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Calculus 1

Increasing and decreasing from the derivative

The derivative gives the slope at a single point. How can it tell us, over a whole stretch of inputs, that a function is going up or going down?

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Critical points and local versus absolute extremaCalculus 1
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Approximating area with Riemann sumsCalculus 1Position, velocity and accelerationCalculus 1The first and second derivative testsCalculus 1
Where you are 4 in Maxima, minima and critical points
Calculus8
Applications of derivatives11
Maxima, minima and critical points4
Critical points and local versus absolute extremaExtrema on a closed interval
Increasing and decreasing from the derivative
The first and second derivative tests

When you reach for it

A graph read by eye shows where a function rises and falls (Increasing, decreasing, constant, positive and negative intervals did this from graphs and tables), but an eye cannot prove that a curve never turns back between two plotted points. The derivative can. If f'(x) > 0 at every point of an interval, f is increasing on that whole interval, with no exceptions hiding between sample points.

Reach for this whenever the question is “where does f go up, and where does it go down?” for a function given by a formula or by the graph of its derivative. The answer is a list of intervals, and the tool for finding them is a sign chart: the line cut at a few special inputs, with the sign of f' recorded on each piece.

The procedure

Day 1: the test and the sign chart

Words. A function f is increasing on an interval I if x_1 < x_2 in I always gives f(x_1) < f(x_2), and decreasing on I if x_1 < x_2 in I always gives f(x_1) > f(x_2). Both are statements about an interval, never about a single point.

The test. Let f be continuous on an interval I.

  • If f'(x) > 0 at every point inside I (every point but the ends), then f is increasing on I.
  • If f'(x) < 0 at every point inside I, then f is decreasing on I.

This was proved in The mean value theorem from the mean value theorem. In one line: for x_1 < x_2 in I there is a c between them with f(x_2) - f(x_1) = f'(c)(x_2 - x_1), and the sign of the right side is the sign of f'(c). The proof needs f'(c) only at points strictly between x_1 and x_2, which is why nothing is asked of f' at the ends of I.

Where to split. The sign of f' can change only at an input where f' is 0, where f' does not exist, or where f itself is undefined. For every function in this lesson f' is continuous wherever it exists, so by the intermediate value theorem it cannot go from positive to negative on an interval without passing through 0. (A derivative keeps this property even when it is not continuous, a result called Darboux’s theorem, stated here without proof.) So we mark on the number line

These cut the domain into open pieces on which f' exists and is never 0, so f' keeps one sign on each piece. One test value in a piece tells the sign for the whole piece. A sign chart records this: the marked inputs on a line, a 0 above each zero of f', and a + or - above each piece (Figure 1).

Closed or open ends. The test asks for f to be continuous on I, including its ends. So if f is defined and continuous at a split point, that point may be included at the end of an interval: f is increasing on [3, \infty), not just on (3, \infty). Both statements are true; the closed one says a little more, and a point such as 3 can sit at the end of a decreasing interval and the start of an increasing one at once. An input where f is undefined is never included.

Day 2: more kinds of function

Inputs outside the domain split the line too. A rational function can change sign at a vertical asymptote (Infinite limits and vertical asymptotes) without f' ever being 0 there. Such an input goes on the chart with an open circle. The intervals on its two sides stay separate, even when f falls on both, because no interval in the answer may contain an input outside the domain.

Factors that never change sign. In e^x(x - 2) the factor e^x is positive for every x (The number e and the natural exponential function), so the sign of the product is the sign of x - 2. The same goes for x^2 + 4 and for any square that is not 0.

Joining intervals. If f is increasing on [a, c] and on [c, b], then it is increasing on [a, b]. Indeed, for x_1 < c < x_2 we have f(x_1) < f(c) < f(x_2), and the other cases are inside one piece. So when f' has the same sign on both sides of a split point c where f is continuous, the two intervals merge.

A zero of f' need not change its sign. If f'(x) = (x - c)^2 g(x) with g continuous and g(c) \ne 0, the square is positive on both sides of c, so f' has the sign of g on both sides: the sign does not change at c. A factor (x - c)^m flips sign at c when m is odd and keeps it when m is even. This is why the sign chart tests a value in each piece rather than assuming the signs alternate.

