From Förberedande kurs i matematik 1

Revision as of 09:03, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 2.3:7c moved to Solution 2.3:7c: Robot: moved page) ← Previous diff		Current revision (11:58, 29 September 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	If we complete the square,
-	<center> [[Image:2_3_7c.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{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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		+	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.

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Current revision

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.