Concavity and Curve Sketching

What is concavity?

The graph of a differentiable function y=f(x) is concave up on an open interval I if f' is increasing on I, and concave down on I if f' is decreasing on I. A function whose graph is concave up is also often called convex. The picture is of how the curve turns. Along y=x^3 coming into the origin from the left, the curve bends to the right and falls below its tangent lines, and the slopes of those tangent lines decrease: concave down. Leaving the origin to the right, the curve bends to the left and rises above its tangent lines, and the slopes increase: concave up.

Since f' is a function in its own right, the monotonic functions of the previous article apply to it: f' increases where its own derivative f'' is positive, and decreases where f'' is negative.

Second Derivative Test for concavity. Let y=f(x) be twice-differentiable on an interval I. 1. If f''>0 on I, the graph of f over I is concave up. 2. If f''<0 on I, the graph of f over I is concave down.

We write f'' or y'' for the second derivative. Concavity is not rising or falling: a concave-up graph can rise, fall or do both, as x^2 does, while what it cannot do is wave.

Example 1. (a) The curve y=x^3 has y''=6x, negative for x<0 and positive for x>0, so it is concave down on (-∞,0) and concave up on (0,∞). (b) The curve y=x^2 has y''=2>0 everywhere, so it is concave up on the whole line.

Example 2. Determine the concavity of y=3+\sin x on 0\le x\le2π. Here y'=\cos x and y''=-\sin x, which is negative on (0,π) and positive on (π,2π). The graph is concave down on (0,π) and concave up on (π,2π).

Type a function and a window below, or choose an example. The graph is dark where it is concave up and orange where it is concave down, and the graph of f'' shows why.

What is a point of inflection?

A point (c,f(c)) where the graph of a function has a tangent line and where the concavity changes is a point of inflection. At such a point the graph crosses its own tangent line. In Example 2 the curve y=3+\sin x changes concavity at (π,3), where the tangent has slope y'(π)=\cosπ=-1.

Since the tangent line exists, either f'(c) exists (is finite), or the graph has a vertical tangent there. At a vertical tangent neither f' nor f'' exists. So at a point of inflection one of two things happens: f''(c)=0 or f''(c) fails to exist. When f'' is continuous the first follows from the Intermediate Value Theorem, because f'' changes sign. The condition is necessary and not sufficient: f''=0 or not existing only marks a candidate, and the sign has to change.

Example 3. Find the concavity and inflection points of f(x)=x^3-3x^2+2. Here f'(x)=3x^2-6x and f''(x)=6x-6, which is negative for x<1, zero at x=1 and positive for x>1. The graph is concave down on (-∞,1), concave up on (1,∞) and has the inflection point (1,0).

Example 4: f'' does not exist. The graph of f(x)=x^5/3 has a horizontal tangent at the origin, since f'(x)=\frac53x^2/3=0 at x=0. The second derivative f''(x)=\frac109x^-1/3 does not exist at 0, but it is negative for x<0 and positive for x>0, so it changes sign: an inflection point at the origin.

Example 5: f''=0 and no inflection. The curve y=x^4 has y''=12x^2, which is 0 at x=0 but does not change sign: the curve is concave up everywhere.

Example 6: a vertical tangent. The graph of y=x^1/3 has an inflection point at the origin because y''=-\frac29x^-5/3 is positive for x<0 and negative for x>0. But y'=\frac13x^-2/3 and y'' both fail to exist at 0, and there is a vertical tangent.

A caution. When the second derivative does not exist at c, an inflection point may or may not occur. x^2/3 has no second derivative at 0 and no inflection: its second derivative -\frac29x^-4/3 is negative on both sides, and the graph has a cusp, with no tangent line at all. Another case: f(x)=x|x| has f'(x)=2|x|, which exists at 0, although f''=2 sign x does not, and it is an inflection point. At such places the graph can have vertical tangents, corners, cusps or breaks, so read the sign of f'' on both sides and ask whether there is a tangent line. Choose each case below to see the signs, the tangent and the verdict.

What does the second derivative say about motion?