Reading a graph of f'. When the graph of f' is drawn, the sign of f' is read off directly: above the axis means f' > 0 and f rises; below means f falls. Where the graph of f' crosses the axis is where f can change direction. Whether the graph of f' itself goes up or down says how the slope of f is changing, not whether f rises.

Which of the turning points are peaks and which are valleys is the next lesson, The first and second derivative tests; this one stops at the intervals.

Worked examples

Example 1

Day 1: Bishal’s cubic

Bishal wants the intervals on which f(x) = 2x^3 - 3x^2 - 36x + 5 is increasing and decreasing.

Solution

A polynomial is defined and differentiable everywhere (The power rule and basic differentiation rules), so the only split points are the zeros of f':

\begin{aligned} f'(x) &= 6x^2 - 6x - 36 \\ &= 6(x^2 - x - 6) = 6(x + 2)(x - 3). \end{aligned}

The zeros -2 and 3 cut the line into (-\infty, -2), (-2, 3) and (3, \infty). On each piece f' is continuous and never 0, so one test value gives the sign:

PieceTest valuef'(x)Sign
(-\infty, -2)-36(-1)(-6) = 36+
(-2, 3)06(2)(-3) = -36-
(3, \infty)46(6)(1) = 36+
−2 0 3 0 + − + f′ f
Figure 1The sign chart for f(x) = 2x^3 - 3x^2 - 36x + 5 in Example 1. The critical points -2 and 3 split the line; one test value in each piece fixes the sign of f' there, and the sign says whether f rises or falls.

f is continuous at -2 and at 3, so those points may close the intervals. By the test, f is increasing on (-\infty, -2] and on [3, \infty), and decreasing on [-2, 3]. As a check, f(-2) = 49 and f(3) = -76: the value dropped across the decreasing interval.

Example 2

Day 1: Sushila’s cusp

Sushila studies g(x) = x^{2/3}(x + 5). Find where g increases and decreases.

Solution

The domain is all real numbers. Distribute to get g(x) = x^{5/3} + 5x^{2/3}, and for x \ne 0 use the power rule:

\begin{aligned} g'(x) &= \frac{5}{3}x^{2/3} + \frac{10}{3}x^{-1/3} \\ &= \frac{5(x + 2)}{3x^{1/3}}. \end{aligned}

The second form puts both terms over 3x^{1/3}, using x^{2/3} \cdot x^{1/3} = x. At x = 0 the difference quotient \dfrac{g(h) - g(0)}{h} = \dfrac{h + 5}{h^{1/3}} grows without bound, so g'(0) does not exist. The split points are the critical points -2 (where g' = 0) and 0 (where g' does not exist).

Signs: at x = -3, the numerator 5(-1) and the denominator 3(-3)^{1/3} are both negative, so g' > 0. At x = -1, \displaystyle g'(-1) = \frac{5}{-3} < 0. At x = 1, g'(1) = 5 > 0.

g is continuous everywhere, including at 0, where only the derivative fails. So g is increasing on (-\infty, -2], decreasing on [-2, 0] and increasing on [0, \infty). The point 0 closes two intervals even though g'(0) does not exist: the test asks for continuity at the ends, not a derivative.

Example 3

Day 1: Prakriti’s square root

Prakriti asks where \displaystyle k(x) = x\sqrt{6 - x} rises and falls.

Solution

The domain is x \le 6, so 6 is an end of the domain and goes on the chart. For x < 6, the product rule and the chain rule (The chain rule), with inner derivative -1, give

\begin{aligned} k'(x) &= \sqrt{6 - x} - \frac{x}{2\sqrt{6 - x}} \\ &= \frac{2(6 - x) - x}{2\sqrt{6 - x}} = \frac{3(4 - x)}{2\sqrt{6 - x}}. \end{aligned}

The denominator is positive for x < 6, so the sign of k' is the sign of 4 - x. The only critical point is 4, and 6 is an end of the domain, not a critical point. Pieces: on (-\infty, 4) the test value 0 gives \displaystyle k'(0) = \frac{12}{2\sqrt{6}} > 0; on (4, 6) the test value 5 gives \displaystyle k'(5) = -\frac{3}{2} < 0.

k is continuous on (-\infty, 6], including from the left at 6. So k is increasing on (-\infty, 4] and decreasing on [4, 6]. The end 6 is included, because k(6) = 0 is defined and the test only needs k' inside the interval. Check: \displaystyle k(4) = 4\sqrt{2}, k(5) = 5, k(6) = 0, falling as promised.

