$2x+\cos x = 0$
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 this 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 this 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.