Question

Show that the equation \({x^3} - 15x + c = 0\) has at most one root in the interval \(( - 2\;,\;2)\).

Short answer

The given equation has at most one real roots in \(( - 2,2)\).

Step-by-step solution

Given data

The equation is \({x^3} - 15x + c = 0\).

Concept of Rolle’s Theorem

“Let a function\(f\)satisfies the following conditions,

1. A function\(f\)is continuous on the closed interval\((a\;,\;b)\).

2. A function\(f\)is differentiable on the open interval\((a\;,\;b)\).

3.\(f(a) = f(b)\)

Then, there is a number\(c\)in open interval\((a\;,\;b)\)such that\({f^\prime }(c) = 0\).”

Calculation to show the given equation has at most one real root

Suppose, \(f(x)\) has two real distinct roots \(a\) and \(b\) where \(a < b\) on the interval \(( - 2,2)\).

Then, \(f(a) = f(b) = 0\).

Since the polynomial is continuous on the closed interval \((a,b)\) and differentiable on open interval \((a,b)\), then by Rolle's Theorem, there exist a number \(c\) in \((a,b)\) such that \({f^\prime }(c) = 0\).

Obtain the derivative of \(f(x)\).

\(\begin{aligned}{c}{f^\prime }(x) &= \frac{d}{{dx}}\left( {{x^3} - 15x + c} \right)\\{f^\prime }(x) &= 3{x^2} - 15\end{aligned}\)

Substitute, \(x = c\).

\({f^\prime }(c) = 3{c^2} - 15\)

Since, \(c \in (a,b)\), we have \(|c| < 2\). So \(|c{|^2} < {(2)^2}\) that is \({c^2} < 4\).

\(\begin{aligned}{c}{f^\prime }(c) &= 3{c^2} - 15\\{f^\prime }(c) &< 3(4) - 15\\{f^\prime }(c) &< 12 - 15\\{f^\prime }(c) &< - 3\end{aligned}\)

Thus,\({f^\prime }(c) \ne 0\).
This contradicts the fact of Rolle's Theorem that \({f^\prime }(c) = 0\).

Therefore, it can be concluded that the given equation cannot have two real roots.

Hence, the given equation has at most one real roots in \(( - 2,2)\).