Application of Derivatives · Lesson 2 of 7

Increasing and Decreasing Functions

“Use derivative signs and interval reasoning to explain where a function rises, falls, or stays constant.”

Learning Objectives

• State increasing, decreasing, and constant behaviour using comparisons between inputs. • Explain the derivative sign test using the Mean Value Theorem. • Construct sign charts while respecting domain restrictions. • Find monotonicity intervals for polynomial and trigonometric functions. • Distinguish strict increase from a nonnegative derivative condition.

Reading a Function from Left to Right

A graph tells a story about how an output changes as its input increases. If the graph rises as you move from left to right, larger inputs give larger outputs. If it falls, larger inputs give smaller outputs. The important comparison is between input values in a stated interval, not simply whether the graph lies above or below the horizontal axis.

For $f (x) = x^{2}$ , moving from $x = - 2$ to $x = - 1$ changes the output from 4 to 1: the function decreases. Moving from 1 to 2 changes the output from 1 to 4: it increases. The formula is the same on both sides; its behaviour depends on the interval.

Inputs to the left of zero	Outputs	Inputs to the right of zero	Outputs
$- 2, - 1, 0$	$4, 1, 0$	$0, 1, 2$	$0, 1, 4$

One function, two kinds of behaviour— The left branch of the parabola falls towards the origin; the right branch rises away from it.

Definition

Strictly increasing on an interval

A function $f$ is strictly increasing on an interval $I$ if, for every pair $x_{1}, x_{2}$ in $I$ , $x_{1} < x_{2}$ implies $f (x_{1}) < f (x_{2})$ . The condition must hold for every such pair, not just selected sample values.

Definition

Strictly decreasing on an interval

A function $f$ is strictly decreasing on $I$ if $x_{1} < x_{2}$ implies $f (x_{1}) > f (x_{2})$ for every pair in $I$ . A constant function instead has the same value $f (x) = c$ throughout the interval.

Some writing uses “increasing” for the weaker condition $f (x_{1}) \leq f (x_{2})$ when $x_{1} < x_{2}$ ; this is also called nondecreasing. We will say “strictly increasing” when the output must be strictly larger, and “nondecreasing” when equality is allowed. This distinction matters when a derivative is zero at some points or across an interval.

Definition

Monotonic and behaviour at a point

A function is monotonic on an interval if it consistently increases or consistently decreases there, with the intended strictness stated. Saying it is increasing near an interior point means there is an open interval containing that point on which it is increasing.

Example — Using the definition directly

Problem
Show that $f (x) = 7 x - 3$ is strictly increasing on the real line.

1.Take any real numbers $x_{1} < x_{2}$ . Multiplication by positive 7 preserves the inequality: $7 x_{1} < 7 x_{2}$ .
2.Subtracting 3 preserves it again: $7 x_{1} - 3 < 7 x_{2} - 3$ . Thus $f (x_{1}) < f (x_{2})$ .
3.Because the argument works for every pair of real inputs, it proves strict increase on the whole real line.

Why the Derivative Sign Controls Behaviour

The derivative gives a local rate. Positive rates throughout an interval suggest that the function rises throughout it. The Mean Value Theorem makes the connection precise: the difference between two function values equals their input separation multiplied by a derivative at some point between them.

Mean Value Theorem comparisonLaTeX

f (x_{2}) - f (x_{1}) = f^{'} (c) (x_{2} - x_{1}), x_{1} < c < x_{2}

For this statement,

f

must be continuous on

[x_{1}, x_{2}]

and differentiable on

(x_{1}, x_{2})

Since $x_{2} - x_{1} > 0$ , the sign of this difference is the sign of $f^{'} (c)$ . If every derivative inside the interval is positive, any later function value must be larger. If every derivative is negative, any later value must be smaller. If every derivative is zero, every pair of values is equal, so the function is constant.

Derivative condition inside the interval	Conclusion
$f^{'} (x) > 0$ everywhere	Strict increase
$f^{'} (x) < 0$ everywhere	Strict decrease
$f^{'} (x) = 0$ everywhere	Constant
$f^{'} (x) \geq 0$ everywhere	Nondecreasing; further reasoning may establish strict increase

For a closed interval $[a, b]$ , the sign test requires continuity on the closed interval and differentiability inside it. A derivative need not be defined at the endpoints. Once behaviour on the interior is established, continuity can allow those endpoints to be included in the monotonicity interval.

A sufficient condition is not a necessary one

Strict increase does not require $f^{'} (x) > 0$ at every single point. For instance, $x^{3}$ is strictly increasing even though its derivative is zero at 0. Conversely, a derivative that is zero throughout an interval gives a constant function, not a strictly increasing one.

