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Topos Uranos
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Monotonicity and the first derivative test

Summary

This lesson turns the sign of the derivative into information about a whole interval. With Lagrange’s mean value theorem it proves that a function with a positive derivative on an interval is strictly increasing on it, and with a nonnegative derivative, increasing; conversely, the derivative of an increasing function is not negative, although that of a strictly increasing function may vanish, and the lesson characterizes when this happens. From this follow the invertibility of a function with a positive derivative, the name of the candidates for local extrema, the critical points, and the first derivative test, which decides whether there is an extremum from the sign change of the derivative. Darboux’s theorem, proved with those of Weierstrass and Fermat, shows that a derivative, even when it is not continuous, does not skip values: it has no removable or jump discontinuities, and if it does not vanish on an interval, it does not change sign on it. The lesson ends with inequalities proved by comparing functions through their derivatives, and it fulfills what the unit on derivatives announced several times without being able to prove it.

Learning objectives

  1. Prove with the mean value theorem that the sign of the derivative on an interval decides monotonicity, and recognize when a strictly increasing function has a zero derivative at some points.
  2. Find the critical points of a function and decide with the first derivative test which of them are local extrema, including the points without a derivative.
  3. Prove Darboux’s theorem and use it to show that a function with a jump is not the derivative of any function on an interval.
  4. Prove inequalities between functions by studying the sign of the derivative of their difference, and notice when that argument would be circular.

The sign of the derivative and monotonicity

The proposition on the sign of the derivative at a point, from the lesson on rates of change, compares each value near aa with f(a)f(a), and nothing more: the function x+2x2sin⁡1xx + 2x^2 \sin \frac{1}{x}, extended by 00 at 00, has derivative 11 at 00 and is not increasing on any interval containing 00. To obtain information about a whole interval one must know the sign of the derivative at all its points, and the mean value theorem, from the lesson on the theorems of Fermat, Rolle, Lagrange and Cauchy, translates that knowledge into a comparison of values. Recall that, according to the lesson on function operations, from Algebra and Geometry I, a function ff is increasing on a set II if x<yx < y implies f(x)≤f(y)f(x) \leq f(y) for all its points, and strictly increasing if it implies f(x)<f(y)f(x) < f(y):

(∀x∈I)(∀y∈I)(x<y→f(x)≤f(y)),(∀x∈I)(∀y∈I)(x<y→f(x)<f(y))(\forall x \in I)(\forall y \in I)\left(x < y \rightarrow f(x) \leq f(y)\right), \qquad (\forall x \in I)(\forall y \in I)\left(x < y \rightarrow f(x) < f(y)\right)

and decreasing or strictly decreasing if it implies f(x)≥f(y)f(x) \geq f(y) or f(x)>f(y)f(x) > f(y).

TheoremMonotonicity from the sign of the derivative

Let ff be continuous on an interval II and differentiable at each interior point of II.

  • If f′(x)≥0f'(x) \geq 0 at each interior point xx of II, then ff is increasing on II; if f′(x)>0f'(x) > 0 at each of them, it is strictly increasing on II.
  • If f′(x)≤0f'(x) \leq 0 at each interior point xx of II, then ff is decreasing on II; if f′(x)<0f'(x) < 0 at each of them, it is strictly decreasing on II.

Between any two points, the mean value theorem gives the difference of the values the sign of the derivative at an intermediate point.

Proof

  1. f(y)−f(x)=f′(c)(y−x),x<c<yf(y) - f(x) = \dato{c}{f'(c)}(y - x), \qquad x < c < y

    Let x<yx < y be two points of II. Since II is an interval, it contains [x,y]\icc{x}{y}, where ff is continuous, and the points of (x,y)\ioo{x}{y} are interior points of II, where ff is differentiable. By Lagrange’s mean value theorem, there exists cc between xx and yy such that:

  2. f′(c)≥0→f(x)≤f(y),f′(c)>0→f(x)<f(y)f'(c) \geq 0 \rightarrow \resaltar{f(x) \leq f(y)}, \qquad f'(c) > 0 \rightarrow \resaltar{f(x) < f(y)}

    The factor y−xy - x is positive, so the difference has the sign of f′(c)f'(c): if f′(c)≥0f'(c) \geq 0, it is not negative, and if f′(c)>0f'(c) > 0, it is positive. Since xx and yy were any two points of II with x<yx < y, the function is increasing, or strictly increasing.

  3. −f(x)≤−f(y)→f(x)≥f(y)-f(x) \leq -f(y) \rightarrow \resaltar{f(x) \geq f(y)}

    If the derivative is not positive, what precedes is applied to −f-f, which is continuous on II and has derivative −f′(x)-f'(x) at each interior point, by the theorem on the derivative of a linear combination, from the lesson on differentiation rules. That −f-f is increasing means that ff is decreasing.

