This question aims to find the real root of the given equation using the **Intermediate theorem** and **Rolle’s theorem**.

If the function is continuous on the interval **[c,d]** then there should be an **x-value** in the interval for every **y-value** that lies in the **f(a)** and **f(b)**. The graph of this function is a curve that shows the **continuity** of the function.

A **continuous function** is a function that has no discontinuities and unexpected variations in its curve. According to **Rolle’s theorem**, if the function is differentiable and continuous on **[m, n]** such that **f(m) = f(n)** then a **k** exists in (m, n) such that **f’(k) = 0.**

## Expert Answer

According to the Intermediate theorem, if the function is continuous on** [a,b]**, then **c** exists as:

\[ f (b) < f (c) < f (a) \]

It can also be written as:

\[ f (a) < f (c) < f (b) \]

The given function is:

\[ 2 x + cos x = 0 \]

Consider the function f (x):

\[ f (x) = 2 x + cos x \]

If we put **+1** and **-1** in the given function:

\[ f (-1) = -2 + cos (-1) < 0 \]

\[ f (1) = 2 + cos (1) > 0 \]

There exists c in **( -1, 1)** when **f(c) = 0** according to intermediate theorem. It means that f(x) has a root.

By taking the derivative of the function:

\[ f’ (x) = 2 – sin (x) \]

For all values of x, the derivative f’(x) must be greater than 0.

If we assume the given function has **two roots,** then according to **Rolle’s theorem**:

\[ f (m) = f (n) = 0 \]

There exists k in ( m, n ) such that f’ (k) = 0

f’ (x) = 2 – sin (x) is always positive so there exists no k such that f’ (k) = 0.

**There cannot be two or more roots.**

**Numerical Results**

The given function $ 2 x + cos x $ has only **one root**.

## Example

Find the real root of 3 x + cos x = 0.

Consider the function f(x):

\[ f(x) = 3 x + cos x \]

If we put +1 and -1 in the given function:

\[ f(-1) = -3 + cos (-1) < 0 \]

\[ f(1) = 3 + cos (1) > 0 \]

By taking the derivative of the function:

\[ f’(x) = 3 – sin (x) \]

For all values of x, the derivative f’(x) must be greater than 0.

If we assume the given function has two roots then:

\[f(m) = f(n) = 0\]

f’(x) = 3 – sin (x) is always positive so there exists no k such that f’(k) = 0.

There cannot be two or more roots.

The given function $ 3 x + cos x $ has only **one root**.

*Image/Mathematical drawings are created in Geogebra.*