Example 4

Day 2: a vertical asymptote

Find the intervals on which r(x) = \dfrac{x^2 + 3}{x + 1} is increasing and decreasing.

Solution

The domain is every x \ne -1. By the quotient rule,

\begin{aligned} r'(x) &= \frac{2x(x + 1) - (x^2 + 3)}{(x + 1)^2} \\ &= \frac{x^2 + 2x - 3}{(x + 1)^2} = \frac{(x + 3)(x - 1)}{(x + 1)^2}. \end{aligned}

The critical points are -3 and 1. The input -1 is not in the domain, so it is not critical, but it still splits the line. The four pieces and test values:

PieceTest valuer'(x)Sign
(-\infty, -3)-4\displaystyle \frac{(-1)(-5)}{9} = \frac{5}{9}+
(-3, -1)-2\displaystyle \frac{(1)(-3)}{1} = -3-
(-1, 1)0\displaystyle \frac{(3)(-1)}{1} = -3-
(1, \infty)2\displaystyle \frac{(5)(1)}{9} = \frac{5}{9}+
−3 0 −1 1 0 + − − + f′ f
Figure 2The sign chart for r(x) = \dfrac{x^2 + 3}{x + 1} in Example 4. The open circle at -1 marks an input outside the domain. It splits the line just as a critical point does, but it never belongs to an interval in the answer.

So r is increasing on (-\infty, -3] and [1, \infty), and decreasing on [-3, -1) and on (-1, 1]. The two decreasing intervals cannot be joined. Joining them would claim, for instance, that -2 < 0 forces r(-2) > r(0), but r(-2) = -7 and r(0) = 3. Across the asymptote the graph jumps from far below to far above.

Example 5

Day 2: Sushila’s exponential product

Sushila asks where f(x) = x^3 e^{-x} increases and decreases.

Solution

By the product rule, with \dfrac{d}{dx}\big[e^{-x}\big] = -e^{-x} from the chain rule,

\begin{aligned} f'(x) &= 3x^2 e^{-x} - x^3 e^{-x} \\ &= x^2(3 - x)e^{-x}. \end{aligned}

The factor e^{-x} is positive for every x, so it never affects the sign. f' exists everywhere and is 0 at x = 0 and x = 3. Test values: f'(-1) = 4e > 0, \displaystyle f'(1) = \frac{2}{e} > 0, f'(4) = -16e^{-4} < 0.

So f is increasing on (-\infty, 0] and on [0, 3], and these two join (they share the point 0, where f is continuous): f is increasing on (-\infty, 3] and decreasing on [3, \infty).

The zero at 0 comes from the factor x^2, which is positive on both sides of 0. The slope is momentarily 0 there, yet f never stops rising: f(x) < 0 = f(0) for x < 0 and f(x) > 0 for x > 0. A zero of f' is a split point, not a promise that the sign changes.

Example 6

Day 2: Prakriti’s tangent on an interval

Prakriti considers f(x) = \tan x - 4x on \displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2}\right). Where is it increasing and decreasing?

Solution

On this interval \tan x is defined and \dfrac{d}{dx}\big[\tan x\big] = \sec^2 x (Derivatives of the trigonometric functions), so

f'(x) = \sec^2 x - 4.

This is 0 when \sec^2 x = 4, that is, \displaystyle \cos^2 x = \frac{1}{4}. On \displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) cosine is positive, so \displaystyle \cos x = \frac{1}{2} and \displaystyle x = \pm\frac{\pi}{3}.

On \displaystyle \left(-\frac{\pi}{3}, \frac{\pi}{3}\right) the test value 0 gives f'(0) = 1 - 4 = -3 < 0. For \displaystyle \frac{\pi}{3} < \lvert x \rvert < \frac{\pi}{2}, \displaystyle 0 < \cos x < \frac{1}{2}, so \sec^2 x > 4 and f' > 0.

So f is increasing on \displaystyle \left(-\frac{\pi}{2}, -\frac{\pi}{3}\right] and on \displaystyle \left[\frac{\pi}{3}, \frac{\pi}{2}\right), and decreasing on \displaystyle \left[-\frac{\pi}{3}, \frac{\pi}{3}\right]. The outer ends stay open because \tan x is undefined at \displaystyle \pm\frac{\pi}{2}. For a sense of size, \displaystyle f\left(\frac{\pi}{3}\right) = \sqrt{3} - \frac{4\pi}{3} \approx -2.46.

