## Task 1 - 221241

a) Investigate whether there exist real numbers $a \neq 0, b$, and $c$ such that the function $f$ defined for all real $x$ by

$$
f(x)=a x^{4}+b x^{2}+c
$$

has the function values $f(0)=1, f(2)=1$ and has a local extremum at $x=1$!

b) Given two arbitrary real numbers $x_{1}$ and $x_{2}$ with $0<x_{1}<x_{2}$.

Determine (in dependence on $x_{1}$ and $x_{2}$) all real $a \neq 0, b, c$ with the property that the function $f$ defined by (1) has the function values $f(0)=1, f\left(x_{2}\right)=1$ and has a local extremum at $x=x_{1}$!
For part a), from $f(0)=1$ we immediately obtain the value $c=1$, and the further equations $f(2)=16 a+4 b+1=1$ and $f^{\prime}(1)=4 a+2 b=0$ then yield $a=b=0$, in contradiction to the assumption that $a \neq 0$. Thus, there is no solution here.

For b), from $\mathrm{f}(0)=1$ we again immediately obtain $c=1$. The further two equations for $a$ and $b$ then look more generally as follows:

$$
\begin{aligned}
& f\left(x_{2}\right)=a x_{2}^{4}+b x_{2}^{2}+1=1 \\
& f^{\prime}\left(x_{1}\right)=4 a x_{1}^{3}+2 b x_{1}=0
\end{aligned}
$$

Given the condition $0 < x_1 < x_2$, if $a \neq 0$ is satisfied, then $a$ can actually be arbitrary, and for $b=-a x_{2}^{2}$, the conditions of the problem are then satisfied.