To find \(f(x)\), integrate
\(f'(x) = (3x - 1)^{-\frac{1}{3}}\).
\(f(x) = \int (3x - 1)^{-\frac{1}{3}} \, dx\)
Using substitution, let \(u = 3x - 1\), then
\(du = 3 \, dx\), so \(dx = \frac{du}{3}\).
\(f(x) = \int u^{-\frac{1}{3}} \cdot \frac{1}{3} \, du\)
\(f(x) = \frac{1}{3} \cdot \frac{u^{\frac{2}{3}}}{\frac{2}{3}} + c\)
\(f(x) = \frac{1}{2} (3x - 1)^{\frac{2}{3}} + c\)
Given \(f(3) = 1\), substitute to find \(c\):
\(1 = \frac{1}{2} (3 \cdot 3 - 1)^{\frac{2}{3}} + c\)
\(1 = \frac{1}{2} \cdot 8^{\frac{2}{3}} + c\)
\(1 = \frac{1}{2} \cdot 4 + c\)
\(1 = 2 + c\)
\(c = -1\)
Thus, \(f(x) = \frac{1}{2} (3x - 1)^{\frac{2}{3}} - 1\).
To find the y-coordinate of \(B\), set \(x = 0\):
\(y = \frac{1}{2} (3 \cdot 0 - 1)^{\frac{2}{3}} - 1\)
\(y = \frac{1}{2} (-1)^{\frac{2}{3}} - 1\)
Since \((-1)^{\frac{2}{3}} = \left((-1)^{\frac{1}{3}}\right)^2 = (-1)^2 = 1\),
\(y = \frac{1}{2} \cdot 1 - 1\)
\(y = -\frac{1}{2}\)
Therefore, the y-coordinate of \(B\) is
\(-\frac{1}{2}\).
