$2x+\cos x = 0$

This **article aims** to find the** roots** of the **given function.** The article uses the concept of the **mean value theorem** and **Rolle’s theorem.** The readers should know the **definition** of the **mean value theorem** and **Rolle’s theorem**.

**Expert Answer**

First, remember the **mean value theorem**, which states that given a function $f(x)$** continuous** on $[a, b]$ then there exists $c$ such that: $f(b) < f(c) < f(a) \:or \: f(a) < f(c) < f(b)$.

\[2x+\cos x =0\]

**Let:**

\[f(x) = 2x +\cos x = 0\]

**Notice that:**

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

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

Using the **mean value theorem**, there exists a $c$ in $(-1, 1)$ such that $f(c) = 0$. This represents that $f(x)$** has a root**.

Now realized that:

\[f'(x) = 2 – \sin x\]

Notice that the $f'(x) > 0 $ for all values of $x$. Keep in mind that **Rolle’s theorem** states that if a** function is continuous on** an interval $[m, n]$ and **differentiable** on

$(m, n)$ where $f(m) = f(n)$ then there exists $k$ in $(m, n)$ such that $f'(k) = 0$.

Let’s assume that t**his function has $2$ roots**.

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

Then there exists $k$ in $(m, n)$ such that $f'(k) = 0$.

But notice how I said:

$f'(x) = 2-\sin x $ is** always positive**, so there is no $k$ such that $f'(k) = 0$. So this proves that there **cannot be two or more roots**.

Hence $ 2x +\cos x$ has **only one root.**

**Numerical Result**

Hence, $ 2x +\cos x$ has **only one root**.

**Example**

**Show that equation has exactly one real root.**

**$4x – \cos \ x = 0$**

**Solution**

First, remember the **mean value theorem**, which states that given a function $f(x)$** continuous** on $[a, b]$ then there exists $c$ such that: $f(b) < f(c) < f(a) \:or \: f(a) < f(c) < f(b)$

\[4x-\cos x =0\]

**Let:**

\[f(x) = 4x -\cos x = 0\]

**Notice that:**

\[ f(-1) = -4 -\cos (-1) < 0 \]

\[ f(1) = 4 – \cos(1) > 0 \]

Using the **mean value theorem**, there exists a $c$ in $(-1, 1)$ such that $f(c) = 0$. This shows that $f(x)$** has a root**.

Now realized that:

\[ f'(x) = 4 + \sin x \]

Notice that $ f'(x) > 0 $ for all values of $ x $. Remember that **Rolle’s theorem** states that if a** function is continuous on** $ [m, n] $ and **differentiable** on:

$(m, n)$ where $f(m) = f(n)$ then there exists $k$ in $(m, n)$ such that $f'(k) = 0$.

Assume that t**his function has $2$ roots**.

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

Then there exists $k$ in $(m, n)$ such that $ f'(k) = 0 $.

But notice how we said:

$ f'(x) = 4+\sin x $ is** always positive**, so there is no $k$ such that $ f'(k) = 0 $. So this proves that there **cannot be two or more roots**.

Hence, $ 4x -\cos x $ has **only one root.**