Lösung 2.3:6c

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Lösning 2.3:6c moved to Solution 2.3:6c: Robot: moved page)
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
+
If we complete the square of the expression, we have that
-
<center> [[Image:2_3_6c.gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
 
 +
<math>\begin{align}
 +
& x^{2}-5x+7=\left( x-\frac{5}{2} \right)^{2}-\left( \frac{5}{2} \right)^{2}+7 \\
 +
& =\left( x-\frac{5}{2} \right)^{2}-\frac{25}{4}+\frac{28}{4}=\left( x-\frac{5}{2} \right)^{2}+\frac{3}{4} \\
 +
\end{align}</math>
 +
 
 +
 
 +
and because
 +
<math>\left( x-\frac{5}{2} \right)^{2}</math>
 +
is a quadratic, this term is at least equal to zero when
 +
<math>x={5}/{2}\;</math>. This shows that the polynomial's smallest value is
 +
<math>\frac{3}{4}</math>.

Version vom 11:03, 21. Sep. 2008

If we complete the square of the expression, we have that


\displaystyle \begin{align} & x^{2}-5x+7=\left( x-\frac{5}{2} \right)^{2}-\left( \frac{5}{2} \right)^{2}+7 \\ & =\left( x-\frac{5}{2} \right)^{2}-\frac{25}{4}+\frac{28}{4}=\left( x-\frac{5}{2} \right)^{2}+\frac{3}{4} \\ \end{align}


and because \displaystyle \left( x-\frac{5}{2} \right)^{2} is a quadratic, this term is at least equal to zero when \displaystyle x={5}/{2}\;. This shows that the polynomial's smallest value is \displaystyle \frac{3}{4}.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.3:6c“