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# Calculus

Applications of the Derivative

# Monotonic Functions

Page 1
Problems 1-2
Page 2
Problems 3-15

### Definition of an Increasing and Decreasing Function

Let $$y = f\left( x \right)$$ be a differentiable function on an interval $$\left( {a,b} \right).$$ If for any two points $${x_1},{x_2} \in \left( {a,b} \right)$$ such that $${x_1} < {x_2},$$ there holds the inequality $$f\left( {{x_1}} \right) \le f\left( {{x_2}} \right),$$ the function is called increasing (or non-decreasing) in this interval.

If this inequality is strict, i.e. $$f\left( {{x_1}} \right) \lt f\left( {{x_2}} \right),$$ then the function $$y = f\left( x \right)$$ is said to be strictly increasing on the interval $$\left( {a,b} \right).$$

Similarly, we define a decreasing (or non-increasing) and a strictly decreasing function.

These concepts can be formulated in a more compact form. A function $$y = f\left( x \right)$$ is called

• increasing (non-decreasing) on the interval $$\left( {a,b} \right)$$ if
${\forall\;{x_1},{x_2} \in \left( {a,b} \right):\;}\kern-0.3pt {{x_1} \lt {x_2}} \Rightarrow {f\left( {{x_1}} \right) \le f\left( {{x_2}} \right);}$
• strictly increasing on the interval $$\left( {a,b} \right)$$ if
${\forall\;{x_1},{x_2} \in \left( {a,b} \right):\;}\kern-0.3pt {{x_1} \lt {x_2}} \Rightarrow {f\left( {{x_1}} \right) \lt f\left( {{x_2}} \right);}$
• decreasing (non-increasing) on the interval $$\left( {a,b} \right)$$ if
${\forall\;{x_1},{x_2} \in \left( {a,b} \right):\;}\kern-0.3pt {{x_1} \lt {x_2}} \Rightarrow {f\left( {{x_1}} \right) \ge f\left( {{x_2}} \right);}$
• strictly decreasing on the interval $$\left( {a,b} \right)$$ if
${\forall\;{x_1},{x_2} \in \left( {a,b} \right):\;}\kern-0.3pt {{x_1} \lt {x_2}} \Rightarrow {f\left( {{x_1}} \right) \gt f\left( {{x_2}} \right).}$

It is clear that a non-decreasing function can contain strictly increasing intervals and intervals where the function is constant. This is schematically illustrated in Figures $$1-4.$$

Figure 1.

Figure 2.

Figure 3.

Figure 4.

If a function $$f\left( x \right)$$ is differentiable on the interval $$\left( {a,b} \right)$$ and belongs to one of the four considered types (i.e. it is increasing, strictly increasing, decreasing, or strictly decreasing), this function is called monotonic on this interval.

The concept of increasing and decreasing functions can also be defined for a single point $${x_0}.$$ In this case, we consider a small $$\delta$$-neighborhood $$\left( {{x_0} – \delta ,{x_0} + \delta } \right)$$ of this point. A function $$y = f\left( x \right)$$ is strictly increasing at $${x_0}$$ if there exists a number $$\delta \gt 0$$ such that

${\forall\;x \in \left( {{x_0} – \delta ,{x_0}} \right)} \Rightarrow {f\left( x \right) \lt f\left( {{x_0}} \right);}$
${\forall\;x \in \left( {{x_0}, {x_0} + \delta} \right)} \Rightarrow {f\left( x \right) \gt f\left( {{x_0}} \right).}$

Similarly, we can define a function $$y = f\left( x \right)$$, which is strictly decreasing at the point $${x_0}.$$

### Criteria for Increasing and Decreasing Functions

Again consider a function $$y = f\left( x \right)$$ assuming it is differentiable on an interval $$\left( {a,b} \right).$$ To determine if the function is increasing or decreasing on the interval, we use the sign of the first derivative of the function.