For a particle with position s(t) the acceleration is a=s'', and the inflection points of the position graph are where the acceleration changes sign. Together with the motion along a line of the rates article, the velocity v=s' says which way the particle moves, and the sign of a against the sign of v says whether it speeds up or slows down: it speeds up when v and a have the same sign, and slows down when they have opposite signs.

Example 7. A particle moves along a horizontal coordinate line (positive to the right) with position s(t)=2t^3-14t^2+22t-5 for t\ge0. The velocity is v(t)=s'(t)=6t^2-28t+22 v(t)=2(t-1)(3t-11) and the acceleration is a(t)=s''(t)=12t-28=4(3t-7). The velocity is zero at t=1 and t=\frac113, and the acceleration at t=\frac73. Putting the signs together:

The particle starts moving right while the leftward acceleration slows it, turns round at t=1 and speeds up to the left. At t=\frac73 the acceleration changes direction, which is the inflection point of the position graph, but the particle keeps moving left, now slowing down. At t=\frac113 it turns round again and, moving the same way as the acceleration, speeds up to the right.

What is the Second Derivative Test for local extrema?

Instead of looking for a sign change of f' at a critical point, we can sometimes read the concavity there: a graph that is concave down at a point where the tangent is horizontal is at the top of a hill, and one that is concave up is at the bottom of a valley.

Second Derivative Test for local extrema. Suppose f'' is continuous on an open interval that contains x=c. 1. If f'(c)=0 and f''(c)<0, then f has a local maximum at x=c. 2. If f'(c)=0 and f''(c)>0, then f has a local minimum at x=c. 3. If f'(c)=0 and f''(c)=0, then the test fails: f may have a local maximum, a local minimum, or neither.

Proof of part 1. If f''(c)<0 then f''<0 on some open interval I around c, because f'' is continuous. So f' is decreasing on I. Since f'(c)=0, the sign of f' changes from positive to negative at c, and f has a local maximum at c by the First Derivative Test. Part 2 is similar. ∎

Part 3. Take y=x^4, y=-x^4 and y=x^3. For each of them the first and second derivatives are zero at x=0. Yet x^4 has a local minimum there, -x^4 has a local maximum, and x^3 is increasing on every open interval around 0 and has neither. The test cannot tell them apart.

The test needs f'' only at c itself, which makes it easy to apply: one number answers the question. The price is that it is inconclusive when f''(c)=0 or when f'' does not exist at c (as for x^2/3 at its cusp); then the First Derivative Test is used. Choose a case below and read the two tests side by side.

How do you sketch the graph of a function?

To sketch y=f(x), collect what f, f' and f'' say and combine it: where the graph rises and falls, how it bends, where it turns, and what it approaches.

Procedure for graphing y=f(x). 1. Identify the domain of f and any symmetries the curve may have. 2. Find the derivatives y' and y''. 3. Find the critical points of f, if any, and identify the behavior of f at each one. 4. Find where the curve is increasing and where it is decreasing. 5. Find the points of inflection, if any, and determine the concavity of the curve. 6. Identify any asymptotes that may exist. 7. Plot key points, such as the intercepts and the points found in steps 3 to 5, and sketch the curve together with any asymptotes.

Asymptotes. The line x=a is a vertical asymptote if f(x)→±∞ as x→ a from one side or both, which for a rational function happens where the denominator is zero and the numerator is not. The line y=L is a horizontal asymptote if f(x)→ L as x→∞ or x→-∞. An oblique asymptote y=mx+b appears when f(x)-(mx+b)→0; for a rational function it is what long division leaves when the numerator is one degree higher than the denominator. A graph may cross a horizontal or oblique asymptote at finite values of x; what it does is approach it at the ends.

Example 8. Sketch f(x)=x^4-4x^3+10. It is continuous with domain the whole line. The derivative f'(x)=4x^3-12x^2=4x^2(x-3) is zero at x=0 and x=3, with signs -,-,+: so f decreases on (-∞,3] and increases on [3,∞), there is no extreme value at 0 and a local minimum f(3)=-17 at 3. The second derivative f''(x)=12x^2-24x=12x(x-2) is zero at 0 and 2, with signs +,-,+: the graph is concave up on (-∞,0), concave down on (0,2) and concave up on (2,∞), with inflection points (0,10) and (2,-6). The curve cuts the y-axis at 10. Notice that x=0 is a critical point and an inflection point, where the curve flattens out while still falling.