Example 7

Day 2: reading a graph of the derivative

Figure 3 shows the graph of the derivative f' of a function f that is continuous on [-3, 5]; the graph meets the axis at -2, 1 and 4. (a) Where is f increasing and decreasing? (b) Which is larger, f(0) or f(3)? (c) Bishal notices that the graph of f' climbs from its low point near x = -1 up to x = 1. Does f rise there?

x y −3 −2 4 5 1 2 4 y = f′(x)
Figure 3The graph of y = f'(x) in Example 7. Where this graph is above the axis, f rises; where it is below, f falls. The height of the graph matters, not whether the graph is going up or down.
Solution

(a) Read the sign of f' from the height of its graph. It is above the axis on [-3, -2), below on (-2, 1) and (1, 4), touching the axis at 1 without crossing, and above on (4, 5]. So f is increasing on [-3, -2] and on [4, 5]. On [-2, 1] and [1, 4] it is decreasing, and these join, since they share the point 1: f is decreasing on [-2, 4]. The touch at 1 is a zero of f' with no sign change, so f does not change direction there.

(b) 0 and 3 both lie in [-2, 4], where f is decreasing, and 0 < 3. So f(0) > f(3).

(c) No. On (-1, 1) the graph of f' is rising but still below the axis, so f' < 0 there and f is falling. The climb only says that f falls less and less steeply.

Boundaries

  • The test needs an interval. f(x) = -\dfrac{1}{x} has f'(x) = \dfrac{1}{x^2} > 0 at every point of its domain, but f(-1) = 1 > f(1) = -1. It is increasing on (-\infty, 0) and on (0, \infty) separately, never on their union, because the mean value theorem cannot be applied across 0.
  • It needs continuity on I. A function that jumps can have f' > 0 everywhere it is differentiable and still drop. An end may be closed only where f is defined and continuous.
  • The test goes one way. An increasing function can have f'(c) = 0 at isolated points (x^3 at 0), and f' > 0 is not needed at every point. What cannot happen is f' < 0 at some point inside an interval where f is increasing, since the difference quotients there are all positive.
  • One test value works only between split points. It relies on f' having no zeros, no undefined points and no domain gaps inside the piece. Skip a split point and the chart can be wrong.
  • A sign chart gives directions, not values. It says where f rises and falls, not how high it gets.

Where this goes next

A change of sign in f' at a critical point is a change of direction, which is how The first and second derivative tests sorts critical points into peaks, valleys and neither. The same sign chart for velocity tells when a moving object goes forward and when it reverses, in Position, velocity and acceleration. Applied to f'' instead of f', the method finds where a graph bends up or down: Concavity and inflection points.

Exercises

Day 1: intervals from a sign chart

For Exercises 1 to 6, find the intervals on which the function is increasing and on which it is decreasing. Show the split points and a test value in each piece.

  1. f(x) = x^3 - 12x + 1
  2. f(x) = -x^3 + 6x^2 + 15x
  3. f(x) = 3x^4 - 4x^3 - 12x^2 + 2
  4. f(x) = x^{1/3}(x - 4)
  5. \displaystyle f(x) = x\sqrt{x + 3}
  6. f(x) = x^3 - 3x^2 - 24x on [-3, 5]

Day 1: reasoning

  1. Using the mean value theorem, prove that if f is continuous on an interval I and f'(x) < 0 at every point inside I, then f is decreasing on I.
  2. Bishal writes that f(x) = x^2 is increasing on [0, \infty); Prakriti writes (0, \infty). Who is right? Can 0 also belong to an interval on which f is decreasing?
  3. Suppose f is continuous on [a, b] and f'(x) > 0 on (a, b). Show that f(a) < f(x) < f(b) for every x with a < x < b.
  4. Show that x^3 + 2x + 7 = 0 has exactly one real solution, and find two consecutive integers it lies between.

Day 2: rational, exponential and trigonometric functions

For Exercises 11 to 20, find the intervals on which the function is increasing and on which it is decreasing.