Theorem $$1$$.
In order for the function $$y = f\left( x \right)$$ to be increasing on the interval $$\left( {a,b} \right),$$ it is necessary and sufficient that the first derivative of the function be non-negative everywhere in this interval:

$f’\left( x \right) \ge 0\;\forall\;x \in \left( {a,b} \right).$

A similar criterion applies to the case of a function that is decreasing on the interval $$\left( {a,b} \right):$$

$f’\left( x \right) \le 0\;\forall\;x \in \left( {a,b} \right).$

We prove both (necessary and sufficient) parts of the theorem for the case of an increasing function.

Necessary condition.
Consider an arbitrary point $${x_0} \in \left( {a,b} \right).$$ If the function $$y = f\left( x \right)$$ is increasing on $$\left( {a,b} \right),$$ then by definition, we can write:

${\forall\;x \in \left( {a,b} \right):x \gt {x_0}} \Rightarrow {f\left( x \right) \gt f\left( {{x_0}} \right);}$
${\forall\;x \in \left( {a,b} \right):x \lt {x_0}} \Rightarrow {f\left( x \right) \lt f\left( {{x_0}} \right).}$

It can be seen that in both cases the following inequality holds:

${\frac{{f\left( x \right) – f\left( {{x_0}} \right)}}{{x – {x_0}}} \ge 0,}\;\;\;\kern-0.3pt {\text{where}\;\;x \ne {x_0}.}$

In the limit as $$x \to {x_0}$$, the left-hand side of the inequality is equal to the derivative of the function at the point $${x_0},$$ that is by the limit sign preservation property:

${\lim\limits_{x \to {x_0}} \frac{{f\left( x \right) – f\left( {{x_0}} \right)}}{{x – {x_0}}} } = {f’\left( {{x_0}} \right) \ge 0.}$

This relation is valid for any $${x_0} \in \left( {a,b} \right).$$

Consider the sufficient condition, i.e. the converse statement. Suppose that the derivative $$f’\left( x \right)$$ of a function $$y = f\left( x \right)$$ is non-negative in the interval $$\left( {a,b} \right):$$

$f’\left( {{x_0}} \right) \ge 0\;\forall\; x \in \left( {a,b} \right).$

If $${x_1}$$ and $${x_2}$$ are two arbitrary points of the interval such that $${x_1}< {x_2},$$ then by Lagrange’s theorem we can write:

${f\left( {{x_2}} \right) – f\left( {{x_1}} \right)} = {f’\left( c \right)\left( {{x_2} – {x_1}} \right),}$

where $$c \in \left[ {{x_1},{x_2}} \right],$$ $$\Rightarrow c \in \left( {a,b} \right).$$

Since $$f’\left( c \right) \ge 0,$$ then the right-hand side of the equality is non-negative. Consequently,

$f\left( {{x_2}} \right) \ge f\left( {{x_1}} \right).$

i.e. the function $$y = f\left( x \right)$$ is increasing in the interval $$\left( {a,b} \right).$$

Consider now the cases of a strictly increasing and strictly decreasing function. There exists a similar theorem that describes the necessary and sufficient conditions. Omitting the proof, we state it for the case of a strictly increasing function.

Theorem $$2$$.
Suppose that a function $$y = f\left( x \right)$$ is differentiable on an interval $$\left( {a,b} \right).$$ In order for the function to be strictly increasing in this interval, it is necessary and sufficient that the following conditions are satisfied:

1. $$f’\left( x \right) \ge 0\;\forall\;x \in \left( {a,b} \right);$$
2. $$f’\left( x \right)$$ is not identically equal to zero at any interval $$\left[ {{x_1},{x_2}} \right] \in \left( {a,b} \right).$$

The condition $$1$$ is contained in Theorem $$1$$ and is an indication of a non-decreasing function. The additional condition $$2$$ is required in order to exclude the intervals of constancy, in which the derivative of $$f\left( x \right)$$ is identically zero.

