Curve Sketching Using Calculus
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Direct answer
A curve reveals itself through a fixed interrogation: domain first, then intercepts, symmetry, asymptotes, monotonicity from f'(x), and concavity plus inflection points from f''(x). Where f'(x) > 0 the curve rises, where f'(x) < 0 it falls, and where f' changes sign it has a local extremum; where f'' changes sign the curve flexes through a point of inflection. Asymptotes complete the frame — vertical where the denominator dies, horizontal from lim f(x) as x → ±∞, and oblique by dividing f(x) by x when the function grows linearly. Assembled in this order, any JEE sketching task becomes procedure.
What you must remember
- The checklist: domain, x- and y-intercepts, symmetry (even/odd/periodic), asymptotes, f' sign table for rise/fall and extrema, f'' sign table for concavity and inflection.
- Extrema discipline: local max/min require f' to change sign (or f'' ≠ 0 at the critical point); f'(c) = 0 alone certifies nothing — the flat-point trap.
- Inflection discipline: f''(c) = 0 with f'' changing sign; y = x⁴ has f''(0) = 0 but no inflection there.
- Vertical asymptotes: x = a where f(x) → ±∞, typically where a denominator vanishes without cancellation; check both one-sided limits for the blow-up direction.
- Horizontal and oblique: y = L if f(x) → L as x → ±∞ (check both ends separately — curves can have two different horizontal asymptotes); y = mx + c if f(x)/x → m ≠ 0, with c = lim [f(x) − mx].
- Symmetry shortcuts: even function → mirror in y-axis; odd → rotational symmetry about origin; periodic (trigonometric) → sketch one period and stamp. A curve may cross a horizontal or oblique asymptote but never a vertical one.
Sketching y = x³ − 3x completely
Domain: all reals; intercepts: x(x² − 3) = 0 gives (0, 0), (√3, 0), (−√3, 0). Odd function, so the sketch for x < 0 mirrors rotationally. f'(x) = 3x² − 3 = 3(x − 1)(x + 1): positive outside [−1, 1], negative inside, so the curve rises to x = −1, falls to x = 1, rises after; local maximum at (−1, 2), local minimum at (1, −2). f''(x) = 6x, negative for x < 0 (concave down) and positive after (concave up), so the origin is a point of inflection where the bending reverses. No asymptotes — a cubic grows without bound, and lim f(x)/x = x² diverges, ruling out oblique ones. Plot the seven facts: three intercepts, two extrema, one inflection, and the end behaviour −∞ to +∞; the sketch draws itself.
The same checklist on a rational function shifts weight to asymptotes: y = x/(x² − 1) has vertical asymptotes x = ±1, horizontal asymptote y = 0 (degree of numerator below denominator), an odd symmetry, and it crosses its horizontal asymptote at the origin — the fact students find most counterintuitive. Each branch falls monotonically since f'(x) = −(x² + 1)/(x² − 1)² is negative wherever it exists — no turning points, just four falling branches.
Sketching as an exam weapon
JEE Main rarely demands a drawn sketch; it demands sketch-level judgments — the number of points where f' vanishes, the number of inflections, the number of solutions of f(x) = k read from a mental graph. That last type is the money question: the number of real roots of x³ − 3x = c is three for |c| < 2 (the horizontal line cuts between the local extrema), two exactly at c = ±2, one for |c| > 2 — pure sketching in disguise. Advanced escalates to implicit curves: y² = x(x − 1)² needs two branches meeting at the double root x = 1. The standard trap is asymptote arithmetic: computing lim f(x) at only one end (missing that x → −∞ can behave differently) and dividing by x for oblique asymptotes without checking the constant term exists. Draw sign tables before plotting anything; a sign table error propagates through the entire sketch.
Frequently asked questions
What is the correct order of steps for curve sketching?
Domain, intercepts, symmetry, asymptotes, then f'(x) sign analysis for monotonicity and extrema, finally f''(x) for concavity and inflection — each step feeds the next.
When is f'(c) = 0 not a local extremum?
When f' does not change sign at c, as at x = 0 for y = x³ — the tangent is horizontal but the curve keeps rising through the point.
How do you find an oblique asymptote?
Compute m = lim f(x)/x; if m is finite and non-zero, find c = lim [f(x) − mx], and y = mx + c is the asymptote.
Can a curve cross its own asymptote?
It can cross a horizontal or oblique asymptote (y = x/(x² + 1) crosses y = 0 at the origin) but never a vertical asymptote.
How does a sketch count the solutions of f(x) = k?
Sketch y = f(x), slide the horizontal line y = k, and count intersections — the answer changes exactly when k passes a local extremum value.