Curve sketching
“Algebra is but written geometry and geometry is but figured algebra.”
A sketch is a collection of constraints
A few plotted points can miss a narrow turn or connect branches across a missing input. Calculus supplies stronger information: where a branch exists, which way it travels, how it bends, and where it goes near a boundary.
The purpose is not to calculate every possible feature. It is to collect enough evidence that the overall shape is forced, then check that the pieces agree.
- Find the domain, intercepts and useful symmetry.
- Examine limits at finite domain boundaries and at infinity.
- Chart the first derivative for growth and local extrema.
- Chart the second derivative for concavity and inflections.
- Place key points and draw each domain branch consistently.
A complete rational example
Consider \(f(x)=x/(x^2+1)\).
Example 1
Domain, symmetry and tails
The denominator is positive for every real input, so the domain is \(\mathbb R\). The only intercept is \((0,0)\). Since \(f(-x)=-f(x)\), the graph has symmetry about the origin: rotating by half a turn leaves it unchanged.
For \(x\ne0\), dividing numerator and denominator by \(x^2\) gives
\[\begin{aligned} &\lim_{x\to\pm\infty}\frac{x}{x^2+1}\\ &\quad=\lim_{x\to\pm\infty}\frac{1/x}{1+1/x^2}=0. \end{aligned}\]Thus \(y=0\) is a horizontal asymptote. Since \(f(x)\) has the sign of \(x\), the left tail approaches it from below and the right tail from above. There are no vertical asymptotes: the function is continuous at every finite input.
Example 1 · continued
The turning points
The quotient rule gives
\[\begin{aligned} f^{\prime}(x)&=\frac{(x^2+1)-2x^2}{(x^2+1)^2}\\ &=\frac{1-x^2}{(x^2+1)^2}. \end{aligned}\]The denominator is positive; only the numerator determines the sign.
| Interval | \((-\infty,-1)\) | \((-1,1)\) | \((1,\infty)\) |
|---|---|---|---|
| \(f^{\prime}\) | \(-\) | \(+\) | \(-\) |
| \(f\) | decreasing | increasing | decreasing |
There is a local minimum at \((-1,-1/2)\) and local maximum at \((1,1/2)\). They are also absolute: on the middle interval all values lie between the two, while the outer branches run from an extremum toward the limit zero. This conclusion uses the whole domain, not the local tests alone.
Example 1 · continued
The bends between and beyond the turns
Differentiating the first derivative and simplifying gives
\[f^{\prime\prime}(x)=\frac{2x(x^2-3)}{(x^2+1)^3}.\]The candidate inflection inputs are \(-\sqrt3,0,\sqrt3\). The denominator remains positive.
| Interval | \((-\infty,-\sqrt3)\) | \((-\sqrt3,0)\) | \((0,\sqrt3)\) | \((\sqrt3,\infty)\) |
|---|---|---|---|---|
| \(f^{\prime\prime}\) | \(-\) | \(+\) | \(-\) | \(+\) |
| Concavity | down | up | down | up |
Continuity and the three sign changes establish the inflection points
\[\left(-\sqrt3,-\frac{\sqrt3}{4}\right),\quad(0,0),\] \[\left(\sqrt3,\frac{\sqrt3}{4}\right).\]Notice the order \(-\sqrt3<-1<0<1<\sqrt3\). The inflections and extrema are different landmarks.
Starting at the far left, the graph lies below zero and initially falls while concave down. It changes concavity before reaching its minimum, rises through the origin, reaches its maximum, and then falls toward zero from above. The origin symmetry is a check on all these features.
A sketch is consistent only when its slopes, bends and boundary behaviour all match the calculations.
Domain breaks change how a sign chart is read
Example 2
Two decreasing branches do not make one decreasing interval
For \(g(x)=1/x\), the domain excludes zero. Its derivative \(-1/x^2\) is negative on each of \((-\infty,0)\) and \((0,\infty)\), so each branch decreases.
Yet \(g(-1)=-1<1=g(1)\). There is no single interval connecting these inputs within the domain. The mean value theorem and derivative sign test cannot bridge the missing point. The vertical asymptote \(x=0\) separates the branches.
Although \(g^{\prime\prime}(x)=2/x^3\) changes sign across zero, there is no inflection there: no point of the graph exists at that input.
Example 3
Cancellation leaves a hole
For \(h(x)=(x^2-1)/(x-1)\), the original domain excludes \(1\). For every other input, \(h(x)=x+1\). Thus the graph is the line \(y=x+1\) with \((1,2)\) removed.
The finite limit \(\lim_{x\to1}h(x)=2\) shows a removable hole, not a vertical asymptote. The simplified formula helps draw the graph but does not restore the missing input.
Warning
An asymptote describes limiting behaviour, not a barrier a graph may never cross. The curve in Example 1 crosses its horizontal asymptote at the origin. Keep exact locations such as \(\sqrt3\) in the analysis; decimal approximations are only aids to placement.