Lösung 2.3:7b
Aus Online Mathematik Brückenkurs 1
We rewrite the expression by completing the square:
\displaystyle \begin{align}
& -x^{2}+3x-4=-\left( x^{2}-3x+4 \right)=-\left( \left( x-\frac{3}{2} \right)^{2}-\left( \frac{3}{2} \right)^{2}+4 \right) \\
& =-\left( \left( x-\frac{3}{2} \right)^{2}-\frac{9}{4}+\frac{16}{4} \right)=-\left( \left( x-\frac{3}{2} \right)^{2}+\frac{7}{4} \right)=-\left( x-\frac{3}{2} \right)^{2}-\frac{7}{4} \\
\end{align}
Now, we see that the first term
\displaystyle -\left( x-\frac{3}{2} \right)^{2}
is a quadratic with a minus sign in front, so that term is always less than or equal to zero. This means that the polynomial's largest value is
\displaystyle -{7}/{4}\;
and that occurs when
\displaystyle x-\frac{3}{2}=0, i.e.
\displaystyle x=\frac{3}{2}.