Now try the procedure on any function below, or on the textbook's five graphed examples. The tool follows the seven steps and draws the result: dark where f rises, orange where it falls, orange dots at local extremes, black dots at inflection points and dashed lines for asymptotes.

What do the sketches of rational, exponential and trigonometric functions look like?

Four more graphs show the steps with asymptotes, a break, symmetry and a trigonometric term.

Example 9: f(x)=((x+1)^2)/(1+x^2). The domain is the whole line, with no symmetry. Since f(x)=1+(2x)/(1+x^2), f'(x)=(2(1-x^2))/((1+x^2)^2) f''(x)=(4x(x^2-3))/((1+x^2)^3). The critical points are x=±1. Since f''(-1)=1>0 there is a local minimum f(-1)=0, and since f''(1)=-1<0 a local maximum f(1)=2. The sign of f' is -,+,-, so f decreases, increases, then decreases again. The denominator of f'' is always positive, and the numerator is zero at -√3, 0 and √3 with signs -,+,-,+: each is an inflection point, \left(-√3,1-\frac√32\right), (0,1) and \left(√3,1+\frac√32\right), with the graph concave down, up, down, up. Dividing by x^2 shows f(x)→1 as x→±∞: the line y=1 is a horizontal asymptote, and there is no vertical one. The graph meets its asymptote at x=0 and approaches it from below on the left, concave down, and from above on the right, concave up.

Example 10: f(x)=(x^2+4)/(2x)=\frac x2+\frac2x. The domain is x≠0, there are no intercepts, and f(-x)=-f(x): an odd function, symmetric about the origin. The derivatives are f'(x)=\frac12-\frac2x^2=(x^2-4)/(2x^2) f''(x)=\frac4x^3. The critical points x=±2 give a local maximum f(-2)=-2 and a local minimum f(2)=2, from the sign of f''. The function increases on (-∞,-2), decreases on (-2,0) and (0,2), and increases on (2,∞). The second derivative is negative for x<0 and positive for x>0 and never zero, so there is no inflection point. Because \frac2x→±∞ as x→0^±, the y-axis is a vertical asymptote, and because f(x)-\frac x2=\frac2x→0 at both ends, y=\frac x2 is an oblique asymptote.

Example 11: f(x)=e^2/x. The domain is x≠0 and there is no symmetry. The derivatives are f'(x)=-(2e^2/x)/(x^2) f''(x)=(4(x+1)e^2/x)/(x^4). Since f'<0 everywhere on its domain, the graph decreases on (-∞,0) and on (0,∞). The second derivative has the sign of x+1, so it changes sign at x=-1: the point \left(-1,e^-2\right) is an inflection point, with concave down to the left of it and concave up after. As x→0^+, \frac2x→∞ and f→∞: the y-axis is a vertical asymptote. From the left f→0. As x→±∞, \frac2x→0 and f→1, so y=1 is a horizontal asymptote. f is never 0, and it has no absolute extremes.

Example 12: f(x)=\cos x-\frac√22x on 0\le x\le2π. Here f'(x)=-\sin x-\frac√22 is zero where \sin x=-\frac√22, at x=\frac5π4 and x=\frac7π4. Since f''(x)=-\cos x, f''\left(\frac5π4\right)=\frac√22>0 gives a local minimum f\left(\frac5π4\right)≈-3.48, and f''\left(\frac7π4\right)=-\frac√22<0 a local maximum f\left(\frac7π4\right)≈-3.18. The second derivative is zero at \fracπ2 and \frac3π2 and changes sign there: inflection points \left(\fracπ2,-1.11\right) and \left(\frac3π2,-3.33\right). At the ends f(0)=1 and f(2π)≈-3.44, so on this closed interval the absolute maximum is f(0)=1 and the absolute minimum is f\left(\frac5π4\right)≈-3.48.

How do f, f′ and f″ fit together, and what can they decide?

