Monotonic Functions and the First Derivative Test

What are increasing and decreasing functions?

A function f is increasing on an interval I if f(x_1)<f(x_2) whenever x_1<x_2 are in I, and decreasing on I if f(x_1)>f(x_2) whenever x_1<x_2 are in I. In words: the graph rises from left to right, or falls from left to right. A function that is increasing on an interval or decreasing on an interval is called monotonic there. When a book writes "increasing" it sometimes means the weaker f(x_1)\le f(x_2), which allows flat stretches; this article uses the strict form, and says "constant" for a flat stretch.

The words belong to an interval. We do not say that a function is increasing or decreasing at a single point, because the definition compares two different points. f(x)=x^2 is decreasing on (-∞,0] and increasing on [0,∞), and it is not monotonic on the whole line. Whether the end points are included changes nothing for a continuous function: if it rises on (0,1) it also rises on [0,1].

A strictly monotonic function never takes the same value twice, because two different inputs are ordered and so are their outputs. That is the reason monotonic functions have inverses, which the last sections come back to. The practical question is how to find the intervals; testing the definition for every pair of points is impossible, and the derivative does it for us.

How does the sign of the derivative show where a function rises and falls?

Corollary 3 of the Mean Value Theorem: suppose f is continuous on [a,b] and differentiable on (a,b). If f'(x)>0 at each point of (a,b), then f is increasing on [a,b]; if f'(x)<0 at each point, then f is decreasing on [a,b].

Proof. Let x_1<x_2 be any two points of [a,b]. The Mean Value Theorem applied to f on [x_1,x_2] gives a point c between them with f(x_2)-f(x_1)=f'(c) (x_2-x_1). The factor x_2-x_1 is positive, so the right side has the sign of f'(c). If f' is positive on (a,b) then f(x_2)>f(x_1), and if f' is negative then f(x_2)<f(x_1). ∎

The corollary also holds on open and infinite intervals, and the derivative need not exist at an end. f(x)=√ x has f'(x)=\frac12√ x>0 on (0,b), so it is increasing on [0,b] for every b>0 even though f'(0) does not exist. Nor is a zero of f' at a single point a problem: x^3 has f'=3x^2=0 at 0 and is increasing on the whole line.

The method. To find the intervals where f rises and falls: 1. find the critical points of f, the places where f' is zero or does not exist; 2. these cut the domain into open intervals; on each one f' is continuous and never zero, so by the Intermediate Value Theorem it has one sign throughout; 3. read that sign by evaluating f' at one convenient point of each interval, and apply the corollary.

Example 1. Find the critical points of f(x)=x^3-12x-5 and the intervals where it rises and falls. The derivative is f'(x)=3x^2-12=3(x+2)(x-2), which is zero at x=-2 and x=2. The intervals (-∞,-2), (-2,2) and (2,∞) are tested at -3, 0 and 3:

Since the corollary also allows \le, f is increasing on (-∞,-2], decreasing on [-2,2] and increasing on [2,∞). Type any function below, or choose an example: the graph is dark where it rises and orange where it falls, and the graph of f' shows why, above the axis or below.

What is the First Derivative Test for local extrema?

At a point where f has a local minimum, f'<0 immediately to the left and f'>0 immediately to the right, so f falls and then rises; at a local maximum it is the other way round. At a local extreme point the sign of f' changes. Together with the critical points of the previous article, which say where an extreme value can be, this gives a test for what happens there.

First Derivative Test for local extrema. Suppose that c is a critical point of a continuous function f, and that f is differentiable at every point of some interval containing c except possibly at c itself. Moving across the interval from left to right: 1. if f' changes from negative to positive at c, then f has a local minimum at c; 2. if f' changes from positive to negative at c, then f has a local maximum at c; 3. if f' does not change sign at c (f' is positive on both sides, or negative on both sides), then f has no local extreme value at c.

Proof of part 1. The sign of f' changes from negative to positive at c, so there are numbers a<c<b with f'<0 on (a,c) and f'>0 on (c,b). If x is in (a,c), then f(c)<f(x) because f'<0 makes f decreasing on [a,c]. If x is in (c,b), then f(c)<f(x) because f'>0 makes f increasing on [c,b]. So f(x)\ge f(c) for every x in (a,b), which is the definition of a local minimum at c. ∎ Parts 2 and 3 are proved in the same way.

