Question

Show that the equation has exactly one real root.

17. \({x^3} + {e^x} = 0\)

Short answer

The required proof \({x^3} + {e^x} = 0\) is obtained.

Step-by-step solution

Given data

The given equation is \({x^3} + {e^x} = 0\).

Concept of Rolle’s Theorem and Intermediate value 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\({f^\prime }(c) = 0\).”

Intermediate Value Theorem:

“Suppose that\(f\)is continuous on the closed interval\((a\;,\;b)\)and let\(N\)be any number between\(f(a)\)and\(f(b)\), where\(f(a) \ne f(b)\).

Then, there exists a number\(c\)in\((a\;,\;b)\)such that\(f(c) = N\).”

Calculation to solve the given equation

Note that the given equation is a combination of polynomial and exponential functions.

Since the given function is polynomial, it is continuous and differentiable.

By intermediate value theorem, \(f\) will take all values in \(( - \infty ,\infty )\).

Therefore, the given equation \({x^3} + {e^x} = 0\) has at least one root.

Suppose that, if the equation has distinct real roots \(a\) and \(b\), then \(f(a) = f(b) = 0\).

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

But, here \({f^\prime }(c) = 3{c^2} + {e^c} > 0\) as \({f^\prime }(x) = 3{x^2} + {e^x}\).

Since the derivative of the function is positive, the function is increasing and do not passes the \(x\)-axis twice.

This leads to a contradiction to the assumption that the given equation has distinct real roots.

Therefore, it can be concluded that the given equation has exactly one root.

Hence the required proof is obtained.