  1. f(x) = \dfrac{x}{x^2 + 4}
  2. f(x) = \dfrac{x^2}{x - 3}
  3. f(x) = \dfrac{1}{x^2 - 4}
  4. f(x) = xe^{2x}
  5. f(x) = (x^2 - 3)e^{-x}
  6. f(x) = e^{-x^2/2}
  7. f(x) = \sin x + \cos x on [0, 2\pi]
  8. f(x) = \cos 2x + 2\sin x on [0, \pi]
  9. f(x) = \sin^2 x on [0, 2\pi]
  10. f(x) = x^3 + 3x^2 + 3x - 7
  11. A function g has g'(x) = (x + 1)^2(x - 3)(x - 5)^3. On which intervals is g increasing, and on which decreasing?

Day 2: reasoning

  1. Use g(x) = x - \sin x to prove that \sin x < x for every x > 0.
  2. Sushila looks at Figure 3 and says f is decreasing on [-3, -2] because the graph there goes down. Explain what is wrong.
  3. Find every number k for which f(x) = x^3 - kx^2 + 3x is increasing on (-\infty, \infty).
  4. Prakriti claims: “if f'(x) > 0 at every point of the domain of f, then f is increasing on its domain.” Give a counterexample, and state a correct version.
Answers
  1. f'(x) = 3x^2 - 12 = 3(x + 2)(x - 2). Signs +, -, + (test -3, 0, 3). Increasing on (-\infty, -2] and [2, \infty), decreasing on [-2, 2].
  2. f'(x) = -3x^2 + 12x + 15, which is -3(x + 1)(x - 5). Signs -, +, - (test -2, 0, 6). Decreasing on (-\infty, -1] and [5, \infty), increasing on [-1, 5].
  3. f'(x) = 12x^3 - 12x^2 - 24x, which is 12x(x + 1)(x - 2). Signs -, +, -, + (test -2, \displaystyle -\frac{1}{2}, 1, 3). Decreasing on (-\infty, -1] and [0, 2], increasing on [-1, 0] and [2, \infty).
  4. f(x) = x^{4/3} - 4x^{1/3}, so for x \ne 0, f'(x) = \dfrac{4(x - 1)}{3x^{2/3}}; f'(0) does not exist. Since x^{2/3} > 0 for x \ne 0, the signs are -, -, + on (-\infty, 0), (0, 1), (1, \infty). f is continuous at 0, so the first two join: decreasing on (-\infty, 1], increasing on [1, \infty). The critical point 0 does not change the sign.
  5. Domain [-3, \infty). For x > -3, \displaystyle f'(x) = \sqrt{x + 3} + \dfrac{x}{2\sqrt{x + 3}}, which is \displaystyle \dfrac{3(x + 2)}{2\sqrt{x + 3}}. Signs -, + (test -2.5, 0). Decreasing on [-3, -2], increasing on [-2, \infty).
  6. f'(x) = 3x^2 - 6x - 24, which is 3(x + 2)(x - 4). Signs +, -, + on the pieces of [-3, 5]. Increasing on [-3, -2] and [4, 5], decreasing on [-2, 4].
  7. Take x_1 < x_2 in I. f is continuous on [x_1, x_2] and differentiable inside it, so the mean value theorem gives c between them with f(x_2) - f(x_1) = f'(c)(x_2 - x_1). Here f'(c) < 0 and x_2 - x_1 > 0, so f(x_2) < f(x_1).
  8. Both. f'(x) = 2x > 0 on (0, \infty) and f is continuous on [0, \infty), so the test gives increasing on [0, \infty), which contains the open statement. Yes: f' < 0 on (-\infty, 0), so f is decreasing on (-\infty, 0]; increasing and decreasing describe intervals, and 0 is an end of each.
  9. By the test, f is increasing on [a, b]. Then a < x gives f(a) < f(x), and x < b gives f(x) < f(b).
  10. f(x) = x^3 + 2x + 7 has f'(x) = 3x^2 + 2 \ge 2 > 0, so f is increasing on (-\infty, \infty) and takes each value at most once. f(-2) = -5 < 0 and f(-1) = 4 > 0, and f is continuous, so by the intermediate value theorem the one solution lies between -2 and -1.
  11. f'(x) = \dfrac{4 - x^2}{(x^2 + 4)^2}, signs -, +, -. Decreasing on (-\infty, -2] and [2, \infty), increasing on [-2, 2].
