(b) To find the stationary point, we first find the derivative \(\frac{dy}{dx}\) and set it to zero. The derivative is:
\(\frac{dy}{dx} = 3(3x+4)^{-0.5} - 1\)
Setting \(\frac{dy}{dx} = 0\):
\(3(3x+4)^{-0.5} - 1 = 0\)
Solving for \(x\):
\(x = \frac{5}{3}\)
Substitute \(x = \frac{5}{3}\) back into the original equation to find \(y\):
\(y = 2\left(3\left(\frac{5}{3}\right) + 4\right)^{0.5} - \frac{5}{3}\)
\(y = 2\left(5 + 4\right)^{0.5} - \frac{5}{3}\)
\(y = 2 \times 3 - \frac{5}{3}\)
\(y = 6 - \frac{5}{3}\)
\(y = \frac{18}{3} - \frac{5}{3}\)
\(y = \frac{13}{3}\)
Thus, the coordinates of the stationary point are \(\left( \frac{5}{3}, \frac{13}{3} \right)\).
(c) To determine the nature of the stationary point, we find the second derivative \(\frac{d^2y}{dx^2}\):
\(\frac{d^2y}{dx^2} = -\frac{9}{2}(3x+4)^{-1.5}\)
At \(x = \frac{5}{3}\), \(\frac{d^2y}{dx^2}\) is negative, indicating that the point is a maximum.