At an end of the domain there is only one side. If f rises from the left end a (that is, f'>0 just to the right of a), then f(a) is a local minimum; if it falls, a local maximum. At a right end, the reverse.

The test uses the sign, not the size: a critical point where f'=0 is not an extreme value if the sign stays the same, and a point where f' does not exist can be one, as the corner of |x| and the cusp of x^2/3 show. Choose each of the eight cases below and read the sign on the left and on the right.

How do you work through a complete example?

A complete analysis finds the critical points, the sign of f' between them, the intervals of increase and decrease, the local extreme values and which of them are absolute.

Example 2. f(x)=x^1/3(x-4)=x^4/3-4x^1/3. The function is continuous everywhere, and f'(x)=\tfrac43x^1/3-\tfrac43x^-2/3=(4(x-1))/(3x^2/3) is zero at x=1 and undefined at x=0. Those are the only critical points.

The test says there is no extreme value at 0 (f' does not change sign: f falls through the vertical tangent there, since f'→-∞) and a local minimum at x=1, with value f(1)=-3. Because f decreases on (-∞,1) and increases on (1,∞), this is also the absolute minimum.

Example 3. f(x)=(x^2-3)e^x. It is differentiable everywhere, so the critical points are the zeros of f'(x)=2xe^x+(x^2-3)e^x f'(x)=(x^2+2x-3)e^x f'(x)=(x+3)(x-1)e^x. Since e^x is never zero, they are x=-3 and x=1. The sign of f' is + for x<-3, - for -3<x<1 and + for x>1. So f increases on (-∞,-3] and [1,∞), decreases on [-3,1], has a local maximum f(-3)=6e^-3≈0.299 and a local minimum f(1)=-2e≈-5.437. The minimum is absolute, because f(x)>0 for x>√3 and f(x)→0 as x→-∞; the maximum is not, because f grows without bound. This is the situation the closed interval method cannot handle alone: on a whole line the signs of f' say which values are absolute.

Example 4. f(x)=x^4-8x^2+16=(x^2-4)^2. Here f'(x)=4x^3-16x=4x(x-2)(x+2), zero at -2, 0 and 2, with signs -,+,-,+ in the four intervals. So f has local minima f(±2)=0 and a local maximum f(0)=16.

Example 5. f(x)=\sin x-\cos x on [0,2π]. Here f'(x)=\cos x+\sin x is zero at x=(3π)/(4) and x=(7π)/(4), and f'(0)=1>0. So f rises on \left[0,\frac3π4\right], falls on \left[\frac3π4,\frac7π4\right] and rises on \left[\frac7π4,2π\right], with a local maximum √2 at (3π)/(4) and a local minimum -√2 at (7π)/(4). At the ends f(0)=f(2π)=-1, which are local extremes of the window: a local minimum at 0 (it rises from there) and a local maximum at 2π.

How do you read a function from its derivative alone?

When only f' is given, the sign chart of f' is all that is needed: where f' is above the axis f rises, where it is below the axis f falls, and the critical points where it crosses the axis are the local extremes. The formula for f does not matter, nor do the actual values of f'; only its signs do.

A factored derivative makes the chart easy. Each factor (x-r)^m changes sign at r when the exponent m is odd, and keeps its sign when m is even. For f'(x)=(x-1)^2(x+2): - at x=-2 the factor x+2 changes sign, and (x-1)^2 is positive: f' goes from negative to positive, a local minimum; - at x=1 the factor (x-1)^2 never changes sign: f' stays positive, so f keeps rising and there is no extreme value, although f'(1)=0.

For f'(x)=(x-1)^2(x+2)^2 the derivative is never negative, so f rises on the whole line and has no extreme value at all, though it has two critical points. A point where f' does not exist counts as critical only if f is defined there; for f'(x)=1-(4)/(x^2) the point x=0 is not in the domain of f(x)=x+\frac4x, so it is a break, not an extreme value, even though f' has the same sign on both sides of it.

