Step-by-step Solution:

Given function: \[ f(x) = x^2 - bx + c \] Let \(\alpha\) and \(\beta\) be the roots of \( f(x) = 0 \). From Vieta's formulas: \[ \alpha + \beta = b, \quad \alpha \beta = c \] Since the roots are prime numbers and \( b \) is an odd positive integer, one of the roots must be \( 2 \). Let \( \alpha = 2 \), then \( \beta = b - 2 \). Since \( \beta \) is also a prime number, and \( b \) is odd, the only possible value for \( \beta \) is \( 11 \). Thus, \[ b = 2 + 11 = 13, \quad c = 2 \times 11 = 22 \] Finding the Global Minimum Since \( f(x) \) is a quadratic function, its minimum value occurs at \[ x = \frac{-(-b)}{2(1)} = \frac{b}{2} = \frac{13}{2} \] Substituting \( x = \frac{13}{2} \) into \( f(x) \): \[ f\left(\frac{13}{2}\right) = \left(\frac{13}{2}\right)^2 - 13 \cdot \frac{13}{2} + 22 \] \[ = \frac{169}{4} - \frac{169}{2} + 22 \] \[ = \frac{169}{4} - \frac{338}{4} + \frac{88}{4} \] \[ = \frac{169 - 338 + 88}{4} = \frac{-81}{4} \] Thus, the global minimum value of \( f(x) \) is \( \frac{-81}{4} \).