Question

Show that the equation \({\bf{3x + 2cosx + 5 = 0}}\) has exactly one real root.

Short answer

Thus, the equation has exactly one real root.

Step-by-step solution

Definition of Rolles’ Theorem:

Consider by the Rolle’s theorem for the closed intervals \(\left( {a,b} \right)\) the function is differentiable and at open interval \(\left( {a,b} \right)\)the function is again differentiable where \({\bf{f}}\left( {\bf{a}} \right){\bf{ = f}}\left( {\bf{b}} \right){\bf{ = 0}}\). If the following condition are define then the first derivative of the function at point \(x\) is zero where \(a \le x \le b\).

Show the equation has a real root.

Let \(f\left( x \right) = 3x + 2\cos x + 5\) and substitute \(x = 0\) in the given function gives,

\(\begin{aligned}{c}f\left( 0 \right) &= 0 + 2\cos 0 + 5\\ &= 2 + 5\\ &= 7 > 0\end{aligned}\)

Thus, \(f\left( x \right)\) is positive when \(x = 0\).

Now, substitute \(x = - 100\) in the given function gives,

\(f\left( { - 100} \right) = - 300 + 2\cos \left( { - 100} \right) + 5\)

Thus, \(\cos \left( { - 100} \right)\) has value between \( - 1\) and 1 then \(f\left( { - 100} \right)\) will be negative.

Thus, the interval is \(\left( { - 100,0} \right)\).

Check the hypotheses of the IVT.

1. \(f\left( x \right)\) is certainly continuous on \(\left( { - 100,0} \right)\) because it is the sum of continuous functions.

2. \(f\left( { - 100} \right) \ne f\left( 0 \right)\)

3. If let \(N = 0\), \(f\left( { - 100} \right) < N < f\left( 0 \right)\)

Thus, there is a number \(c\) between -100 and 0 such that \(f\left( c \right) = N = 0\).

Hence, the equation has a root.

Show that the equation does not have more than one real root.

Use Rolle’s theorem, Let the two roots be \(a\) and \(b\) then \(f\left( a \right) = 0 = f\left( b \right)\).

Since, \(f\left( x \right)\) is continuous everywhere then it continuous on \(\left( {a,b} \right)\) and the derivative \(f'\left( x \right) = 3 - 2\sin x\) exits for all \(x\), thus \(f\left( x \right)\) is differentiable everywhere and in particular on the open interval \(\left( {a,b} \right)\).

Thus, by Rolle’s theorem, there is a number \(c\) such that \(f'\left( c \right) = 0\).

But \(f'\left( x \right) = 3 - 2\sin x\) is never zero, because \(\sin x\) is always between \( - 1\) and 1. So the equation cannot have more than one root.

Hence, the equation has exactly one real root.