Example 2.28. Find the intervals of convexity and concavity and the points of inflection of the graph of the function $f(x)=3 x^{5}+5 x^{4}-20 x^{3}+60 x-5$.
Solution. The function is defined on the entire number line. Differentiating it twice, we get

$$
\begin{gathered}
f^{\prime}(x)=15 x^{4}+20 x^{3}-60 x^{2}+60 \\
f^{\prime \prime}(x)=60 x^{3}+60 x^{2}-120 x=60 x(x-1)(x+2)
\end{gathered}
$$

The second derivative exists on the entire number line and is zero at $x=-2, x=0$, and $x=1$. These points divide the domain into four intervals: $(-\infty,-2)$, $(-2,0)$, $(0,1)$, and $(1,+\infty)$. In each of these intervals, $f^{\prime \prime}(x)$ maintains its sign. By determining the sign of the second derivative at an arbitrary point in each interval, we get the sign of the second derivative in the corresponding interval:

| in | the interval | $(-\infty,-2)$ | we have | $f^{\prime \prime}(-3)0$, |
| » | » | $(0,1)$ | » | $f^{\prime \prime} \frac{1}{2}0$. |

Thus, in the intervals $(-\infty,-2)$ and $(0,1)$, the curve is convex, and in the intervals $(-2,0)$ and $(1, \infty)$, it is concave.

The boundary points $x_{1}=-2, x_{2}=0, x_{3}=1$ of these intervals are the abscissas of the inflection points. We calculate the values of the function $y=f(x)$ at these points:

$$
f(-2)=19, \quad f(0)=-5, \quad f(1)=43
$$

Thus, the function has three inflection points: $(-2 ; 19)$, $(0 ;-5)$, and $(1 ; 43)$.

An asymptote of a curve is a line to which a point on the curve approaches infinitely as the point moves along the curve to infinity.

I. If $\lim _{x \rightarrow a} f(x)= \pm \infty$, then the line $x=a$ is an asymptote of the curve $y=f(x)$. For example, the curve $y=\frac{a}{x-a}$ has the asymptote $x=a$.

II. If the right-hand side of the equation of the curve $y=f(x)$ can be separated into a linear part $y=f(x)=k x+b+\alpha(x)$ such that the remaining part $\alpha(x) \rightarrow 0$ as $x \rightarrow \pm \infty$, then the line $y=k x+b$ is an asymptote of the curve.

Examples:

1) The curve $y=\frac{x^{3}+x^{2}+1}{x^{2}}=x+1+\frac{1}{x^{2}}$ has the asymptote $y=x+1$ (and the asymptote $x=0$);
2) The curve $y=\frac{a}{x-a}=0+\frac{a}{x-a}$ has the asymptote $y=0$.

![](https://cdn.mathpix.com/cropped/2024_05_22_e915d089cc29fd681be5g-097.jpg?height=96&width=1043&top_left_y=830&top_left_x=219)
$\lim _{x \rightarrow+\infty}[f(x)-k x]=b$, then the line $y=k x+b$ is an asymptote.

To find the maximum and minimum values of a function that is continuous on some interval $[a, b]$, we need to compute the values of the function at the endpoints of the interval and at all its critical points that belong to this interval (such points in this case are points where the first derivative of the function is zero or does not exist). The largest and smallest of the obtained values are, respectively, the maximum and minimum values of the function on the interval.

In the case where the function has discontinuities at some points in the interval $[a, b]$ or is defined on an infinite interval, it is necessary to additionally consider its behavior in the neighborhood of the points of discontinuity and as $x \rightarrow \pm \infty$.