Example — A cubic that always increases

Problem
Show that $f (x) = x^{3} - 3 x^{2} + 4 x$ is strictly increasing on the real line.

1.Differentiate: $f^{'} (x) = 3 x^{2} - 6 x + 4$ .
2.Complete the square: $f^{'} (x) = 3 (x - 1)^{2} + 1$ . A square is nonnegative, so this derivative is always at least 1.
3.Thus $f^{'} (x) > 0$ for every real $x$ . Apply the derivative sign test on any interval to conclude strict increase on the real line.

Example — Cosine on different intervals

Problem
Determine the behaviour of $f (x) = cos x$ on $(0, π)$ , $(π, 2 π)$ , and $(0, 2 π)$ .

1.The derivative is $f^{'} (x) = - sin x$ . On $(0, π)$ , sine is positive, so $f^{'} (x) < 0$ and cosine decreases.
2.On $(π, 2 π)$ , sine is negative, so $f^{'} (x) > 0$ and cosine increases.
3.The function first falls and then rises on $(0, 2 π)$ . It is therefore neither increasing throughout nor decreasing throughout that larger interval.

Finding Intervals by a Sign Chart

When a derivative changes sign, one global answer is usually impossible. Find the places where its sign can change, split the domain there, and test the derivative on each resulting interval. A sign chart records the reasoning so that no interval is missed.

For a polynomial, zeros of the derivative divide the real line into intervals. For a function with restrictions, also separate points where the function is not defined or the derivative does not exist. Test only points in the domain, and report separate intervals rather than claiming monotonicity across a gap.

Example — A quadratic with one dividing point

Problem
Find where $f (x) = x^{2} - 4 x + 6$ increases and decreases.

1.Differentiate: $f^{'} (x) = 2 x - 4 = 2 (x - 2)$ . Its only zero is $x = 2$ .
2.For $x < 2$ , the derivative is negative; for $x > 2$ , it is positive.
3.Thus $f$ decreases on $(- \infty, 2]$ and increases on $[2, \infty)$ . Including 2 is valid by continuity and does not claim a nonzero derivative there.

A derivative changing from negative to positive— The sign applies throughout an interval, not just at one test point.

Example — A cubic with two dividing points

Problem
Find intervals of increase and decrease for $f (x) = 4 x^{3} - 6 x^{2} - 72 x + 30$ .

1.Differentiate and factor: $f^{'} (x) = 12 x^{2} - 12 x - 72 = 12 (x - 3) (x + 2)$ .
2.Its zeros $- 2$ and 3 split the real line. For $x < - 2$ , both factors are negative, so their product is positive.
3.For $- 2 < x < 3$ , the factors have opposite signs, so the derivative is negative. For $x > 3$ , both are positive.
4.The function increases on $(- \infty, - 2]$ and $[3, \infty)$ , and decreases on $[- 2, 3]$ . It is not monotonic on the entire real line.

Interval	Sign of $(x - 3) (x + 2)$	Behaviour
$(- \infty, - 2)$	$(-) (-) = +$	Increasing
$(- 2, 3)$	$(-) (+) = -$	Decreasing
$(3, \infty)$	$(+) (+) = +$	Increasing

Example — A trigonometric function on a restricted domain

Problem
Find the increasing and decreasing intervals of $f (x) = sin 3 x$ for $0 \leq x \leq π /2$ .

1.Its derivative is $3 cos 3 x$ . Because $0 \leq 3 x \leq 3 π /2$ , the possible zeros in the given domain are $x = π /6$ and $x = π /2$ .
2.On $0 < x < π /6$ , $cos 3 x > 0$ ; on $π /6 < x < π /2$ , $cos 3 x < 0$ . The zero at the right endpoint does not create another interval inside the domain.
3.By continuity, $f$ increases on $[0, π /6]$ and decreases on $[π /6, π /2]$ . Restricting the angles before solving prevents extra periodic roots from entering the answer.

Restrict the sign chart to the given domain— The sign applies throughout an interval, not just at one test point.

Example — Combining sine and cosine

Problem
Determine the behaviour of $f (x) = sin x + cos x$ on $[0, 2 π]$ .

1.Differentiate: $f^{'} (x) = cos x - sin x$ . Its zeros in the domain are $π /4$ and $5 π /4$ .
2.Test the intervals: the derivative is positive on $(0, π /4)$ , negative on $(π /4, 5 π /4)$ , and positive on $(5 π /4, 2 π)$ .
3.Therefore the function increases on $[0, π /4]$ and $[5 π /4, 2 π]$ , and decreases on $[π /4, 5 π /4]$ .
4.The two increasing intervals are separate: do not join them across the decreasing portion.

