Let $I \subseteq \mathbb{R}$ be an open interval. The function $f: I \to \mathbb{R}$ is:

1.) **increasing** on $I$ if ($\forall x_1, x_2 \in I$) $(x_1 < x_2) \Longrightarrow( f(x_1) \le f(x_2))$,

2.) **strictly increasing** on $I$ if ($\forall x_1, x_2 \in I$) ($x_1 < x_2) \Longrightarrow (f(x_1) < f(x_2))$,

3.) **decreasing** on $I$ if ($\forall x_1, x_2 \in I$ $x_1 < x_2) \Longrightarrow f(x_1) \ge f(x_2)$,

4.) **strictly decreasing** on $I$ if ($\forall x_1, x_2 \in I$) $(x_1 < x_2) \Longrightarrow( f(x_1) > f(x_2))$.

Functions above are called **monotonic functions**, that is, **strictly monotonic functions**.

Let $f$ be the function such that is differentiable on $\left \langle a, b \right \rangle$ and continuous on $[a, b]$.

a) The function $f$ is increasing on $[a,b]$ if $f'(x) > 0, \forall x \in \left \langle a, b \right \rangle$.

b) The function $f$ is decreasing on $[a,b]$ if $f'(x) < 0, \forall x \in \left \langle a, b \right \rangle$.

c) The function $f$ is constant on $[a, b]$ if $f'(x) = 0, \forall x \in \left \langle a, b \right \rangle$.

Let $I \in \mathbb{R}$ be an open interval, $x_0 \in I$ and $f: I \to \mathbb{R}$. Then $x_0$ is a **critical number **(or** stationary point**) of the function $f$ if $f'(x_0) = 0$, whereby the function $f$ is defined at $x_0$.

To determine intervals in which the function $f$ is either always increasing or decreasing, first we need to find critical points by solving the equation $f'(x) =0.$

The domain of the function is divided into open intervals with critical numbers and on each of these intervals we need to determine the sign of the first derivative.

**Example 1**. Determine critical numbers and monotone intervals of the function $f$ if:

$$f(x) = x^5 – 5x^4 +4.$$

**Solution**:

The first derivative of the function $f$ is equal to:

$$f'(x) = 5 x^4 – 20x^3= 5x^3 (x – 4).$$

Critical numbers we obtain by solving the equation:

$$f'(x) = 0 \Longrightarrow 5x^3 (x – 4) = 0.$$

Solutions of the equation above are:

$$x_1 = 0, \quad \quad x_2 =4,$$

that is, $x_1=0$ and $x_2=4$ are critical numbers.

Therefore, monotone intervals are:

$$\left \langle – \infty, 0 \right \rangle, \left \langle 0, 4 \right \rangle, \left \langle 4, + \infty \right \rangle.$$

Now we choose any point from the each interval and determine the sign of the first derivative:

$$ x= -1 \in \left \langle – \infty, 0 \right \rangle,\quad \quad f'(-1) = 20 > 0,$$

$$x=1 \in \left \langle 0, 4 \right \rangle, \quad \quad f'(1) = -15 < 0,$$

$$x = 5 \in \left \langle 4, + \infty \right \rangle, \quad \quad f'(5) = 625 > 0.$$

We can conclude that the function $f$ is increasing on intervals $\left \langle – \infty, 0 \right \rangle$ and $\left \langle 4, + \infty \right \rangle$ and decreasing on $\left \langle 0, 4 \right \rangle$.

**Local extrema**

Let $I \subseteq \mathbb{R}$ be an open interval, $x_0 \in I$ and $f: I \to \mathbb{R}$.

Then the function $f$ has the **local minimum** at point $x_0$ if $f(x_0) \le f(x), \forall x \in I$ and the **local maximum** at point $x_0$ if $f(x_0) \ge f(x), \forall x \in I$.

The local minimum and local maximum of a function are called **local extreme values** or **local extrema** of a function.

**The necessary condition for local extrema (Fermat’s theorem)**

If the function $f$ has the local minimum or maximum at point $x_0$ and if the function $f$ has the first derivative at point $x_0$, then $f'(x_0) = 0$.

**Theorem**. (First derivative test)

Let $f: I \to \mathbb{R}, I \subseteq \mathbb{R}$ be differentiable function on $I$ and $x_0 \in I$ a critical number.

1.) If $f'(x)$ changes sign from positive to negative around number $x_0$, then $(x_0, f(x_0))$ is a local maximum.

