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Show that the equation has exactly one Real root 2x+cosx=0.

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.

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