Lösung 2.3:7c

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If we complete the square,
If we complete the square,
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{{Displayed math||<math>x^{2}+x+1=\Bigl(x+\frac{1}{2}\Bigr)^{2}-\Bigl(\frac{1}{2} \Bigr)^{2}+1 = \Bigl(x+\frac{1}{2}\Bigr)^{2} + \frac{3}{4}\,,</math>}}
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{{Abgesetzte Formel||<math>x^{2}+x+1=\Bigl(x+\frac{1}{2}\Bigr)^{2}-\Bigl(\frac{1}{2} \Bigr)^{2}+1 = \Bigl(x+\frac{1}{2}\Bigr)^{2} + \frac{3}{4}\,,</math>}}
we see on the right-hand side that we can make the expression arbitrarily large simply by choosing <math>x+\tfrac{1}{2}</math> sufficiently large. Hence, there is no maximum value.
we see on the right-hand side that we can make the expression arbitrarily large simply by choosing <math>x+\tfrac{1}{2}</math> sufficiently large. Hence, there is no maximum value.

Version vom 08:34, 22. Okt. 2008

If we complete the square,

\displaystyle x^{2}+x+1=\Bigl(x+\frac{1}{2}\Bigr)^{2}-\Bigl(\frac{1}{2} \Bigr)^{2}+1 = \Bigl(x+\frac{1}{2}\Bigr)^{2} + \frac{3}{4}\,,

we see on the right-hand side that we can make the expression arbitrarily large simply by choosing \displaystyle x+\tfrac{1}{2} sufficiently large. Hence, there is no maximum value.

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