Question

Show that the equation has exactly one real solution.

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

Short answer

It is proved that there exists only one real solution for the given equation.

Step-by-step solution

Mean Value Theorem and Rolle’s Theorem

The Mean Value Theorem for any function \(f\left( x \right)\) signifies that there exists a number \(c\) in a domain \(\left( {a,b} \right)\) such that:

\({\rm{for }}c \in \left( {a,b} \right) \to f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\). Whereas Rolle’s theorem states there exist a number \(r\) in a domain \(\left( {a,b} \right)\) such that:

\({\rm{for r}} \in \left( {a,b} \right) \to f'\left( r \right) = 0\).

Solving the given function as per the theorems:

The given equation is:

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

Let, \(f\left( x \right) = {x^3} + {e^x}\),

We see the function is the summation of a polynomialand an exponentialfunction.

Then, it is differentiable as well as continuous for all values of \(x\).

Suppose the function is having two distinct real solutions\(a{\rm{ and }}b\,\,\left( {a < b} \right)\).

Then:

\(f\left( a \right) = f\left( b \right) = 0\)

And if they are continuous as well differentiable in the interval \(\left( {a,b} \right)\), then we have \(r \in \left( {a,b} \right)\) such that:

\(f'\left( r \right) = 0\)

But,

\(f'\left( r \right) = 3{r^2} + {e^r} > 0\)

This contradictsthe function is having two distinct real solutions.

Hence, there exists only one real solution for the given equation.