Lösung 2.3:7b
Aus Online Mathematik Brückenkurs 1
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Version vom 14:13, 22. Okt. 2008
We rewrite the expression by completing the square,
| \displaystyle \begin{align}
-x^{2}+3x-4 &= -\bigl(x^{2}-3x+4\bigr)\\[5pt] &= -\Bigl(\Bigl(x-\frac{3}{2}\Bigr)^{2}-\Bigl(\frac{3}{2}\Bigr)^{2}+4\Bigr)\\[5pt] &= -\Bigl(\Bigl(x-\frac{3}{2}\Bigr)^{2}-\frac{9}{4}+\frac{16}{4}\Bigr)\\[5pt] &= -\Bigl(\Bigl(x-\frac{3}{2}\Bigr)^{2}+\frac{7}{4}\Bigr)\\[5pt] &= -\Bigl(x-\frac{3}{2}\Bigr)^{2}-\frac{7}{4}\,\textrm{.} \end{align} |
Now, we see that the first term \displaystyle -(x-\tfrac{3}{2})^{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-\tfrac{3}{2}=0\,, i.e. \displaystyle x=\tfrac{3}{2}\,.