The derivative is required only at the interior points; at the endpoints that belong to the interval, continuity suffices. Thus x\sqrt{x} is strictly increasing on [0,∞)\ico{0}{\infty}, although it is not differentiable at 00. The theorem fulfills three announcements of the unit on derivatives. The lesson on the derivatives of the elementary functions observed that the derivative of the tangent, 1+tan⁡2x1 + \tan^2 x, is positive, although the tangent is not increasing on its domain, which is not an interval; on each interval (−π2+kπ,π2+kπ)\ioo{-\frac{\pi}{2} + k\pi}{\frac{\pi}{2} + k\pi}, with kk an integer, the theorem makes it strictly increasing. The lesson on the variable in the exponent computed the derivative x1x⋅1−ln⁡xx2x^{\frac{1}{x}} \cdot \frac{1 - \ln x}{x^2}, which is positive on (0,e)\ioo{0}{e} and negative on (e,∞)\ioo{e}{\infty}; since the function is continuous on (0,∞)\ioo{0}{\infty}, it is strictly increasing on (0,e]\ioc{0}{e} and strictly decreasing on [e,∞)\ico{e}{\infty}. This recovers the proposition on the maximum of xx to the power 1x\frac{1}{x}, from that lesson, and moreover compares two numbers on the same side of ee: since e<3<4e < 3 < 4, we have 313>4143^{\frac{1}{3}} > 4^{\frac{1}{4}}, that is, 34>433^4 > 4^3. The same lesson proved that the derivative of (1+1x)x\left(1 + \frac{1}{x}\right)^x is positive at every x>0x > 0: the function is strictly increasing on (0,∞)\ioo{0}{\infty}.

PropositionThe sign of the derivative of a monotonic function

Let ff be increasing on an interval II and differentiable at an interior point aa of II. Then f′(a)≥0f'(a) \geq 0. If ff is decreasing, f′(a)≤0f'(a) \leq 0. If the derivative were negative, the proposition on the sign of the derivative at a point would place to the right of aa values smaller than f(a)f(a).

Proof

  1. x∈J, a<x<a+δ→f(x)<f(a)x \in J, \ a < x < a + \delta \rightarrow \dato{menor}{f(x) < f(a)}

    Let r>0r > 0 be such that J=(a−r,a+r)J = \ioo{a - r}{a + r} is contained in II, and let FF be the restriction of ff to JJ; by the lemma on the local character of the derivative, from the lesson on differentiation rules, FF is differentiable at aa and F′(a)=f′(a)F'(a) = f'(a). Suppose that f′(a)<0f'(a) < 0. By the proposition on the sign of the derivative at a point, applied to FF on the open interval JJ, there exists δ>0\delta > 0 such that:

  2. f(a)≤f(a+ρ2)<f(a)→f′(a)≥0f(a) \leq f\left(a + \frac{\rho}{2}\right) < f(a) \qquad \rightarrow \qquad \resaltar{f'(a) \geq 0}

    Let ρ\rho be the smaller of δ\delta and rr. The point a+ρ2a + \frac{\rho}{2} lies in JJ and is greater than aa; since ff is increasing, its value is not smaller than f(a)f(a), contradicting the first move. Therefore f′(a)≥0f'(a) \geq 0. If ff is decreasing, −f-f is increasing, and −f′(a)≥0-f'(a) \geq 0.

For a function differentiable at all the points of an open interval, being increasing is therefore equivalent to having a nonnegative derivative. The converse does not extend to strict monotonicity: x3x^3 is strictly increasing on R\mathbb{R}, and its derivative vanishes at 00. The proposition also explains the example at the beginning: the lesson on rates of change found, arbitrarily close to 00, points at which the derivative of x+2x2sin⁡1xx + 2x^2 \sin \frac{1}{x} is negative, and on no interval containing them can the function be increasing. The defect lies in the fact that this derivative is not continuous at 00; if it is, the sign at a point extends to an interval.

CorollaryPositive derivative continuous at a point

Let ff be differentiable at each point of an open interval II, and let aa be a point of II at which f′f' is continuous, with f′(a)>0f'(a) > 0. Then there exists a δ>0\delta > 0 such that ff is strictly increasing on (a−δ,a+δ)\ioo{a - \delta}{a + \delta}. The derivative keeps its sign near aa, and the theorem applies on the interval where it keeps it.

Proof

  1. ∣x−a∣<δ→f′(x)>0\abs{x - a} < \delta \rightarrow \resaltar{f'(x) > 0}

    Since f′f' is continuous at aa, its limit at aa is f′(a)>0f'(a) > 0. By the proposition on local boundedness and preservation of sign, from the lesson on the limit of a function, there exists a δ>0\delta > 0, which can be taken so that (a−δ,a+δ)\ioo{a - \delta}{a + \delta} is contained in II, with f′(x)>0f'(x) > 0 on that interval. There ff is differentiable, and therefore continuous; by the theorem, it is strictly increasing.

When is a function with a nonnegative derivative strictly increasing? The example of x3x^3 shows that the derivative may vanish at some point; what it cannot do is vanish on a whole interval.

