Derivative graphs and increasing or decreasing functions

For f(x) = x³ − 3x² − 9x + 5:

f′(x) = 3x² − 6x − 9
= 3(x + 1)(x − 3)

The derivative is positive for x < −1 and x > 3, and negative for −1 < x < 3. Thus f increases on (−∞,−1] and [3,∞), and decreases on [−1,3].

The strict inequalities describe where f′ has a strict sign. The endpoints can be included when describing these monotonic intervals: the derivative may be zero there without spoiling the increase or decrease across the interval.

The function x³ is strictly increasing across all real inputs, although its derivative 3x² is zero at x = 0. A single flat point does not create a decreasing interval or a constant stretch.

On an interval, f′ ≥ 0 guarantees a differentiable function is non-decreasing. If f′ is positive apart from isolated zeros, it is strictly increasing. A whole interval with f′ = 0 is a constant part.

For f(x) = 1/x, f′(x) = −1/x² < 0 whenever x ≠ 0. The function is decreasing on each interval (−∞,0) and (0,∞).

Do not merge them into one interval across zero. For example −1 < 1 but f(−1) = −1 < f(1) = 1. This pair does not fit a globally decreasing claim on the disconnected domain.

Mark excluded inputs and roots of the derivative separately on your sign chart.

1. Keep the same x-scale.
2. Mark zeros of f′ at differentiable stationary points of f.
3. Place f′ above the axis where f rises and below it where f falls.
4. Use steeper parts of f to suggest larger |f′|.
5. Check how the slopes change as you move along the curve.

At a usual smooth local maximum, f′ changes from positive to negative; at a local minimum, from negative to positive. At a stationary inflection like x³ at zero, f′ touches zero and keeps its sign.

An x-intercept of f does not automatically give an x-intercept of f′. Nor does every inflection lie on the x-axis of f′: x³ + x has an inflection at zero but derivative 3x² + 1 is 1 there.

Let f(x) = (x + 2)(x − 1)². Its zeros are −2 and 1, with a repeated root at 1. Expanding and differentiating gives:

f(x) = x³ − 3x + 2
f′(x) = 3x² − 3 = 3(x − 1)(x + 1)

The gradient graph is an upward parabola crossing the x-axis at (−1,0),(1,0) and the y-axis at (0,−3). The original curve has stationary points (−1,4) and (1,0). Its other zero, x = −2, is not a zero of the derivative.

When only a qualitative sketch of f is supplied, you may know the derivative’s zeros and signs without knowing its exact y-intercept or maximum height. Do not invent exact coordinates from a rough drawing.

These correspond to concave-up and concave-down parts of a twice-differentiable curve. They do not tell you whether f itself is increasing: y = x² has f″ = 2 > 0 even on its decreasing left-hand half.

For f(x) = x³ − 3x² − 9x + 5, f″(x) = 6x − 6. Its concavity changes at x = 1, where f′(1) = −12 rather than zero. This is a non-stationary inflection.

For f(x) = 2 + 3/(x − 1), the derivative is −3/(x − 1)². Both have a vertical asymptote x = 1. As |x| grows, f approaches y = 2 while f′ approaches y = 0 from below.

This follows from the formulas and their limits. Familiar smooth curves flatten as they approach a horizontal asymptote, but “differentiate the asymptote” is not a universal theorem for every possible function. For a sketch question, read the depicted local slopes and state only the behaviour supported by the diagram.