In practice (when finding the intervals of monotonicity), the sufficient condition for a strictly increasing or a strictly decreasing function is commonly used. Theorem $$2$$ implies the following wording of the sufficient criterion:

If the condition $$f’\left( x \right) \gt 0$$ is satisfied for all $$x \in \left( {a,b} \right)$$, except perhaps only a few distinct points where $$f’\left( x \right) = 0,$$ then the function $$f\left( x \right)$$ is strictly increasing in this interval.

Accordingly, the condition $$f’\left( x \right) \lt 0$$ defines a strictly decreasing function.

The number of points where $$f’\left( x \right) = 0$$ is usually finite. According to Theorem $$2,$$ they can not tightly fill any subinterval of the interval $$\left( {a,b} \right).$$

We also give a criterion for increasing/decreasing functions at a point:

Theorem $$3$$.
Let $${x_0} \in \left( {a,b} \right).$$

• If $$f’\left( {{x_0}} \right) \gt 0$$, then the function $$f\left( x \right)$$ is strictly increasing at the point $${x_0};$$
• If $$f’\left( {{x_0}} \right) \lt 0$$, then the function $$f\left( x \right)$$ is strictly decreasing at the point $${x_0}.$$

### Properties of Monotonic Functions

Increasing and decreasing functions have certain algebraic properties, which may be useful in the investigation of functions. Here are some of them:

1. If the functions $$f$$ and $$g$$ are increasing (decreasing) on the interval $$\left( {a,b} \right),$$ then the sum of the functions $$f + g$$ is also increasing (decreasing) on this interval.
2. If the function $$f$$ is increasing (decreasing) on the interval $$\left( {a,b} \right),$$ then the opposite function $$-f$$ is decreasing (increasing) on this interval.
3. If the function $$f$$ is increasing (decreasing) on the interval $$\left( {a,b} \right),$$ then the inverse function $$\large\frac{1}{f}\normalsize$$ is decreasing (increasing) on this interval.
4. If the functions $$f$$ and $$g$$ are increasing (decreasing) on the interval $$\left( {a,b} \right)$$ and moreover, $$f \ge 0$$, $$g \ge 0$$, then the product of the functions $$fg$$ is also increasing (decreasing) on this interval.
5. If the function $$g$$ is increasing (decreasing) on the interval $$\left( {a,b} \right)$$ and the function $$f$$ is increasing (decreasing) on $$\left( {c,d} \right)$$ where $$g:\left( {a,b} \right) \to \left( {c,d} \right),$$ then the composition of functions $$f \circ g$$ (i.e. the composite function $$y = f\left( {g\left( x \right)} \right)$$ is also increasing (decreasing) on the interval $$\left( {a,b} \right).$$

## Solved Problems

Click on problem description to see solution.

### ✓Example 1

Using the definition of monotonicity prove that the function $$f\left( x \right) = {x^2} + 1$$
is strictly increasing for $$x \ge 0.$$

### ✓Example 2

Using the definition of monotonicity prove that the cubic function $$f\left( x \right) = {x^3}$$ is strictly increasing for all $$x \in \mathbb{R}.$$

### ✓Example 3

Using the properties of monotonic functions prove that the function $$f\left( x \right) = {x^4} + 3{x^2}$$ is strictly increasing for $$x \ge 0.$$

### ✓Example 4

Using the definition of monotonicity prove that the function $$f\left( x \right) = \cos x$$
is strictly decreasing on the interval $$\left[ {0,\pi } \right].$$

### ✓Example 5

Find the intervals of monotonicity of the function $$f\left( x \right) = {x^3} – 12x + 5.$$

### ✓Example 6

Find the intervals of monotonicity of the function $$f\left( x \right) = x + \sin x.$$

### ✓Example 7

Find the intervals of monotonicity of the function

$f\left( x \right) = \frac{x}{{{x^2} + 1}}.$

### ✓Example 8

Find the intervals of monotonicity of the function

${f\left( x \right) = \frac{{\sqrt x }}{{x – 1}}\;\;}\kern-0.3pt{\left( {x \ge 0} \right).}$