2.) If $f'(x)$ changes sign from negative to positive around number $x_0$, then $(x_0, f(x_0))$ is a local minimum.

3.) If $f'(x)$ has the same sign from both sides of $x_0$, then the local extremum does not exists at $x_0$.

**Example 2**. Determine local extrema of the function:

$$f(x) = -x^2 + 2x +3.$$

**Solution**:

Calculate the first derivative:

$$f'(x) = – 2x +2,$$

which is equal to $0$ for $x_0 = 1$. On the interval $\left \langle – \infty, 1 \right \rangle$ the first derivative is positive, that is, in interval $\left \langle – \infty, 1 \right \rangle$ the given function is increasing. The first derivative is negative on the interval $\left \langle 1, + \infty \right \rangle$, that is, the function is decreasing. Therefore, the first derivative changes sign from positive to negative. The value of the function $f$ at point $x_0 = 1$ is equal to:

$$f(1) = -1 + 2 + 3 = 4$$

Finally, the point $T(1,4)$ is the local maximum of the given function $f$ (see the picture below).

**Global extrema**

Let $f$ be the function defined on $[a, b], a< b$, and $x_0 \in [a,b]$. Then the function $f$ has the **global minimum** at point $x_0$ if $f(x_0) \le f(x), \forall x \in [a, b]$ and the **global maximum** at point $x_0$ if $f(x_0) \ge f(x), \forall x \in [a, b]$.

Difference between local and global extrema is that the global extrema of a function is the largest or the smallest value on its entire domain, and local extrema of a function is the largest or the smallest value in a given range of a function.

How to find global extrema of the function $f$ on the closed interval $[a,b]$?

1.) Find all critical numbers of the function $f$ on $\left \langle a, b \right \rangle$.

2.) Evaluate the function $f$ at the endpoints $a$ and $b$ and at all critical numbers from the step 1.)

3.) The largest of obtained values is the absolute maximum and the smallest is the absolute minimum.

**Example 3**. Find the global minimum and global maximum of the function $f$ on the interval $[-4,4]$ if

$$f(x) = 2x^3 – 3x^2 – 36x.$$

**Solution**:

We finding the first derivative:

$$f'(x) = 6x^2 – 6x – 36 = 6(x^2 – x – 6).$$

Critical numbers we find by solving the equation $f'(x) = 0$, that is $x^2 – x – 6=0$. We obtain that critical numbers are $x_1 = – 2$ and $x_2 = 3$.

Now we need evaluate the function $f$ at the endpoints of the interval $[a,b]=[-4, 4]$ and at all critical numbers. Therefore, we have:

$$f(a) = f( -4) = -128 – 48 + 144 = -32,$$

$$f(x_1) = f(-2) = -16 – 12 + 72 = 44 ,$$

$$f(x_2) = f(3) =54 – 27 – 108 = -81 ,$$

$$f(b) = f(4) = 128 – 48 – 144 = -64.$$

The largest value of the function $f$ is $44$ at the point $x_1 = – 2$, and the smallest value of the function $f$ is $-81$ at the point $x_2 = 3$. Therefore, the point $m(3, -81)$ is the absolute minimum of the function $f$ and point $M(-2, 44)$ is the absolute maximum of the function $f$ on the interval $[-4, 4]$.

The sign of the first derivative in the neighborhood of a critical number it is not always easy to determine. Therefore, there is one more criteria for determining the character of a critical number, by using the second derivative.

**Concavity and points of inflection**

The function $f:I \to \mathbb{R}, I \subseteq \mathbb{R}$, is **concave up** if $f’$ is increasing on $I$.

The function $f: I \to \mathbb{R}, I \subseteq \mathbb{R}$, is **concave down** if $f’$ is decreasing on $I$.

**Theorem**. Let $I \subseteq \mathbb{R}$ be an open interval and $f: I \to \mathbb{R}$ a function which is twice differentiable on $I$.

1.) If $f” (x) >0, \forall x \in I$, then the function $f$ is concave up on $I$.

2.) If $f”(x) < 0, \forall x \in I$, then the function $f$ is concave down on $I$.

3. ) The point $(x_o, f(x_0)$, $x_0 \in I$ is called an **inflection point** of the function $f$ if $f'(x_0)$ exists and if concavity changes the sign at $(x_o, f(x_0)$ from positive to negative and from negative to positive.

That is, if is it the second derivative on the intervals $\left \langle a, b \right \rangle$ and $\left \langle b, c \right \rangle$ of different signs, then the function $f$ at point $b$ has an inflection point, where $f$ changes from concave up to concave down or from concave down to concave up.

