Question 50

Code Analysis:

double f(double x){
    if( abs(x*x - 3) < 0.01) return x;
    else return f(x/2 + 1.5/x);
}

Condition for $f(q) = q$:

• For $f(q)$ to return $q$ immediately (or as a fixed point), the initial value $q$ must satisfy the base condition:
  $$|q^2 - 3| < 0.01.$$
• This inequality translates to:
  $$-0.01 < q^2 - 3 < 0.01 \iff 2.99 < q^2 < 3.01.$$

Evaluating $q$:

• Taking positive roots:
  $$\sqrt{2.99} \approx 1.72916, \qquad \sqrt{3.01} \approx 1.73493.$$
• Since $\sqrt{3} \approx 1.73205$, setting $q = 1.73$ satisfies:
  $$1.73^2 = 2.9929 \implies |2.9929 - 3| = 0.0071 < 0.01.$$
• Thus, $f(1.73)$ immediately returns $1.73$.
• Any value $q$ in the range $[1.72, 1.74]$ (or $[-1.74, -1.72]$) is accepted by the official answer key.