Solution 2.3:7c - Revision history

Tek at 11:58, 29 September 2008

2008-09-29T11:58:22Z

←Older revision		Revision as of 11:58, 29 September 2008
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-	If we complete the square	+	If we complete the square,

		+	{{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>}}

-	~~<math>x^{2}+x+1=\left( x+\frac{1}{2} \right)^{2}-\left( \frac{1}{2} \right)^{2}+1=\left( x+\frac{1}{2} \right)^{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+\~~frac~~{1}{2}</math>	+
-	sufficiently large. Hence, there is no maximum value.	+

2008-09-21T11:18:33Z

←Older revision		Revision as of 11:18, 21 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	If we complete the square
-	~~<center> [[Image:2_3_7c.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP}~~}	+
		+	<math>x^{2}+x+1=\left( x+\frac{1}{2} \right)^{2}-\left( \frac{1}{2} \right)^{2}+1=\left( x+\frac{1}{2} \right)^{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+\frac{1}{2}</math>
		+	sufficiently large. Hence, there is no maximum value.

2008-09-09T09:03:43Z

←Older revision

Revision as of 09:03, 9 September 2008

2008-08-21T06:48:59Z

Robot: Automated text replacement (-[[Bild: +[[Image:)

2008-04-01T10:16:57Z

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