; $\lim_{x\to\infty}\dfrac{x+\sin x}{x}=1$ even though $\lim_{x\to\infty}\dfrac{1+\cos x}{1}$ does not exist. Never differentiate the quotient with the quotient rule; differentiate top and bottom separately.

5. First derivative: monotonicity and extrema

Critical number: a number $c$ in the domain of $f$ with $f'(c)=0$ or $f'(c)$ undefined. Points outside the domain are never critical numbers.

Increasing/Decreasing Test. If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ with $f'>0$ throughout $(a,b)$, then $f$ is increasing on $[a,b]$; if $f'<0$ throughout, $f$ is decreasing. Continuity on the whole interval is essential — monotonicity conclusions can never be carried across a vertical asymptote or a gap.

First Derivative Test. Let $c$ be a critical number of $f$, let $f$ be continuous at $c$, and let $f$ be differentiable on an open interval containing $c$ except possibly at $c$ itself (this is what makes "the sign of $f'$ on each side" meaningful). If $f'$ changes from $+$ to $-$ at $c$, then $f(c)$ is a local maximum; from $-$ to $+$, a local minimum; if $f'$ keeps the same sign on both sides, $c$ gives neither.

Second Derivative Test. If $f''$ is continuous on an open interval containing $c$ and $f'(c)=0$, then $f''(c)>0\Rightarrow$ local minimum, $f''(c)<0\Rightarrow$ local maximum, and $f''(c)=0\Rightarrow$ no conclusion (fall back on the First Derivative Test).

Building the sign chart: mark on a number line every critical number and every point missing from the domain (asymptotes, holes, endpoints). Test one point in each resulting open interval using the factored form of $f'$, tracking only the sign of each factor.

6. Second derivative: concavity and inflection

Let $f$ be differentiable on an interval $I$. Then $f$ is concave up on $I$ if $f'$ is increasing on $I$ (equivalently, the graph lies above each of its tangent lines on $I$), and concave down on $I$ if $f'$ is decreasing on $I$.

Concavity Test. Suppose $f''$ exists on the interval $I$. If $f''(x)>0$ for all $x\in I$ then $f$ is concave up on $I$; if $f''(x)<0$ for all $x\in I$ then $f$ is concave down on $I$. (The converse is only one-way: $f(x)=x^{4}$ is concave up on $(-\infty,\infty)$ yet $f''(0)=0$.)

Inflection point: a point $(c,f(c))$ on the graph where $f$ is continuous at $c$ and the concavity changes at $c$. If $f''(c)$ exists then $f''(c)=0$ is necessary, but it is not sufficient: $f(x)=x^{4}$ has $f''(0)=0$ and no inflection. Concavity can also change where $f''$ fails to exist ($f(x)=x^{1/3}$ at $0$). A sign change of $f''$ across a vertical asymptote is not an inflection point, because $c$ is not in the domain.

Vertical tangents and cusps. Suppose $f$ is continuous at $c$, differentiable on both sides of $c$ near $c$, and $|f'(x)|\to\infty$ as $x\to c$. If $f'$ has the same sign on both sides there is a vertical tangent at $c$; if the signs are opposite there is a cusp.

7. Assembling the sketch

Draw the asymptotes as dashed lines, plot the intercepts and every critical and inflection point with coordinates, mark the direction of each tail, then join with arcs whose slope sign and concavity match the two charts. Sanity check: between consecutive critical numbers the curve cannot turn around.

Worked example 1: a rational function with a slant asymptote

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

Domain: $x\ne1$, i.e. $(-\infty,1)\cup(1,\infty)$.

Intercepts: $f(0)=0$; $f(x)=0\Rightarrow x^{2}=0\Rightarrow x=0$. Only $(0,0)$.

Symmetry: $f(-x)=\dfrac{x^{2}}{-x-1}=-\dfrac{x^{2}}{x+1}$, which is neither $f(x)$ nor $-f(x)$. None.

