Derivative Sign and Monotonicity

The sign of the derivative decides where the curve rises and falls. Drag the point and watch the tangent's color track the sign.

Where the slope is positive, the curve climbs. Where it is negative, the curve drops. Reading the sign alone reveals the monotone stretches. The mean value theorem proves this link.

For example, while you climb a 🪜 staircase your height keeps growing. While you go down, it keeps shrinking. The tilt alone tells you whether you rise or fall.

f′≥0⇒f increasingf' \ge 0 \Rightarrow f \text{ increasing}
f′(x0)=lim⁡h→0f(x0+h)−f(x0)hf'(x_0) = \lim_{h\to 0}\frac{f(x_0+h)-f(x_0)}{h}

Examples & Exercises

Let ff be continuous on [a,b][a, b] and differentiable on (a,b)(a, b), with f′≥0f' \ge 0 at every point of (a,b)(a, b).

Prove completely, using the Mean Value Theorem, that ff is increasing on [a,b][a, b].

f(y)−f(x)=f′(c)(y−x)≥0(x<y, c∈(x,y))f(y) - f(x) = f'(c)(y - x) \ge 0 \quad (x < y,\ c \in (x, y))
f(x)=x3−3xf(x) = x^3 - 3x, x0=1x_0 = 1: here f′(1)=0f'(1) = 0 (gray tangent), yet ff is still increasing across the whole interval — f′≥0f' \ge 0 allows equality.
Show proof

  1. Take any two points x<yx < y in [a,b][a, b]. The restriction of ff to [x,y][x, y] is continuous on [x,y][x, y] and differentiable on (x,y)(x, y), so it satisfies the hypotheses of the Mean Value Theorem (the original hypotheses on [a,b][a,b] pass down to every subinterval).

  2. The Mean Value Theorem gives a point c∈(x,y)c \in (x, y) with f(y)−f(x)=f′(c)(y−x)f(y) - f(x) = f'(c)(y - x). Since c∈(a,b)c \in (a, b), the hypothesis yields f′(c)≥0f'(c) \ge 0, and y−x>0y - x > 0, so the product f′(c)(y−x)≥0f'(c)(y - x) \ge 0.

  3. Hence f(y)−f(x)≥0f(y) - f(x) \ge 0, i.e. f(x)≤f(y)f(x) \le f(y).

    Since x<yx < y were arbitrary, ff is increasing on [a,b][a, b].

    Note the non-strict inequality (≤\le) — this is what distinguishes this result from the strict version in the next problem.

Under the same hypotheses, if f′>0f' > 0 at every point of (a,b)(a, b), prove that ff is strictly increasing on [a,b][a, b].

f(y)−f(x)=f′(c)(y−x)>0(x<y, c∈(x,y))f(y) - f(x) = f'(c)(y - x) > 0 \quad (x < y,\ c \in (x, y))
f(x)=exf(x) = e^x, x0=0x_0 = 0: f′(0)=1>0f'(0) = 1 > 0 (green tangent), and since the derivative never vanishes the whole background is green — the archetype of strict increase.
Show proof

  1. Take arbitrary x<yx < y in [a,b][a, b] and, as in problem 1, apply the Mean Value Theorem to ff restricted to [x,y][x, y]: there is c∈(x,y)c \in (x, y) with f(y)−f(x)=f′(c)(y−x)f(y) - f(x) = f'(c)(y - x).

  2. This time the hypothesis is strict: c∈(a,b)c \in (a, b) gives f′(c)>0f'(c) > 0, and y−x>0y - x > 0, so the product f′(c)(y−x)>0f'(c)(y - x) > 0. Hence f(y)−f(x)>0f(y) - f(x) > 0, i.e. f(x)<f(y)f(x) < f(y) — ff is strictly increasing.

  3. The only difference from problem 1 is the strictness of the inequality: f′≥0f' \ge 0 permits flat segments — e.g. f(x)=(max⁡(0,x))2f(x) = (\max(0, x))^2 on [−1,1][-1, 1] is C1C^1 with f′=0f' = 0 on [−1,0][-1, 0] and f′=2x≥0f' = 2x \ge 0 on [0,1][0, 1] (at x=0x = 0 both one-sided difference quotients tend to 00, so f′(0)=0f'(0) = 0); ff is nondecreasing but flat on [−1,0][-1, 0], hence not strictly increasing. By contrast, f′>0f' > 0 forces every secant slope to be positive, so the graph never pauses for an instant.

Prove completely that if ff is differentiable and increasing on an interval II, then f′≥0f' \ge 0 at every point of II (a partial converse).

f′(x0)=lim⁡h→0f(x0+h)−f(x0)h≥0f'(x_0) = \lim_{h \to 0}\frac{f(x_0+h) - f(x_0)}{h} \ge 0
f(x)=x3−3xf(x) = x^3 - 3x, x0=−1.25x_0 = -1.25: f′(−1.25)=1.6875>0f'(-1.25) = 1.6875 > 0 (green tangent) and the function is increasing around here — the derivative cannot be negative on an increasing stretch.
Show proof

  1. Fix x0∈Ix_0 \in I and look at the difference quotient f(x0+h)−f(x0)h\frac{f(x_0+h) - f(x_0)}{h} as h→0h \to 0. For h>0h > 0, the numerator f(x0+h)−f(x0)≥0f(x_0+h) - f(x_0) \ge 0 (since ff is increasing) — and the denominator is positive, so the quotient is ≥0\ge 0.

  2. The same holds for h<0h < 0: monotonicity gives f(x0+h)≤f(x0)f(x_0+h) \le f(x_0), so the numerator is ≤0\le 0, while the denominator h<0h < 0, and a nonpositive divided by a negative is again ≥0\ge 0. So the difference quotient is nonnegative for every nonzero hh.

  3. Limits preserve weak inequalities: if g(h)≥0g(h) \ge 0 and lim⁡h→0g(h)=L\lim_{h \to 0} g(h) = L exists, then L≥0L \ge 0 (if L<0L < 0, then g(h)g(h) would eventually stay near L/2<0L/2 < 0, a contradiction).

    Since ff is differentiable, the limit f′(x0)f'(x_0) exists, so f′(x0)≥0f'(x_0) \ge 0.

    As x0x_0 was arbitrary, f′≥0f' \ge 0 throughout II.

  4. This is not a full converse: f(x)=x3f(x) = x^3 is strictly increasing on R\mathbb{R} yet f′(0)=0f'(0) = 0, so an increasing function need not satisfy f′>0f' > 0. All we get is f′≥0f' \ge 0.

10. Differentiation