Consider the following functiondouble f(double x) { if ( abs(x*x - 3) < 0.01) return x; else return f(x/2 + 1.5/x); }Give a value $q$ (to 2 decimals) such that $f (q)$ will return $q$: _____.

The base condition checks whether $|x^2 - 3| < 0.01$. If it is satisfied, the function returns $x$ directly without making any further recursive calls.

For $f(q)$ to immediately return $q$, the condition must hold for $q$:

Notice that the recurrence update formula $x_{n+1} = \frac{x_n}{2} + \frac{1.5}{x_n} = \frac{1}{2}\left(x_n + \frac{3}{x_n}\right)$ is the standard Newton-Raphson method for finding $\sqrt{3}$.

The exact fixed point is:

Checking $q = 1.73$:

Thus, any value in the range 1.72 to 1.74 is valid.