For the function \(f(x) = 2e^x + x^3 - 1\text{,}\) determine the number of roots and find a bracketing interval for each of them.
Solution.
The derivative \(f'(x) = 2e^x + 3x^2\) is always positive. Therefore, the function \(f\) increases on the real line. Such a function either has no roots (if its graph never crosses the \(y\)-axis), or has one root. Since \(f(x)\to \infty\) as \(x\to \infty\) and \(f(x) \to -\infty\) as \(x\to-\infty\text{,}\) it follows that the graph crosses the \(y\)-axis. We need a finite bracketing interval, since for the bisection method to work, both \(a\) and \(b\) must be finite. Since \(f(0) = 2 + 0 - 1 = 1 > 0\) it remains to find a negative value. For example, \(f(-1) = 2e^{-1} - 1 - 1 = 2(1-e)/e \lt 0\text{.}\)
Answer: one root, with a bracketing interval \([-1, 0]\text{.}\)
