110
Skrytý text:First of all, g has to be zero. Indeed, by substitution (for
, otherwise
\end{align*} "kopírovat do textarea")
), we get
dt=\frac1x\int_0^xg(s)ds\end{align*} "kopírovat do textarea")
,
hence
ds=0\end{align*} "kopírovat do textarea")
and by taking the derivative, we obtain
=0\end{align*} "kopírovat do textarea")
for all
.
Thus, the exercise can be reformulated as a problem of finding a minimum of
=\int_2^4|x^4-4x-k|dx\end{align*} "kopírovat do textarea")
.
Note that
is a continuous function that satisfies
=\infty\end{align*} "kopírovat do textarea")
. Moreover, since
\end{align*} "kopírovat do textarea")
for all
and all
, where
are finite sets (the roots), we deduce that
is continusously differentiable (the integral does not see the finite sets
) and
=-\int_2^4\text{sgn}(x^4-4x-k)dx\end{align*} "kopírovat do textarea")
.
This can be zero only if the integrand changes sign in the middle of the interval, that is, if
, i.e.,
. Due to the coercivity and continuous differentiability of
, the point
is the global minimum. Hence the answer is
=-\int_2^3(x^4-4x-69)dx+\int_3^4(x^4-4x-69)=\ldots=110\end{align*} "kopírovat do textarea")
.
Remark: the only property of the polynomial
=x^4-4x\end{align*} "kopírovat do textarea")
that is used is
=69\end{align*} "kopírovat do textarea")
