Problem 3. Solve the inequality

$$
a^{5}-1.3 a^{2}-2.8 a-3.1 \geqslant 0
$$
Solution. We investigate the extremum and monotonicity of the continuous function

$$
f(a)=a^{5}-1.3 a^{2}-2.8 a-3.1
$$

We find:

$$
f^{\prime}(a)=5 a^{4}-2.6 a-2.8
$$

We determine the roots of the equation

$$
5 a^{4}-2.6 a-2.8=0
$$

as follows: we transform it to the form $5 a^{4}=$
$=2.6 a+2.8$. We sketch the graphs of the continuous functions:

$$
\varphi(a)=5 a^{4} ; u(a)=2.6 a+2.8 .
$$

Now it is clear that the equation (3) has only two roots $m$ and $n$. We refine the values of the roots using a calculator

$$
m \approx-0.65 ; n \approx 1.05
$$

Since $(-0.65)^{5}<0$ and $1.05^{5}>0$, it is clear that there exists a value of $a$ for which $f(a)>0$, i.e., $f(2)>f(n)$. Therefore, the continuous function $f(a)$ is monotonically increasing on the interval $(n ;+\infty)$, and its graph intersects the x-axis at a single point $a_{0}\left(1.05<a_{0}<2\right)$. Thus, the solutions to the inequality (1) are all $a \geqslant a_{0}$.

We determine $a_{0}$ with an accuracy of 0.001.

To simplify the calculations on the calculator, we transform $f(a)$ as follows: $f(a)=a\left(a^{4}-1.3 a-2.8\right)-3.1=$ $=a\left(a\left(a^{3}-1.3\right)-2.8\right)-3.1$.

We create a table of values for $f(a)$ using a calculator:

| $a$ | $f(a)$ |
| :---: | :---: |
| 1.5 | $-2.6312507$ |
| 1.6 | $-0.4222404$ |
| 1.7 | 2.5815688 |
| 1.61 | $-0.2379322$ |
| 1.62 | 0.1 |
| 1.615 | $-0.0261103$ |
| 1.616 | 0.0009 |

Thus, $a_{0} \approx 1.616$.