### ✓Example 9

Find the intervals of monotonicity of the function $$f\left( x \right) = x\ln x.$$

### ✓Example 10

Find the intervals of monotonicity of the function $$f\left( x \right) = {x^2}{e^{ – x}}.$$

### ✓Example 11

Find the intervals of monotonicity of the function $$f\left( x \right) = \sqrt {x – {x^2}}.$$

### ✓Example 12

Find the intervals of monotonicity of the function $$f\left( x \right) = {\large\frac{{1 – \sin x}}{{\cos x}}\normalsize}.$$

### ✓Example 13

Find all values of the parameter $$a,$$ for which the function $$f\left( x \right) = {x^3} – 6{x^2} + ax$$ is strictly increasing in the whole domain.

### ✓Example 14

Find all values of the parameter $$a,$$ for which the equation $${x^3} – 6{x^2} + 9x + a = 0$$ has three distinct real roots.

### ✓Example 15

Determine the number of roots of the cubic equation $${x^3} – 12x + a = 0$$ depending on the parameter $$a.$$

### Example 1.

Using the definition of monotonicity prove that the function $$f\left( x \right) = {x^2} + 1$$ is strictly increasing for $$x \ge 0.$$

#### Solution.

We take two arbitrary points $${x_1}$$ and $${x_2}$$ such that

$0 \le {x_1} \lt {x_2}.$

Consider the difference between the values of the function at these points:

${f\left( {{x_2}} \right) – f\left( {{x_1}} \right) } = {\left( {x_2^2 + 1} \right) – \left( {x_1^2 + 1} \right) } = {x_2^2 – x_1^2 } = {\left( {{x_2} – {x_1}} \right)\left( {{x_2} + {x_1}} \right).}$

It is obvious that in the last expression $${{x_2} – {x_1}} \gt 0$$ and $${{x_2} + {x_1}} \gt 0$$ (since, by assumption, only non-negative values of $$x$$ are considered). As a result, we have

${\left( {{x_2} – {x_1}} \right)\left( {{x_2} + {x_1}} \right) > 0,\;\;}\Rightarrow {f\left( {{x_2}} \right) – f\left( {{x_1}} \right) > 0.}$

This means by definition that the function $$f\left( x \right) = {x^2} + 1$$ is strictly increasing on the given interval.

### Example 2.

Using the definition of monotonicity prove that the cubic function $$f\left( x \right) = {x^3}$$ is strictly increasing for all $$x \in \mathbb{R}.$$

#### Solution.

We choose two arbitrary points $${{x_1}}$$ and $${{x_2}}$$ such that $${x_1} \lt {x_2}.$$ Consider the difference:

$f\left( {{x_2}} \right) – f\left( {{x_1}} \right) = x_2^3 – x_1^3.$

Factoring it as the difference of cubes, we obtain:

${x_2^3 – x_1^3 } = {\left( {{x_2} – {x_1}} \right)\left( {x_2^2 + {x_1}{x_2} + x_1^2} \right).}$

In the second bracket we can get a perfect square:

${x_2^2 + {x_1}{x_2} + x_1^2 } = {x_1^2 + 2 \cdot {x_1} \cdot \frac{{{x_2}}}{2} + \frac{{x_2^2}}{4} + \frac{{3x_2^2}}{4} } = {{\left( {{x_1} + \frac{{{x_2}}}{2}} \right)^2} + \frac{{3x_2^2}}{4} \gt 0.}$

Hence it is clear that the quadratic expression is always positive (it is equal to zero only if $${x_1} = {x_2} = 0,$$ which contradicts the condition $${x_1} \lt {x_2}.$$)

Thus, $$f\left( {{x_2}} \right) – f\left( {{x_1}} \right) \gt 0,$$ if $${x_2} – {x_1} \gt 0,$$ i.e. the function $$f\left( x \right) = {x^3}$$ is strictly increasing.

Page 1
Problems 1-2
Page 2
Problems 3-15