Show that the equation has exactly one real root. 2x+cos⁡x=0

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Show that the equation has exactly one real root.  2x+cos⁡x=0

FieniChoonin · Accepted Answer

Keep in mind, first, the intermediate value theorem, which says 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)
$2 x + \cos x = 0$
Let $f (x) = 2 x + \cos x$
$Notice that: f (- 1) = - 2 + \cos (- 1) < 0$
$f (1) = 2 + \cos (1) > 0$
The intermediate value theorem states that there is a c in the range (-1, 1) with f(c) equal to 0.
This shows that f(x) has a root.
Realize this now:
$f^{'} (x) = 2 - \sin (x)$
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.
Note however as I said:
$f^{'} (x) = 2 - \sin (x)$ is always positive so there exists no k such that f'(k) = 0.
This thus prove there cannot be two or more roots.
Hence:
$2 x + \cos x = 0$
only has one real root.

Show that the equation has exactly one real root. 2x+\cos x=0

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