Lösung 2.3:7b

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Lösning 2.3:7b moved to Solution 2.3:7b: Robot: moved page)
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
+
We rewrite the expression by completing the square:
-
<center> [[Image:2_3_7b.gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
 
 +
<math>\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}</math>
 +
 
 +
 
 +
Now, we see that the first term
 +
<math>-\left( x-\frac{3}{2} \right)^{2}</math>
 +
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
 +
<math>-{7}/{4}\;</math>
 +
and that occurs when
 +
<math>x-\frac{3}{2}=0</math>, i.e.
 +
<math>x=\frac{3}{2}</math>.

Version vom 11:14, 21. Sep. 2008

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}.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.3:7b“