  12. f'(x) = \dfrac{x(x - 6)}{(x - 3)^2}, split at 0, 3, 6, signs +, -, -, +. Increasing on (-\infty, 0] and [6, \infty), decreasing on [0, 3) and (3, 6].
  13. f'(x) = \dfrac{-2x}{(x^2 - 4)^2}, split at -2, 0, 2, signs +, +, -, -. Increasing on (-\infty, -2) and (-2, 0], decreasing on [0, 2) and (2, \infty); the pieces do not join across \pm 2.
  14. f'(x) = e^{2x}(1 + 2x). Decreasing on \displaystyle \left(-\infty, -\frac{1}{2}\right], increasing on \displaystyle \left[-\frac{1}{2}, \infty\right).
  15. f'(x) = 2xe^{-x} - (x^2 - 3)e^{-x}, which is -(x + 1)(x - 3)e^{-x}, signs -, +, -. Decreasing on (-\infty, -1] and [3, \infty), increasing on [-1, 3].
  16. f'(x) = -xe^{-x^2/2}, by the chain rule. Increasing on (-\infty, 0], decreasing on [0, \infty).
  17. f'(x) = \cos x - \sin x, zero at \displaystyle \frac{\pi}{4} and \displaystyle \frac{5\pi}{4}. Test 0, \pi, \displaystyle \frac{3\pi}{2}: 1, -1, 1. Increasing on \displaystyle \left[0, \frac{\pi}{4}\right] and \displaystyle \left[\frac{5\pi}{4}, 2\pi\right], decreasing on \displaystyle \left[\frac{\pi}{4}, \frac{5\pi}{4}\right].
  18. f'(x) = -2\sin 2x + 2\cos x, which is 2\cos x\,(1 - 2\sin x), zero at \displaystyle \frac{\pi}{6}, \displaystyle \frac{\pi}{2}, \displaystyle \frac{5\pi}{6}. Signs +, -, +, -. Increasing on \displaystyle \left[0, \frac{\pi}{6}\right] and \displaystyle \left[\frac{\pi}{2}, \frac{5\pi}{6}\right], decreasing on \displaystyle \left[\frac{\pi}{6}, \frac{\pi}{2}\right] and \displaystyle \left[\frac{5\pi}{6}, \pi\right].
  19. f'(x) = 2\sin x\cos x = \sin 2x, zero at 0, \displaystyle \frac{\pi}{2}, \pi, \displaystyle \frac{3\pi}{2}, 2\pi. Increasing on \displaystyle \left[0, \frac{\pi}{2}\right] and \displaystyle \left[\pi, \frac{3\pi}{2}\right], decreasing on \displaystyle \left[\frac{\pi}{2}, \pi\right] and \displaystyle \left[\frac{3\pi}{2}, 2\pi\right].
  20. f'(x) = 3x^2 + 6x + 3, which is 3(x + 1)^2, positive except at -1. Increasing on (-\infty, -1] and [-1, \infty), which join: increasing on (-\infty, \infty).
  21. The square (x + 1)^2 never changes sign; (x - 3) and (x - 5)^3 flip at 3 and 5. Test -2, 0, 4, 6: g' = 1715, 375, -25, 147. Increasing on (-\infty, 3] and [5, \infty), decreasing on [3, 5]; nothing changes at -1.
  22. g'(x) = 1 - \cos x \ge 0, with g' = 0 only at x = 2\pi n for integers n. On each [2\pi n, 2\pi(n + 1)], g' > 0 inside, so g is increasing there; neighboring intervals share ends and join, so g is increasing on [0, \infty). For x > 0, g(x) > g(0) = 0, that is, \sin x < x.
  23. The graph is of f', not f. On [-3, -2) it is above the axis, so f' > 0 and f is increasing on [-3, -2]. The graph going down says f' is decreasing: f rises less steeply.
  24. -3 \le k \le 3. f'(x) = 3x^2 - 2kx + 3 has discriminant 4k^2 - 36. If \lvert k \rvert < 3 it is negative and f' > 0 everywhere. If k = \pm 3, f'(x) = 3(x \mp 1)^2, zero at one point only, and the two sides join. If \lvert k \rvert > 3, f' < 0 between its two zeros, so f decreases there.
  25. f(x) = -\dfrac{1}{x}: f'(x) = \dfrac{1}{x^2} > 0 on the whole domain, but f(-1) = 1 > f(1) = -1. Correct version: if f'(x) > 0 at every point of an interval I (inside it, with f continuous on I), then f is increasing on I.
unattributed · CC-BY-SA-4.0