To locate the maxima and minima following the procedure in the textbook:

• End-points of the domain: Since the domain is (-\infty, +\infty) there are no endpoints.
• Stationary points: We set the derivative equal to zero and solve for x:
  \begin{align*} &y = (x-1)^{2/3} + x-1 \quad \text{so}\\ &\frac{dy}{dx} = \frac{2}{3(x-1)^{1/3}} + 1\\ &\frac{2}{3(x-1)^{1/3}} + 1 = 0 \text { when}\\ &\frac{2}{3(x-1)^{1/3}} = -1, \text { so cross-multiplying gives}\\ &2 = -3(x-1)^{1/3} \text{ or}\\ &(x-1)^{1/3} = -\frac{2}{3}\\ &x-1 = -\frac{8}{27} \quad \text{(cubing both sides)}\\ &x = 1- \frac{8}{27} = \frac{19}{27} \end{align*}

  The corresponding y-coordinate is obtained by substituting this value of x in the function:

  \begin{align*} y &= (x-1)^{2/3} + x-1\\ &= \Bigl(\frac{19}{27}-1\Bigr)^{2/3} + \frac{19}{27}-1\\ &= \frac{4}{27} \end{align*}
  We can see on the graph that this stationary point is a relative maximum.
• Singular points: These are isolated points where the derivative is not defined. We already have the derivative:
  \frac{dy}{dx} = \frac{2}{3(x-1)^{1/3}} + 1
  which is not defined when x = 1, giving this as a singular point. When x = 1,\ y = (1-1)^{2/3} + 1-1 = 0, so we have a singular point at (1, 0), which we see is a relative minimum.

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