 Solution

A stationary point is a point where the function is differentiable and its first derivative vanishes. Interpret the cube root as the real cube root, and put $q(x)=x^2-8x+7$. Then $f(x)=15+(x-3)\sqrt[3]{q(x)}$. For $x\ne 1,7$, the product rule and chain rule giveThe denominator is positive wherever it is defined. Consequently the two stationary abscissae areUsing the stationary equation to simplify the value of $q$ gives $q(x_{\pm })=-2(x_{\pm }+5)/5$. Thus the complete stationary points areNumerically they are approximately $(1.46762,17.10362)$ and $(6.13238,9.84664)$. The numerator of $f^{\prime }$ is positive outside $[x_{-},x_{+}]$ and negative inside, so the first is a local maximum and the second a local minimum.

At $x=1,7$, the real function is continuous but has a vertical tangent; the factor $x-3$ does not cancel either cube-root singularity. Those points are not stationary points. This is the distinction required when finding stationary points with real cube-root factors.