PropositionStrict monotonicity with a nonnegative derivative

Let ff be continuous on an interval II and differentiable at each interior point of II, with f′(x)≥0f'(x) \geq 0 at each of them. Then ff is strictly increasing on II if and only if there is no interval (p,q)\ioo{p}{q}, with p<qp < q, contained in II, at all of whose points f′f' vanishes. In particular, if f′f' vanishes only at finitely many points, ff is strictly increasing on II. An increasing function that is not strictly increasing repeats a value, and between the two points where it repeats it, it is constant.

Proof

  1. f′(x)=0,s<x<t→f(s)=f(t)f'(x) = 0, \quad s < x < t \qquad \rightarrow \qquad \resaltar{f(s) = f(t)}

    Suppose that f′f' vanishes at all the points of an interval (p,q)\ioo{p}{q} contained in II, and let p<s<t<qp < s < t < q. On [s,t]\icc{s}{t}, the function is continuous and has a zero derivative at the points of (s,t)\ioo{s}{t}; by the corollary on the zero derivative on an interval, from the lesson on the theorems of Fermat, Rolle, Lagrange and Cauchy, it is constant on [s,t]\icc{s}{t}, and it is not strictly increasing.

  2. s<x<t→f(s)≤f(x)≤f(t)=f(s)→f(x)=f(s)s < x < t \rightarrow f(s) \leq f(x) \leq f(t) = f(s) \rightarrow \dato{cte}{f(x) = f(s)}

    Conversely, suppose that ff is not strictly increasing. By the theorem, it is increasing, so there exist s<ts < t in II with f(s)=f(t)f(s) = f(t); every intermediate value is enclosed between two equal values, and ff is constant on [s,t]\icc{s}{t}.

  3. s<x<t→f′(x)=0s < x < t \rightarrow \resaltar{f'(x) = 0}

    Let xx be in (s,t)\ioo{s}{t}. The restriction of ff to (s,t)\ioo{s}{t} is constant, with a zero derivative at xx; by the lemma on the local character of the derivative, f′(x)=0f'(x) = 0. The derivative vanishes on the whole interval (s,t)\ioo{s}{t}, contained in II. If f′f' vanishes only at finitely many points, there is no such interval, because every interval (p,q)\ioo{p}{q} with p<qp < q has infinitely many points.

The function x+sin⁡xx + \sin x, from solved problem 1 of the lesson on the derivative of the inverse function, has derivative 1+cos⁡x≥01 + \cos x \geq 0, which vanishes at the infinitely many points π+2kπ\pi + 2k\pi; between two consecutive ones the derivative is positive, and no interval consists of zeros. The proposition makes it strictly increasing on R\mathbb{R}, which that lesson obtained with the formula for the difference of two sines and which seemed surprising there. The same reasoning admits finitely many points without a derivative: if ff is continuous on II and has a positive derivative at all its interior points except at x1<⋯<xkx_1 < \dots < x_k, the theorem makes it strictly increasing on each interval between two consecutive such points (and on the end intervals), and the inequalities combine. Thus x3\sqrt[3]{x} is strictly increasing on R\mathbb{R}, although it is not differentiable at 00. Continuity cannot be dropped (proposed problem 1).

CorollaryPositive derivative and inverse function

Let ff be continuous on an interval II with more than one point and differentiable at each interior point of II, with f′(x)>0f'(x) > 0 at each of them, or with f′(x)<0f'(x) < 0 at each of them. Then ff is strictly monotonic on II, its image J=f(I)J = f(I) is an interval and the inverse f−1 ⁣:J→If^{-1} \colon J \to I is continuous; at each point b=f(a)b = f(a), with aa interior to II, the inverse is differentiable, and (f−1)′(b)=1f′(a)\left(f^{-1}\right)'(b) = \frac{1}{f'(a)}. The theorem gives strict monotonicity, which is the hypothesis of the theorem on the derivative of the inverse function.

Proof

  1. x<y→f(x)<f(y),J=f(I)x < y \rightarrow \resaltar{f(x) < f(y)}, \qquad J = f(I)

    By the theorem, ff is strictly increasing, or strictly decreasing, on II. By the corollaries on the image of an interval and on the continuity of the inverse, from the lesson on the intermediate value theorem, JJ is an interval and f−1f^{-1} is continuous on JJ.

  2. (f−1)′(b)=1f′(a)\left(f^{-1}\right)'(b) = \resaltar{\frac{1}{f'(a)}}

    If aa is interior, ff is differentiable at aa with f′(a)≠0f'(a) \neq 0, and the theorem on the derivative of the inverse function, from the lesson on the derivative of the inverse function, gives the derivative of f−1f^{-1} at b=f(a)b = f(a).

The assessments of the unit on derivatives took for granted that x5+xx^5 + x and x3+2x+1x^3 + 2x + 1 are strictly increasing on R\mathbb{R}; now it suffices to observe that their derivatives, 5x4+15x^4 + 1 and 3x2+23x^2 + 2, are positive. And the challenge on the inverse of xexx e^x, in the synthesis of that unit, which asked to prove without the sign of the derivative that this function is strictly increasing on [−1,∞)\ico{-1}{\infty}, is now settled in one line: it is continuous there and its derivative, (1+x)ex(1 + x) e^x, is positive on (−1,∞)\ioo{-1}{\infty}.