Interval interior	Derivative sign	Function behaviour
$(0, π /4)$	Positive	Increasing
$(π /4, 5 π /4)$	Negative	Decreasing
$(5 π /4, 2 π)$	Positive	Increasing

More Factors and Composite Functions

The same sign method works when formulas become longer. Factor what you can, identify factors that are always positive, and focus on the remaining signs. A positive multiplier changes the size of a derivative but never its sign.

Example — A quartic with three dividing points

Problem
Find the increasing and decreasing intervals of $f (x) = 3 x^{4} /10 - 4 x^{3} /5 - 3 x^{2} + 36 x /5 + 11$ .

1.Differentiate: $f^{'} (x) = 6 x^{3} /5 - 12 x^{2} /5 - 6 x + 36/5 = (6/5) (x - 1) (x + 2) (x - 3)$ .
2.The dividing points in order are $- 2, 1, 3$ . For an input less than $- 2$ , all three factors are negative, so the derivative is negative.
3.On $(- 2, 1)$ , two factors are negative and one is positive, giving a positive product. On $(1, 3)$ , one factor is negative, giving a negative product. On $(3, \infty)$ all factors are positive.
4.Thus the function decreases on $(- \infty, - 2]$ and $[1, 3]$ , and increases on $[- 2, 1]$ and $[3, \infty)$ . The positive factor $6/5$ has no effect on this classification.

Three zeros produce four sign intervals— The sign applies throughout an interval, not just at one test point.

Example — An inverse trigonometric composition

Problem
Show that $f (x) = tan^{- 1} (sin x + cos x)$ increases on $(0, π /4)$ .

1.Apply the chain rule: $f^{'} (x) = (cos x - sin x) / (1 + (sin x + cos x)^{2})$ .
2.The denominator is positive. It can also be written $2 + sin 2 x$ , which is positive because it is at least 1.
3.On $(0, π /4)$ , $cos x > sin x$ , so the numerator is positive. Hence $f^{'} (x) > 0$ throughout this interval, proving strict increase.

Keep the domain visible

A positive derivative on two separate pieces does not automatically establish monotonicity across their gap. For $f (x) = x + 1/ x$ , the derivative is positive when $∣ x ∣ > 1$ , but statements of increase should be made on intervals contained in the domain, such as $(- \infty, - 1]$ or $[1, \infty)$ .

Check Your Understanding

Use these questions to check both your calculations and your interpretation of the ideas. Read the conditions carefully before choosing an answer.

Quick check

If $f^{'} (x) < 0$ throughout an interval, what follows under the sign-test hypotheses?

Quick check

Where does $f (x) = x^{2} - 6 x$ decrease?

Quick check

What does $f^{'} (x) = 0$ at every point of an interval imply?

Quick check

Which statement about $x^{3}$ is correct?

Quick check

On which interval does $cos x$ increase?

Quick check

The derivative factors as $(x - 1)^{2} (x + 2)$ . Which zero changes its sign?

Practice Problems

Find the increasing and decreasing intervals of $2 x^{3} - 3 x^{2} - 36 x + 7$ . Show a complete sign chart.
Show that $lo g (1 + x) - 2 x / (2 + x)$ is strictly increasing for $x > - 1$ . Discuss the isolated point where its derivative is zero.
Find the intervals where $[x (x - 2)]^{2}$ increases.
Prove that $4 sin θ / (2 + cos θ) - θ$ increases on $[0, π /2]$ .
Find all real $a$ for which $x^{2} + a x + 1$ increases on $[1, 2]$ .
Explain why $lo g x$ increases on $(0, \infty)$ , and determine the behaviour of $lo g (sin x)$ on $(0, π)$ .
Determine the behaviour of $lo g ∣ cos x ∣$ on $(0, π /2)$ and $(3 π /2, 2 π)$ .
Find the increasing intervals of $x^{2} e^{- x}$ , keeping negative inputs in the sign analysis.
Show that $x^{3} - 3 x^{2} + 3 x - 100$ is strictly increasing even though its derivative vanishes at one point.

Key Takeaways

• Increasing and decreasing describe comparisons on an interval, not the sign of the output. • A positive derivative ensures strict increase; a negative derivative ensures strict decrease. • A derivative that is zero everywhere on an interval means the function is constant there. • Derivative zeros, undefined derivatives, and domain gaps guide interval splitting. • Use continuity when including endpoints, and distinguish strict increase from nondecrease.

Previous · Lesson 1

Rates of Change in Real Situations

Next · Lesson 3

Understanding Maximum and Minimum Values

Reading a Function from Left to Right

Why the Derivative Sign Controls Behaviour

Finding Intervals by a Sign Chart

More Factors and Composite Functions

Check Your Understanding

Quiz

Practice Problems

Key Takeaways