(i) To find the \(x\)-coordinate of \(A\), set \(f'(x) = -1\):
\(3x^{\frac{1}{2}} - 2x^{-\frac{1}{2}} = -1\)
Multiply through by \(x^{\frac{1}{2}}\) to clear the fraction:
\(3x - \frac{2}{x} = -1\)
Rearrange to form a quadratic equation:
\(3x + 1 = \frac{2}{x}\)
\(3x^2 + x - 2 = 0\)
Let \(z = x^{\frac{1}{2}}\), then \(z^2 = x\):
\(3z^2 + z - 2 = 0\)
Solving this quadratic gives \(z = \frac{2}{3}\) or \(z = -1\).
Since \(z = x^{\frac{1}{2}}\), \(x = \left(\frac{2}{3}\right)^2 = \frac{4}{9}\).
(ii) To find the \(y\)-coordinate of \(A\), integrate \(f'(x)\) to find \(f(x)\):
\(f(x) = \int (3x^{\frac{1}{2}} - 2x^{-\frac{1}{2}}) \, dx\)
\(f(x) = 3 \cdot \frac{2}{3} x^{\frac{3}{2}} - 2 \cdot 2 x^{\frac{1}{2}} + c\)
\(f(x) = 2x^{\frac{3}{2}} - 4x^{\frac{1}{2}} + c\)
Given \(f(4) = 10\):
\(10 = 2(4)^{\frac{3}{2}} - 4(4)^{\frac{1}{2}} + c\)
\(10 = 16 - 8 + c\)
\(c = 2\)
Thus, \(f(x) = 2x^{\frac{3}{2}} - 4x^{\frac{1}{2}} + 2\).
Substitute \(x = \frac{4}{9}\) to find \(y\):
\(y = 2 \left(\frac{4}{9}\right)^{\frac{3}{2}} - 4 \left(\frac{4}{9}\right)^{\frac{1}{2}} + 2\)
\(y = 2 \cdot \frac{8}{27} - 4 \cdot \frac{2}{3} + 2\)
\(y = \frac{16}{27} - \frac{8}{3} + 2\)
\(y = \frac{16}{27} - \frac{72}{27} + \frac{54}{27}\)
\(y = -\frac{2}{27}\)
