(i) To find the \(x\)-coordinate of \(M\), we need to find the derivative of \(y = (x + 1) e^{-\frac{1}{3}x}\) and set it to zero to find the critical points.
Using the product rule, the derivative is:
\(\frac{dy}{dx} = -\frac{1}{3}(x + 1)e^{-\frac{1}{3}x} + e^{-\frac{1}{3}x}\)
Setting \(\frac{dy}{dx} = 0\), we solve for \(x\):
\(-\frac{1}{3}(x + 1) + 1 = 0\)
\(-\frac{1}{3}x - \frac{1}{3} + 1 = 0\)
\(-\frac{1}{3}x + \frac{2}{3} = 0\)
\(x = 2\)
(ii) To find the area of the shaded region, we integrate \(y = (x + 1) e^{-\frac{1}{3}x}\) from \(x = -1\) to \(x = 0\).
Using integration by parts, we have:
\(\int (x + 1) e^{-\frac{1}{3}x} \, dx = -3(x + 1)e^{-\frac{1}{3}x} - 9e^{-\frac{1}{3}x}\)
Evaluating from \(x = -1\) to \(x = 0\):
\(\left[-3(x + 1)e^{-\frac{1}{3}x} - 9e^{-\frac{1}{3}x}\right]_{-1}^{0}\)
\(= \left[-3(0 + 1)e^{0} - 9e^{0}\right] - \left[-3(-1 + 1)e^{\frac{1}{3}} - 9e^{\frac{1}{3}}\right]\)
\(= [-3 - 9] - [0 - 9e^{-\frac{1}{3}}]\)
\(= -12 + 9e^{-\frac{1}{3}}\)
\(= 9e^{-3} - 12\)