Try the derivatives of the textbook's exercises below and see which critical points are extremes. The graph of f' is dark where f'>0 (so f rises) and orange where f'<0.

How does monotonicity prove an inequality?

To prove f(x)\ge g(x) for x\ge a, look at the difference h=f-g: if h(a)\ge0 and h'\ge0 for x>a, then h never decreases from a value that is at least 0, so h\ge0 and f\ge g. The derivative turns a comparison of two functions into the sign of one derivative.

Example 6. Show that \ln x<x for x>1. Let h(x)=x-\ln x. Then h'(x)=1-\frac1x>0 for x>1, so h is increasing on [1,∞) and h(x)>h(1)=1>0. Hence x>\ln x+1>\ln x. In fact h has its minimum 1 at x=1, so x>\ln x for every x>0.

Example 7. Show that e^x\ge1+x for x\ge0. Let h(x)=e^x-1-x. Then h(0)=0 and h'(x)=e^x-1\ge0 for x\ge0, so h is increasing from 0 and h\ge0. Repeating the idea one step further: with k(x)=e^x-1-x-(x^2)/(2) we have k(0)=0 and k'(x)=e^x-1-x\ge0 by what was just proved, so e^x\ge1+x+(x^2)/(2) for x\ge0. Each inequality is the base for the next.

The proof needs the right starting point: e^x\ge1+x also holds for negative x, but there h is decreasing (it falls to 0 at x=0), so the argument above does not show it; the minimum of h at 0 does. Pick a pair below; the second graph is the difference h, dark where it rises and orange where it falls. The last pair is false, and the widget finds where.

Where else does monotonicity matter: motion, quadratics and inverses?

Wherever a quantity is a function of another, the sign of the derivative says which way it moves.

Motion along a line. For a position s(t) the velocity is v=s': when s is increasing the particle moves right, and when s is decreasing it moves left, as in the motion along a line of the rates article. Example 8. A particle has s(t)=2t^3-14t^2+22t-5 for t\ge0. Then v(t)=6t^2-28t+22=2(t-1)(3t-11), which is zero at t=1 and t=(11)/(3). The sign of v is + for 0\le t<1, - for 1<t<\frac113 and + after: the particle moves right, then left, then right again, and it is momentarily at rest at t=1 (at s=5, a local maximum of the position) and at t=(11)/(3) (at s=-(377)/(27)≈-13.96, a local minimum). The acceleration a=12t-28 changes sign at t=\frac73, where the position changes its bending, a topic of the next article.

Quadratics. For f(x)=ax^2+bx+c with a≠0 we have f'(x)=2ax+b, with one zero at x=-\frac b2a. If a>0, f decreases to the left of that point and increases to the right, so the vertex is a minimum; if a<0 it is the reverse. A single formula covers every parabola.

Inverses. A strictly monotonic function is one-to-one: x_1\ne x_2 forces f(x_1)\ne f(x_2), because one of the two inputs is larger and the outputs are ordered the same way. So a function with f'>0 on an interval, or f'<0, has an inverse there. For f(x)=x^3+x we have f'(x)=3x^2+1>0, so f is increasing on the whole line and has an inverse, although no simple formula for it. Its derivative at the point 2 is still available by the rule for inverses: f(1)=2, so \left(f^-1\right)'(2)=\frac1f'(1)=\frac14.

How do you find where a function rises and falls in code?

In code the method is the same three steps: find the critical points, read the sign of f' at one point of each interval between them, and compare neighbouring signs. SymPy gives the intervals exactly: sp.solve_univariate_inequality(df > 0, x, relational=False) returns the set where f rises, and the same with < 0 where it falls. Without a CAS, take the midpoint of each interval, which is what the JavaScript and C++ programs do; a sign read at one point is enough because f' cannot change sign inside an interval that has no critical point.