For instance, the function $f(x) = x^2$ is concave up on $\mathbb{R}$, because $f”(x) = 2 > 0$, $\forall x \in \mathbb{R}$.

The function $f(x) = \ln x$ is concave down on $\left \langle 0, + \infty \right \rangle$, because $f”(x) = – \frac{1}{x^2} < 0$, for all $x \in \left \langle 0, + \infty \right \rangle$.

How to find open intervals where the function $f$ is concave up or concave down?

- Find the second derivative of the function $f$.
- Potential inflection points find by solving the equation $f”(x) =0$. Check if they are in the domain of the function $f$.
- On those intervals on which $f”(x) > 0$ the function $f$ is concave up, otherwise, concave down. On the border between concavity intervals is the inflection point.

**Example 4**. Find the open intervals where the function $f$ is concave up or concave down if

$$f(x) = x – \sqrt[3]{x-1}.$$

**Solution**:

Firstly, we need to find $f”$.

$$f'(x) = (x – \sqrt[3]{x-1})’ = 1 – \frac{1}{3 \sqrt[3]{(x-1)^2}}.$$

$$f”(x) = (1 – \frac{1}{3 \sqrt[3]{(x-1)^2}})’ = \frac{2}{9 \sqrt[3]{(x-1)^5}}.$$

The second derivative is not equal to $0$ for any $x$, however for $x=1$ the second derivative it is not define. Therefore, for $x \in \left \langle – \infty, 1 \right \rangle$ is $f”(x) < 0$, that is, the function $f$ is concave down and for $x \in \left \langle 1, + \infty \right \rangle$ is $f”(x ) >0$, that is, the function $f$ is concave up.

**Theorem. **(The second derivative test)

Let the function $f: I \to \mathbb{R}, I\subseteq \mathbb{R}$ be twice differentiable on $I$. To find the local extrema of the function $f$ we

1.) calculate $f’$ and $f”$,

2.) find critical numbers by solving the equation $f'(x) =0$,

3.) to each critical number $x_0$ apply the second derivative test; if $f”(x_0) > 0$, then $x_0$ is the local minimum, and if $f”(x_0) < 0$, then $x_0$ is the local maximum. If $f”(x_0) = 0$, then the character of the point $x_0$ we finding by using the sign of the first derivative.

**Example 5.** Determine extrema of the function $f$ if

$$f(x) = \frac{x – 1}{x^2}.$$

**Solution**:

Firstly, we need to find $f'(x)$ and $f”(x)$. We have

$$f'(x) = \frac{(x-1)’ \cdot x^2 – (x-1) \cdot(x^2)’}{(x^2)^2}$$

$$= \frac{x^2 – (x-1) \cdot 2x}{x^4}$$

$$= \frac{-x^2 +2x}{x^4}$$

$$f”(x) = \frac{(-x^2 + 2x)’ \cdot x^2 – (-x^2 +2x) \cdot (x^4)’}{(x^4)^2} $$

$$ = \frac{(-2x + 2) \cdot x^4 – (-x^2 + 2x) \cdot 4x^3}{x^8}$$

$$ =\frac{-2x^5 + 2x^4 + 4x^5 – 8x^4}{x^8}$$

$$= \frac{2x^5 – 6 x^4}{x^8}. $$

Critical numbers we finding by solving the equation $f'(x) = 0$, that is

$$ \frac{-x^2 +2x}{x^4} = 0$$

$$ \Leftrightarrow – x^2 + 2x = 0$$

$$ \Leftrightarrow x(2-x) = 0.$$

Therefore, the critical points are $x_1= 0$ and $x_2=2$. Since the second derivative of the function $f$ is defined on $\mathbb{R}$ and the function $f$ on $\mathbb{R}/ \{0\}$, the second derivative of the function $f$ at point $x_1=0$ is equal to $0$. The second derivative of the function $f$ at point $x_2 =2$ is equal to

$$f”(2) = \frac{2 \cdot 2^5 – 6 \cdot 2^4}{2^8} = \frac{32 – 96}{256} = \frac{-64}{256} = -\frac{1}{4} <0.$$

Now we need to find the value of the function $f$ at point $x_2=2$, that is

$$f(2) =\frac{2-1}{2^2} = \frac{1}{4}.$$

From the second derivative test we conclude that the given function $f$ has the local maximum at point $\left (2, \frac{1}{4} \right)$.