Question

Is there a number that is exactly 1 more than its cube?

Short answer

\(c = {c^3} + 1\) and the number \(c\) is between\( - 2\) and \( - 1\).

Step-by-step solution

Intermediate Value Theorem

Consider thefunction\(f\)iscontinuous on the closed interval\(\left( {a,b} \right)\)and let\(N\)be any number between\(f\left( a \right)\)and\(f\left( b \right)\)with\(f\left( a \right) \ne f\left( b \right)\). Then there exists anumber\(c\)in \(\left( {a,b} \right)\) such that \(f\left( c \right) = N\).

Determine a number that is exactly 1 more than its cube

When there exists such a number, it satisfies the equation.

\(\begin{array}{c}{x^3} + 1 = x\\{x^3} - x + 1 = 0\end{array}\)

Consider \(f\left( x \right)\) as the left-hand side of this equation. Then \(f\left( { - 2} \right) = - 5 < 0\), and \(f\left( { - 1} \right) = 1 > 0\).

It is observed that \(f\left( x \right)\) is a polynomial. Therefore, the function \(f\left( x \right)\) is continuous.

According to the Intermediate Value Theorem, there exists a number \(c\) between \( - 2\) and \( - 1\) such that \(f\left( c \right) = 0\). Hence, \(c = {c^3} + 1\).

Thus, \(c = {c^3} + 1\) and the number \(c\) is between\( - 2\) and \( - 1\).