1. a) Determine the set of points where the function from the family defined by the formula $f(x)=\frac{4 x^{3}}{p^{2}}-3 x+p$, where $p \neq 0$, reaches a local maximum and the set of points where the function from this family reaches a local minimum. Sketch the graph of the function obtained for $p=1$ and determine the points of intersection of this graph with the $x$-axis.
b) Express $\cos 3 t$ as a function of $\cos t$ and, using the substitution $x=\cos t$, determine the zeros of the function $f(x)=4 x^{3}-3 x+\frac{\sqrt{2}}{2}$.
Solution. a) The function $f(x)=\frac{4 x^{3}}{p^{2}}-3 x+p$, where $p \neq 0$, has a local maximum at the point $-\frac{p}{2}$ equal to $2 p$, and a local minimum at the point $\frac{p}{2}$ equal to 0. The set of maximum points is $\{(x, y) \mid y=-4 x, x \neq 0\}$, and the set of minimum points is $\{(x, 0) \mid x \neq 0\}$.

The graph of the function $f(x)=4 x^{3}-3 x+1$ is shown in the diagram on the right.

![](https://cdn.mathpix.com/cropped/2024_06_05_faf575a75202fd173f3cg-06.jpg?height=469&width=495&top_left_y=419&top_left_x=988)

b) Since $\cos 3 t=4 \cos ^{3} t-3 \cos t$, by substituting $x=\cos t$ we get

$$
f(x)=f(\cos t)=4 \cos ^{3} t-3 \cos t+\frac{\sqrt{2}}{2}=\cos 3 t+\frac{\sqrt{2}}{2} .
$$

The set of solutions to the equation $\cos 3 t=-\frac{\sqrt{2}}{2}$ is

$$
S=\left\{\left.\frac{\pi}{4}+\frac{2 k \pi}{3} \right\rvert\, k \in \mathbb{Z}\right\} \cup\left\{\left.\frac{5 \pi}{12}+\frac{2 k \pi}{3} \right\rvert\, k \in \mathbb{Z}\right\}=\bigcup_{j=1}^{6}\left\{t_{j}+2 k \pi \mid k \in \mathbb{Z}\right\},
$$

where $t_{1}=\frac{\pi}{4}, t_{2}=\frac{5 \pi}{12}, t_{3}=\frac{11 \pi}{12}, t_{4}=\frac{13 \pi}{12}, t_{5}=\frac{19 \pi}{12}, t_{6}=\frac{21 \pi}{12}$. It is easily verified that $\cos t_{1}=\cos t_{6}, \cos t_{2}=\cos t_{5}, \cos t_{3}=\cos t_{4}$. Therefore, the equation $4 x^{3}-3 x+\frac{\sqrt{2}}{2}=0$ has three solutions: $x_{1}=\cos \frac{\pi}{4}, x_{2}=\cos \frac{5 \pi}{12}, x_{3}=\cos \frac{11 \pi}{12}$.