import sympy as sp

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


def first_derivative_test(f):
    """Where f rises and falls, and what the First Derivative Test says at each critical point."""
    df = sp.diff(f, x)
    rising = sp.solve_univariate_inequality(df > 0, x, relational=False)
    falling = sp.solve_univariate_inequality(df < 0, x, relational=False)
    eps = sp.Rational(1, 10**6)
    verdict = {}
    for c in sorted(sp.solveset(sp.Eq(df, 0), x, sp.S.Reals), key=float):
        left_down, right_up = falling.contains(c - eps), rising.contains(c + eps)
        left_up, right_down = rising.contains(c - eps), falling.contains(c + eps)
        verdict[c] = 'min' if left_down and right_up else 'max' if left_up and right_down else 'none'
    return rising, falling, verdict


for f in (x**3 - 12 * x - 5, (x**2 - 3) * sp.exp(x), x**4 - 8 * x**2 + 16, x**3, (x - 1)**2 * (x + 2)):
    print(f, '->', first_derivative_test(f))
# x**3 - 12*x - 5 -> (Union(Interval.open(-oo, -2), Interval.open(2, oo)), Interval.open(-2, 2), {-2: 'max', 2: 'min'})
# (x**2 - 3)*exp(x) -> (Union(Interval.open(-oo, -3), Interval.open(1, oo)), Interval.open(-3, 1), {-3: 'max', 1: 'min'})
# x**4 - 8*x**2 + 16 -> (Union(Interval.open(-2, 0), Interval.open(2, oo)), Union(Interval.open(-oo, -2), Interval.open(0, 2)), {-2: 'min', 0: 'max', 2: 'min'})
# x**3 -> (Union(Interval.open(-oo, 0), Interval.open(0, oo)), EmptySet, {0: 'none'})
# (x - 1)**2*(x + 2) -> (Union(Interval.open(-oo, -1), Interval.open(1, oo)), Interval.open(-1, 1), {-1: 'max', 1: 'min'})

// the pieces between the critical points, the sign of f' at the middle of each, and the verdict at every critical point
const signChart = (df, critical, lo, hi) => {
  const edges = [lo, ...critical, hi]
  const pieces = edges.slice(1).map((to, i) => {
    const mid = (edges[i] + to) / 2
    return { from: edges[i], to, slope: df(mid), trend: df(mid) > 0 ? 'rising' : 'falling' }
  })
  const verdicts = critical.map((c, i) => {
    const [l, r] = [pieces[i].trend, pieces[i + 1].trend]
    return [c, l === 'falling' && r === 'rising' ? 'min' : l === 'rising' && r === 'falling' ? 'max' : 'none']
  })
  return { pieces: pieces.map((p) => [p.from, p.to, p.trend]), verdicts }
}
signChart((x) => 3 * x * x - 12, [-2, 2], -5, 5)
// pieces [[-5,-2,'rising'], [-2,2,'falling'], [2,5,'rising']]   verdicts [[-2,'max'], [2,'min']]
signChart((x) => 3 * x * x, [0], -2, 2)
// pieces [[-2,0,'rising'], [0,2,'rising']]   verdicts [[0,'none']]: f' = 0 at 0, but the sign does not change
signChart((x) => 4 * x * (x - 2) * (x + 2), [-2, 0, 2], -3, 3)
// pieces [[-3,-2,'falling'], [-2,0,'rising'], [0,2,'falling'], [2,3,'rising']]   verdicts [[-2,'min'], [0,'max'], [2,'min']]

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

double df(double x) { return 3 * x * x - 12; }       // f(x) = x^3 - 12x - 5, Example 1

int main() {
  double edge[4];
  edge[0] = -5; edge[1] = -2; edge[2] = 2; edge[3] = 5;    // the ends and the critical points
  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] = df(mid) > 0 ? 1 : -1;
    if (sign[i] > 0) printf("from %.1f to %.1f; f' at %.2f is %.2f; rising\n", edge[i], edge[i + 1], mid, df(mid));
    else printf("from %.1f to %.1f; f' at %.2f is %.2f; falling\n", edge[i], edge[i + 1], mid, df(mid));
  }
  for (int i = 0; i < 2; i++) {                             // the First Derivative Test at each critical point
    double c = edge[i + 1];
    if (sign[i] < 0 && sign[i + 1] > 0) printf("x=%.1f local minimum\n", c);
    else if (sign[i] > 0 && sign[i + 1] < 0) printf("x=%.1f local maximum\n", c);
    else printf("x=%.1f no extreme value\n", c);
  }
  printf("f(-2)=%.0f f(2)=%.0f\n", pow(-2.0, 3) - 12 * (-2.0) - 5, pow(2.0, 3) - 12 * 2.0 - 5);
  return 0;
}

/* output:
from -5.0 to -2.0; f' at -3.50 is 24.75; rising
from -2.0 to 2.0; f' at 0.00 is -12.00; falling
from 2.0 to 5.0; f' at 3.50 is 24.75; rising
x=-2.0 local maximum
x=2.0 local minimum
f(-2)=11 f(2)=-21
*/

Where do the signs of the derivative come from?

