Prove that there is exactly one real value of \(x\) satisfying \(x^3 = 29 - x\).
<1>2. The function \(f\) has at least one real zero.
<1>3. The function \(f\) is strictly increasing on \(\mathbb R\).
<1>4. Therefore \(f\) has exactly one real zero, so the original equation has exactly one real solution.
Solution
1. Define\[ f(x)=x^3+x-29. \]Then the equation \(x^3=29-x\) is equivalent to \(f(x)=0\).
The polynomial \(f\) is continuous, and \[ f(2)=8+2-29=-19<0, \qquad f(3)=27+3-29=1>0. \] By the Intermediate Value Theorem, there exists \(c\in(2,3)\) with \(f(c)=0\).
For every \(x\in\mathbb R\), \[ f'(x)=3x^2+1>0. \] Hence \(f\) is strictly increasing.
2. A strictly increasing function can take the value \(0\) at most once, giving uniqueness by
3.