The first derivative says where the graph rises and falls and where it turns; the second says how it bends; and together with the asymptotes and a few points they fix the picture. The signs combine like this:

The same facts answer questions about the constants in a formula.

Quadratics and cubics. For a parabola y=ax^2+bx+c with a≠0, y'=2ax+b is zero at x=-\frac b2a, the vertex, and y''=2a never changes sign: concave up if a>0, down if a<0, and a quadratic curve never has an inflection point. For a cubic y=ax^3+bx^2+cx+d, y''=6ax+2b is zero at exactly one point, x=-\frac b3a, and changes sign there: every cubic has exactly one inflection point. If it has to be at x=1 for y=x^3+bx^2+cx+d, then 6+2b=0 and b=-3.

Finding constants. For which a, b, c does y=ax^3+bx^2+cx have a local maximum at x=3, a local minimum at x=-1 and an inflection point at (1,11)? The inflection at x=1 needs y''(1)=6a+2b=0, so b=-3a. Then y'=3ax^2+2bx+c with y'(3)=27a+6b+c=0 gives c=-9a, and y'(-1)=3a-2b+c=0 holds automatically. The point (1,11) needs a+b+c=-11a=11, so a=-1, b=3, c=9, and y=-x^3+3x^2+9x. Check: y''=-6x+6 is -12<0 at 3 (a maximum y(3)=27) and 12>0 at -1 (a minimum y(-1)=-5).

Given only f''. If y''=(x+1)(x-2), the sign of y'' is +,-,+ with changes at x=-1 and x=2: those are the inflection points, with no need to know f.

How do you find concavity and inflection points in code?

In code, concavity is the same sign test as before, one derivative higher: find the points where f'' is zero or does not exist, read the sign of f'' at one point between each pair, and call a candidate an inflection point when the sign changes. SymPy does it exactly with sp.solve_univariate_inequality(d2f > 0, x, relational=False) and the same with < 0; the Second Derivative Test is one substitution, d2f.subs(x, c). Without a CAS, take midpoints, which is what the JavaScript and C++ programs do. For a numerical f the second derivative is itself approximated by (f(c+h)-2f(c)+f(c-h))/(h^2), which needs a moderate h because the numerator is a difference of nearly equal numbers.

import sympy as sp

x = sp.symbols('x', real=True)


def concavity_report(f):
    """Where the graph of f is concave up and down, its inflection points, and the Second Derivative Test."""
    df, d2f = sp.diff(f, x), sp.diff(f, x, 2)
    up = sp.solve_univariate_inequality(d2f > 0, x, relational=False)
    down = sp.solve_univariate_inequality(d2f < 0, x, relational=False)
    eps = sp.Rational(1, 10**6)
    inflections = [(c, f.subs(x, c)) for c in sorted(sp.solveset(sp.Eq(d2f, 0), x, sp.S.Reals), key=float)
                   if (up.contains(c - eps) and down.contains(c + eps)) or (down.contains(c - eps) and up.contains(c + eps))]
    second = {}
    for c in sorted(sp.solveset(sp.Eq(df, 0), x, sp.S.Reals), key=float):
        k = d2f.subs(x, c)
        second[c] = 'min' if k > 0 else 'max' if k < 0 else 'fails'
    return up, down, inflections, second


for f in (x**4 - 4 * x**3 + 10, (x + 1)**2 / (1 + x**2), x**3):
    print(f, '->', concavity_report(f))
# x**4 - 4*x**3 + 10 -> (Union(Interval.open(-oo, 0), Interval.open(2, oo)), Interval.open(0, 2), [(0, 10), (2, -6)], {0: 'fails', 3: 'min'})
# (x + 1)**2/(x**2 + 1) -> (Union(Interval.open(-sqrt(3), 0), Interval.open(sqrt(3), oo)), Union(Interval.open(-oo, -sqrt(3)), Interval.open(0, sqrt(3))), [(-sqrt(3), (1 - sqrt(3))**2/4), (0, 1), (sqrt(3), (1 + sqrt(3))**2/4)], {-1: 'min', 1: 'max'})
# x**3 -> (Interval.open(0, oo), Interval.open(-oo, 0), [(0, 0)], {0: 'fails'})

