To find \(f(x)\), we need to integrate \(f'(x) = 3x^{\frac{1}{2}} + 3x^{-\frac{1}{2}} - 10\).

Integrating term by term:

\(\int 3x^{\frac{1}{2}} \, dx = 2x^{\frac{3}{2}} + C_1\)

\(\int 3x^{-\frac{1}{2}} \, dx = 6x^{\frac{1}{2}} + C_2\)

\(\int -10 \, dx = -10x + C_3\)

Thus, \(f(x) = 2x^{\frac{3}{2}} + 6x^{\frac{1}{2}} - 10x + c\).

We know the curve passes through the point \((4, -7)\), so substitute \(x = 4\) and \(f(x) = -7\):

\(-7 = 2(4)^{\frac{3}{2}} + 6(4)^{\frac{1}{2}} - 10(4) + c\)

\(-7 = 16 + 12 - 40 + c\)

\(-7 = -12 + c\)

\(c = 5\)

Therefore, \(f(x) = 2x^{\frac{3}{2}} + 6x^{\frac{1}{2}} - 10x + 5\).