
Problem 3. a) Find the maximal value of the function $y=\left|4 x^{3}-3 x\right|$ in the interval $[-1,1]$. b)Let $a, b$ and $c$ be real numbers and $M$ be the maximal value of the function $y=$ $\left|4 x^{3}+a x^{2}+b x+c\right|$ in the interval $[-1,1]$. Prove that $M \geq 1$. For which $a, b, c$ is the equality reached?

Figure 3.


Solution. a) Using that $\left(4 x^{3}-3 x\right)^{\prime}=12 x^{2}-3=$ $12\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)$, we find that the function

![](https://cdn.mathpix.com/cropped/2024_06_03_2b3603673083113190ceg-2.jpg?height=300&width=528&top_left_y=1506&top_left_x=501)

$$
y=\left|4 x^{3}-3 x\right|
$$

has a local maximums when $x= \pm \frac{1}{2}$. Then its maximal value in the interval $[-1,1]$ is the biggest among the numbers $y(-1), y(1), y\left(-\frac{1}{2}\right)$ and $y\left(\frac{1}{2}\right)$ (Figure 3 ).

But $y(-1)=y\left(-\frac{1}{2}\right)=y\left(\frac{1}{2}\right)=y(1)=1$, thus the maximal value is equal to 1.

b) Let $f(x)=4 x^{3}+a x^{2}+b x+c$. Assume that there exist numbers $a, b, c$, for which the maximal value $M$ of the function $y=|f(x)|$ in $[-1,1]$ is less than 1, i.e. $M0$, $g\left(-\frac{1}{2}\right)0$ and $g(1)<0$. Consequently $g(x)$ changes its sign at least 3 times, which means that the quadratic equation $a x^{2}+(b+3) x+c=0$ has at least 3 different roots. This is possible only if $a=b+3=c=0$, i.e. if $f(x)=4 x^{3}-3 x$. According to a) the maximal value of $y=\left|4 x^{3}-3 x\right|$ is $\leq 1$.

The equality $M=1$ is reached only when $a=0, b=-3$ and $c=0$.
