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

chillywilly12a 2021-10-28 Answered
Show that the equation has exactly one real root.
\(\displaystyle{2}{x}+{\cos{{x}}}={0}\)

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Expert Answer

FieniChoonin
Answered 2021-10-29 Author has 14210 answers

First, remember the intermediate 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)
\(\displaystyle{2}{x}+{\cos{{x}}}={0}\)
Let \(\displaystyle{f{{\left({x}\right)}}}={2}{x}+{\cos{{x}}}\)
\(\displaystyle\text{Notice that: }\ {f{{\left(-{1}\right)}}}=-{2}+{\cos{{\left(-{1}\right)}}}{<}{0}\)
\(\displaystyle{f{{\left({1}\right)}}}={2}+{\cos{{\left({1}\right)}}}{>}{0}\)
Using the intermediate value theorem there exists c in (-1, 1) such that f(c) = 0.
This shows that f(x) has a root.
Now realize that:
\(\displaystyle{f}'{\left({x}\right)}={2}-{\sin{{\left({x}\right)}}}\)
Notice that \(\displaystyle{f}'{\left({x}\right)}{>}{0}\) 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 :
\(\displaystyle{f{{\left({m}\right)}}}={f{{\left({n}\right)}}}={0}\)
Then there exists k in (m, n) such that f'(k) = 0.
Note however as I said:
\(\displaystyle{f}'{\left({x}\right)}={2}-{\sin{{\left({x}\right)}}}\) is always positive so there exists no k such that f'(k) = 0.
This thus prove there cannot be two or more roots.
Hence:
\(\displaystyle{2}{x}+{\cos{{x}}}={0}\)
only has one real root.

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