Example 2.27. Find the extrema of the function $y=x \sqrt[33]{(x-1)^{2}}$.
Solution. The function is defined on the entire number line. Let's compute the derivative:

$$
y^{\prime}=3 x^{2} \sqrt[3]{(x-1)^{2}}+x^{3} \cdot \frac{2}{3 \sqrt[3]{x-1}}=\frac{x^{2}}{3 \sqrt[3]{x-1}}(9 x-9+2 x)=\frac{x^{2}(11 x-9)}{3 \sqrt[3]{x-1}}
$$

The derivative $f^{\prime}$ is zero at $x=0$ and $x=\frac{9}{11}$ and does not exist at $x=1$. These points divide the number line into four intervals, in each of which $f^{\prime}$ maintains a certain sign: $(-\infty, 0), 0, \frac{9}{11}, \frac{9}{11}, 1,(1,+\infty)$. Let's find the sign of the derivative $y^{\prime}$ in the obtained intervals:
in the interval $(-\infty, 0)$ we have $f^{\prime \prime}(-1)>0$;
> >
$0, \frac{9}{11}$
>
$y^{\prime} \frac{1}{3}>0$
$>\quad>$
$\frac{9}{11}, 1$
>
$y^{\prime} \frac{10}{11}\quad>$
$(1,+\infty)$
>
$y^{\prime}(5)>0$.

The points $x_{1}=\frac{9}{11}$ and $x_{2}=1$ are extremal points, where $x_{1}=\frac{9}{11}$ is a point of maximum and $x_{2}=1$ is a point of minimum. The extrema of the function are obtained by calculating its values at the extremal points: $y \frac{9}{11}=\frac{729}{1331} \sqrt[3]{\frac{4}{121}} \approx 0.176-$ maximum and $y(1)=0-$ minimum of the function.

A curve is called convex (concave) in some interval if it lies below (above) the tangent line drawn to the curve at any point in that interval. The convexity or concavity of a curve, which is the graph of a function $y=f(x)$, is characterized by the sign of the second derivative $f^{\prime \prime}(x)$, specifically, if in some interval $f^{\prime \prime}(x)0\right]$, then the curve is convex (concave) in that interval.

Thus, finding the intervals of convexity and concavity of the graph of a function $y=f(x)$ reduces to finding the intervals of sign constancy of its second derivative $f^{\prime \prime}(x)$.

A point of inflection of a curve is a point that separates a segment of convexity from a segment of concavity.

Points of inflection of the graph of a function $y=f(x)$ can only be points whose abscissas are critical points of the second kind, i.e., points located within the domain of the function $y=f(x)$, at which the second derivative $f^{\prime \prime}(x)$ is zero or has a discontinuity.

Points of inflection of the graph of a function $y=f(x)$ are only those of the specified points, at which the second derivative $f^{\prime \prime}(x)$ changes sign when passing through them.

From this, we obtain the rule for finding the intervals of convexity and concavity and the points of inflection of the graph of a function:

1. Find the points at which the second derivative $f^{\prime \prime}(x)$ is zero or has a discontinuity.
2. Determine the sign of $f^{\prime \prime}(x)$ in the intervals into which the points obtained in step 1 divide the domain of $f(x)$; intervals where $f^{\prime \prime}(x)0$, - intervals of concavity of the graph of the function $y=f(x)$. If on two adjacent intervals, the boundary point of which is a zero of the second derivative $f^{\prime \prime}(x)$, the sign of $f^{\prime \prime}(x)$ is the same, then they form a single interval of convexity or concavity.
3. From the points obtained in step 1, select those at which the function $f(x)$ is defined and on either side of which the second derivative $f^{\prime \prime}(x)$ has opposite signs - these are the abscissas of the points of inflection of the graph of the function $y=f(x)$.