// concavity between the candidates, and the verdict at each candidate: an inflection point when the bending changes
const bendChart = (d2, candidates, lo, hi) => {
  const edges = [lo, ...candidates, hi]
  const pieces = edges.slice(1).map((to, i) => [edges[i], to, d2((edges[i] + to) / 2) > 0 ? 'up' : 'down'])
  const inflections = candidates.filter((c, i) => pieces[i][2] !== pieces[i + 1][2])
  return { pieces, inflections }
}
// the Second Derivative Test at c, given f' and f'' as functions
const secondTest = (df, d2, c) => {
  if (Math.abs(df(c)) > 1e-9) return 'not critical'
  const k = d2(c)
  return Math.abs(k) < 1e-9 ? 'fails' : k > 0 ? 'min' : 'max'
}
const d2 = (x) => 12 * x * x - 24 * x                        // f(x) = x^4 - 4x^3 + 10
bendChart(d2, [0, 2], -2, 5)
// pieces [[-2,0,'up'], [0,2,'down'], [2,5,'up']]   inflections [0, 2]
const df = (x) => 4 * x ** 3 - 12 * x ** 2
secondTest(df, d2, 0)                                        // 'fails': f'(0) = f''(0) = 0 (the First Derivative Test: no extreme value)
secondTest(df, d2, 3)                                        // 'min': f''(3) = 36 > 0
bendChart((x) => 12 * x * x, [0], -2, 2)                     // x^4: pieces [[-2,0,'up'], [0,2,'up']]   inflections []: f'' = 0 at 0 but no inflection

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

double d2(double x) { return 12 * x * x - 24 * x; }       // f(x) = x^4 - 4x^3 + 10, f'' = 12x(x - 2)
double d1(double x) { return 4 * x * x * x - 12 * x * x; }

int main() {
  double edge[4];
  edge[0] = -2; edge[1] = 0; edge[2] = 2; edge[3] = 5;       // the ends and the places where f'' is zero
  int sign[3];
  for (int i = 0; i < 3; i++) {
    double mid = (edge[i] + edge[i + 1]) / 2;                // one test point in each interval
    sign[i] = d2(mid) > 0 ? 1 : -1;
    if (sign[i] > 0) printf("from %.1f to %.1f; f'' at %.2f is %.2f; concave up\n", edge[i], edge[i + 1], mid, d2(mid));
    else printf("from %.1f to %.1f; f'' at %.2f is %.2f; concave down\n", edge[i], edge[i + 1], mid, d2(mid));
  }
  for (int i = 0; i < 2; i++)                                // an inflection point where the bending changes
    if (sign[i] != sign[i + 1]) printf("x=%.1f inflection point\n", edge[i + 1]);
    else printf("x=%.1f no inflection point\n", edge[i + 1]);
  double crit[2];
  crit[0] = 0; crit[1] = 3;                                  // the zeros of f'
  for (int i = 0; i < 2; i++) {                              // the Second Derivative Test
    double k = d2(crit[i]);
    if (fabs(k) < 1e-9) printf("x=%.1f; f''=%.2f; the test fails\n", crit[i], k);
    else if (k > 0) printf("x=%.1f; f''=%.2f; local minimum\n", crit[i], k);
    else printf("x=%.1f; f''=%.2f; local maximum\n", crit[i], k);
  }
  return 0;
}

/* output:
from -2.0 to 0.0; f'' at -1.00 is 36.00; concave up
from 0.0 to 2.0; f'' at 1.00 is -12.00; concave down
from 2.0 to 5.0; f'' at 3.50 is 63.00; concave up
x=0.0 inflection point
x=2.0 inflection point
x=0.0; f''=0.00; the test fails
x=3.0; f''=36.00; local minimum
*/

Where do inflection points and curve sketching come from?

Studying the shape of a curve with derivatives began with the first rules of calculus and became a routine procedure by the 18th century.