Asymptotes: the numerator at $x=1$ is $1\ne0$, so nothing cancels and $x=1$ is a vertical asymptote: $\lim_{x\to1^-}\frac{x^{2}}{x-1}=-\infty \qquad \lim_{x\to1^+}\frac{x^{2}}{x-1}=+\infty$ (numerator $\to1>0$, denominator $\to0^-$ then $0^+$). Long division: $x^{2}=(x-1)(x+1)+1$, so $f(x)=x+1+\frac{1}{x-1}$ Since $\dfrac{1}{x-1}\to0$ as $x\to\pm\infty$, the slant asymptote is $y=x+1$; the degrees are $2$ and $1$, so there is no horizontal asymptote. Because $f(x)-(x+1)=\dfrac{1}{x-1}$, the graph lies above $y=x+1$ for $x>1$ and below it for $x<1$.

First derivative: $f'(x)=1-\frac{1}{(x-1)^{2}}=\frac{(x-1)^{2}-1}{(x-1)^{2}}=\frac{x^{2}-2x}{(x-1)^{2}}=\frac{x(x-2)}{(x-1)^{2}}$

Critical numbers $x=0$ and $x=2$ ($x=1$ is not one: it is not in the domain). Local maximum $(0,0)$, local minimum $(2,4)$.

Second derivative: $f''(x)=\dfrac{2}{(x-1)^{3}}$, never zero. It is negative for $x<1$ (concave down) and positive for $x>1$ (concave up); since $x=1$ is not in the domain, there is no inflection point. Cross-check: $f''(0)=-2<0$ confirms the maximum, $f''(2)=2>0$ confirms the minimum.

Sketch: the left branch rises along $y=x+1$ from below, peaks at $(0,0)$, then plunges to $-\infty$ at $x=1^-$, concave down throughout. The right branch drops from $+\infty$ at $x=1^+$ to the minimum $(2,4)$ and then rises, staying above $y=x+1$ and concave up throughout.

Worked example 2: an exponential

$f(x)=xe^{-x}$

Domain: $(-\infty,\infty)$. Intercepts: only $(0,0)$. Symmetry: none.

End behaviour: as $x\to\infty$ write $f(x)=\dfrac{x}{e^{x}}$, an $\dfrac{\infty}{\infty}$ indeterminate form with both functions differentiable and $e^{x}\ne0$; by l'Hopital's Rule $\lim_{x\to\infty}\dfrac{x}{e^{x}}=\lim_{x\to\infty}\dfrac{1}{e^{x}}=0$, and since this limit exists the application is legitimate. So $y=0$ is a horizontal asymptote as $x\to+\infty$, approached from above because $f(x)>0$ for $x>0$. As $x\to-\infty$, $x\to-\infty$ and $e^{-x}\to+\infty$, so $f(x)\to-\infty$: no asymptote there (and no indeterminate form, so l'Hopital would be illegal). No vertical asymptotes.

First derivative: $f'(x)=e^{-x}-xe^{-x}=e^{-x}(1-x)$. Since $e^{-x}>0$ always, $f'>0$ for $x<1$ and $f'<0$ for $x>1$: increasing on $(-\infty,1)$, decreasing on $(1,\infty)$. Local (indeed absolute) maximum at $\left(1,\dfrac{1}{e}\right)\approx(1,0.368)$.

Second derivative: $f''(x)=-e^{-x}(1-x)+e^{-x}(-1)=e^{-x}(x-2)$. So $f''<0$ on $(-\infty,2)$ and $f''>0$ on $(2,\infty)$: concave down then up, with inflection point $\left(2,\dfrac{2}{e^{2}}\right)\approx(2,0.271)$. Check: $f''(1)=-e^{-1}<0$, consistent with a maximum at $x=1$.

Sketch: rising steeply out of $-\infty$ on the left, crossing the origin, peaking at $(1,1/e)$, bending at $(2,2/e^{2})$ and then decaying to the axis from above.

Common mistakes