Telling rising from falling by the sign of a rate is as old as the derivative, but proving that the sign does what it should took until the 19th century.

• Fermat, about 1629. Pierre de Fermat located maxima and minima by comparing f(x) with f(x+e). What he found is the condition f'=0; telling a maximum from a minimum was done by looking at the values on both sides.
• Leibniz, 1684. Gottfried Wilhelm Leibniz published the first rules of the differential calculus in his Nova methodus pro maximis et minimis, in the title of which the search for maxima and minima comes first.
• Cauchy, 1823. Augustin-Louis Cauchy, with the derivative defined as a limit, showed through the mean value theorem that a positive derivative on an interval forces a function to increase. Before that, "the derivative is positive, so the function rises" was taken as obvious from the picture.
• Today. The sign chart is the first tool for graph sketching and optimisation, and a numerical method that stops when the slope changes sign is applying the same test.

What mistakes do people make with the sign of the derivative?

The most common mistakes with monotonic functions and the First Derivative Test 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: monotonic functions and the test at a glance

• Definitions: increasing means f(x_1)<f(x_2) for x_1<x_2 in an interval; monotonic means increasing or decreasing; never "at a point".
• Corollary 3: f'>0 on an interval gives increasing, f'<0 gives decreasing; ends and isolated zeros of f' are allowed.
• Method: critical points, one test point in each open interval between them, the sign of f' there.
• First Derivative Test: -→+ is a local minimum, +→- a local maximum, no change means no extreme value; at an end only one side exists.
• Signs, not sizes: a double factor such as (x-1)^2 does not change the sign, and a point where f' does not exist can still be an extreme value.
• Inequalities: f\ge g from h=f-g with h(a)\ge0 and h'\ge0.
• Uses: the direction of a particle, the vertex of a parabola, and inverses of strictly monotonic functions.
• Next: the second derivative says how the graph bends, the subject of the next article.
• Code: sp.solve_univariate_inequality(df > 0, x, relational=False), or the sign of df at the midpoint of each interval.

Glossary: key terms

Increasing function

A function whose value is larger at a larger input anywhere on an interval: f of x one is less than f of x two whenever x one is less than x two.

Decreasing function

A function whose value is smaller at a larger input anywhere on an interval; its graph falls from left to right.

Monotonic function

A function that is either increasing on an interval or decreasing on it; a strictly monotonic function is one-to-one.

First Derivative Test

The test that finds local extremes from the sign of the derivative on both sides of a critical point: negative then positive is a minimum, positive then negative a maximum, no change means none.

Sign chart

A table of the sign of the derivative on each interval between the critical points, with what the function does on each.

One-to-one function

A function that never takes the same value at two different inputs, so that it has an inverse.

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 4.3: Monotonic Functions and the First Derivative Test. Pearson.
2. James Stewart (2015). Calculus (8th ed.), section 4.3: How Derivatives Affect the Shape of a Graph. Cengage Learning.
3. Gottfried Wilhelm Leibniz (1684). Nova methodus pro maximis et minimis, itemque tangentibus. Acta Eruditorum.
4. Augustin-Louis Cauchy (1823). Résumé des leçons données à l'École royale polytechnique sur le calcul infinitésimal. Paris.

Related articles

• 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 Rate of Change: The derivative as a rate of change: velocity, speed, acceleration and jerk, free fall, marginal cost and revenue, sensitivity to change, and rates from data in code.
• Derivatives of Inverse Functions and Logarithms: Derivatives of inverse functions and logarithms: the rule for inverses, d/dx ln x = 1/x, a^x, log base a, logarithmic differentiation, and the number e.