• Leibniz, 1684. In his first paper on the differential calculus, Gottfried Wilhelm Leibniz used the second differential to tell a maximum from a minimum, and noted that the second differential vanishes at an inflection point.
• l'Hôpital, 1696. Guillaume de l'Hôpital's textbook Analyse des infiniment petits, the first on the differential calculus, applies the method to find points of inflection and cusps of curves given by equations.
• Maclaurin, 1742. Colin Maclaurin's Treatise of Fluxions organised the tests for maxima, minima and inflection from the signs of the first and second fluxions, in the form still taught.
• Euler, 1748. Leonhard Euler's Introductio in analysin infinitorum studies curves by their asymptotes and branches, the part of the procedure that comes from algebra rather than from derivatives.
• Today. Graphing software draws the curve point by point; the sketching steps are still how one checks that a plot shows every extreme, every inflection and every asymptote.

What mistakes do people make with concavity?

The most common mistakes with concavity and curve sketching are the nine below, each with the correct statement.

Practice: test your understanding

These questions mix everything above. A wrong answer costs nothing here: read the hint and try again.

Summary: concavity and curve sketching at a glance

• Concavity: concave up where f' increases (f''>0), concave down where f' decreases (f''<0).
• Point of inflection: a point with a tangent line where the concavity changes; f''=0 or does not exist there, but a zero of f'' alone is not enough.
• Cases: x^3 yes, x^4 no, x^5/3 and x^(1/3) yes (with f'' undefined), x^2/3 no (a cusp).
• Motion: a=s''; speeding up when v and a agree in sign, slowing down when they differ.
• Second Derivative Test: f'(c)=0 and f''(c)<0 is a maximum, f''(c)>0 a minimum, f''(c)=0 fails; then use the First Derivative Test.
• Curve sketching: domain and symmetry, f' and f'', critical points, rising and falling, inflection and concavity, asymptotes, key points.
• Asymptotes: vertical where f→±∞, horizontal and oblique from the ends; a graph can cross one.
• Cubics: exactly one inflection point, at x=-\frac b3a; parabolas have none.
• Code: sp.solve_univariate_inequality(d2f > 0, x, relational=False) and d2f.subs(x, c).

Glossary: key terms

Concave up

A graph is concave up on an interval where its slope is increasing, which is where the second derivative is positive; it bends like a cup and lies above its tangent lines.

Concave down

A graph is concave down on an interval where its slope is decreasing, which is where the second derivative is negative; it bends like a cap and lies below its tangent lines.

Point of inflection

A point of a graph that has a tangent line and where the concavity changes from up to down or from down to up.

Second Derivative Test

The test that classifies a critical point by the sign of the second derivative there: negative is a local maximum, positive a local minimum, zero is inconclusive.

Asymptote

A line that the graph of a function approaches without bound: vertical where the function blows up, horizontal or oblique at the ends of the x-axis.

Curve sketching

Drawing the graph of a function from its domain, symmetry, first and second derivatives, critical points, inflection points, asymptotes and a few key points.

Frequently asked questions

References

1. Joel Hass, Christopher Heil, Maurice D. Weir et al. (2023). Thomas' Calculus: Early Transcendentals in SI Units (15th ed.), Section 4.4: Concavity and Curve Sketching. Pearson.
2. James Stewart (2015). Calculus (8th ed.), section 4.3 and 4.5: How Derivatives Affect the Shape of a Graph; Summary of Curve Sketching. Cengage Learning.
3. Guillaume de l'Hôpital (1696). Analyse des infiniment petits pour l'intelligence des lignes courbes. Paris.
4. Colin Maclaurin (1742). A Treatise of Fluxions. Edinburgh.
5. Leonhard Euler (1748). Introductio in analysin infinitorum. Lausanne.

Related articles

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• The Mean Value Theorem: The Mean Value Theorem and Rolle’s Theorem: statements, proofs, how to find c, speed and rates, and the corollaries about constant functions, with examples and code.
• Extreme Values of Functions on Closed Intervals: Extreme values on closed intervals: absolute and local extremes, the Extreme Value Theorem, critical points and the closed interval method, with examples and code.
• The Derivative as a Function: The derivative as a function: f′(x) from the definition, notation, graphing f′ from f, step functions, domains, one-sided derivatives and corners, and code.