Derivative Sign and Monotonicity

The sign of f′f' decides where the curve rises and falls: up on the green background (f′>0f' > 0), down on the red one (f′<0f' < 0). Drag x0x_0 and watch the tangent's color track the sign — one step behind the Mean Value Theorem.

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 spots (like f(x)=x3−3xf(x) = x^3 - 3x near x=1x = 1), while 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.

9. Differentiation