Solution 2.3:10b - Revision history

Tek at 14:45, 29 September 2008

2008-09-29T14:45:22Z

←Older revision		Revision as of 14:45, 29 September 2008
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	\|align="center"\|[[Image:2_3_10_b1-1.gif\|center]]		\|align="center"\|[[Image:2_3_10_b1-1.gif\|center]]
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-	\|align="center"\|[[Image:2_3_10_b1-1.gif\|center]]	+	\|align="center"\|[[Image:2_3_10_b1-2.gif\|center]]
	\|-		\|-
	\|align="center"\|<small>The region y ≤ 1 - ''x''²</small>		\|align="center"\|<small>The region y ≤ 1 - ''x''²</small>

Tek at 14:44, 29 September 2008

2008-09-29T14:44:49Z

←Older revision		Revision as of 14:44, 29 September 2008
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-	The inequality	+	The inequality <math>y\le 1-x^{2}</math> defines the area under and on the curve
-	<math>y\le ~~\text{~~1}-x^~~{\text~~{2}}</math>	+	<math>y=1-x^{2}</math>, which is a parabola with a maximum at (0,1). We can rewrite the other inequality <math>x\ge 2y-3</math> as <math>y\le x/2+3/2</math> and it defines the area under and on the straight line <math>y=x/2+3/2</math>.
-	defines the area under and on the curve	+
-	<math>y=~~\text{~~1}-x^~~{\text~~{2}}</math>, which is a parabola with a maximum at	+
-	~~<math>\left~~( 0 ~~\right.~~,~~\left.~~ 1 ~~\right~~)~~</math>~~. We can rewrite the other inequality	+
-	<math>x\ge ~~\text{ 2}y~~-~~\text{~~3}</math>	+
-	as	+
-	<math>y\le ~~\text{ }{~~x}/{2~~}\;~~+{3}/{2~~}\;~~</math>	+
-	and it defines the area under and on the straight line	+
-	<math>y=~~\text{ }{~~x}/{2~~}\;~~+{3}/{2~~}\;~~</math>.	+


-	[[Image:2_3_10_b1.gif\|center]]	+	{\| align="center"
-	~~Of the figures above, it seems that the region associated with the parabola lies completely under the line~~	+	\|align="center"\|[[Image:2_3_10_b1-1.gif\|center]]
-	<~~math~~>y~~=\text{ }{~~x}/~~{2}\~~;~~+{3}/{~~2}\;</~~math~~>	+	\|width="10px"\|
-	~~and this means that the area under the parabola satisfies both inequalities.~~	+	\|align="center"\|[[Image:2_3_10_b1-1.gif\|center]]
		+	\|-
		+	\|align="center"\|<small>The region y ≤ 1 - ''x''²</small>
		+	\|\|
		+	\|align="center"\|<small>The region x ≥ 2''y'' - 3</small>
		+	\|}

-	[[Image:2_3_10_b2.gif\|center]]

-	~~NOTE: If you feel unsure about whether~~ the parabola ~~really does lie~~ under the line~~, i.e. that it just happens to look as though it does, we can investigate if the~~	+	Of the figures above, it seems that the region associated with the parabola lies completely under the line <math>y=x/2+3/2</math> and this means that the area under the parabola satisfies both inequalities.
-	<math>y~~</math>~~	+
-	~~-values on the line~~	+
-	~~<math>y_{\text{line}}~~=~~\text{ }{~~x}/{2~~}\;~~+{3}/{2~~}\;~~</math>	+
-	~~is always larger than~~ the ~~corresponding~~	+
-	~~<math>y</math>~~	+
-	~~-value on~~ the parabola	+
-	~~<math>y_{\text{parabola}}=\text{1}-x^{\text{2}}</math>~~	+
-	~~by studying the difference between them:~~	+


-	~~<math>y_~~{~~\text{line}}~~-~~y_{\text{parabola}}~~=~~\text{ }\frac{x}{2}+\frac{3}{2}-\left( \text{~~1}-x~~^{\text{~~2~~}} \right)~~</~~math~~>	+	{\| align="center"
		+	\|align="center"\|[[Image:2_3_10_b2.gif\|center]]
		+	\|-
		+	\|align="center"\|<small>The region y ≤ 1 - ''x''² and x ≥ 2''y'' - 3</small>
		+	\|}


-	If ~~this difference is positive regardless of how~~	+	Note: If you feel unsure about whether the parabola really does lie under the line, i.e. that it just happens to look as though it does, we can investigate if the ''y''-values on the line <math>y_{\scriptstyle\text{line}} = x/2+3/2</math> is always larger than the corresponding ''y''-value on the parabola <math>y_{\scriptstyle\text{parabola}} = 1-x^{2}</math> by studying the difference between them
-	~~<math>x</math>~~	+
-	~~is chosen~~, ~~then we know~~ that the ~~line~~'s	+
-	<math>y</math>	+
-	~~-value~~ is always ~~greater~~ than the ~~parabola~~'s	+
-	<math>y</math>	+
-	~~-value. After a little simplification and completing~~ the ~~square, we have~~	+

		+	{{Displayed math\|\|
		+	<math>y_{\scriptstyle\text{line}} - y_{\scriptstyle\text{parabola}} = \frac{x}{2} + \frac{3}{2} - (1-x^{2})\,\textrm{.}</math>}}

-	~~<math>\begin{align}~~	+	If this difference is positive regardless of how ''x'' is chosen, then we know that the line's ''y''-value is always greater than the parabola's ''y''-value. After a little simplification and completing the square, we have
-	~~& y_{\text{~~line}}-~~y_{\text{~~parabola~~}}=\text{ }\frac{x}{2}+\frac{3}{2}~~-~~\left( \text{1}-x^{\text{2}} \right)=x^{2}+\frac{1}{2}x+\frac{1}{2} \\~~	+
-	~~& =\left( x+\frac{1}{4} \right)^{2}-\left( \frac{1}{4} \right)^{2}+\frac{1}{2}=\left( x+\frac{1}{4} \right)^{2}+\frac{7}{16} \\~~	+
-	~~\end{align}</math>~~	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	y_{\scriptstyle\text{line}} - y_{\scriptstyle\text{parabola}}
		+	&= \frac{x}{2} + \frac{3}{2} - (1-x^{2})\\[5pt]
		+	&= x^{2}+\frac{1}{2}x+\frac{1}{2}\\[5pt]
		+	&= \Bigl(x+\frac{1}{4}\Bigr)^{2} - \Bigl(\frac{1}{4}\Bigr)^{2} + \frac{1}{2}\\[5pt]
		+	&= \Bigl(x+\frac{1}{4}\Bigr)^{2} + \frac{7}{16}
		+	\end{align}</math>}}

-	and this expression is always positive because	+	and this expression is always positive because <math>\tfrac{7}{16}</math> is a positive number and <math>\bigl(x+\tfrac{1}{4}\bigr)^{2}</math> is a quadratic which is never negative. In other words, the parabola is completely under the line.
-	<math>\~~frac~~{7}{16}</math>	+
-	is a positive number and	+
-	<math>\~~left~~( x+\~~frac~~{1}{4} \~~right~~)^{2}</math>	+
-	is a quadratic which is never negative. In other words, the parabola is completely under the line.	+

Ian at 12:39, 21 September 2008

2008-09-21T12:39:36Z

←Older revision		Revision as of 12:39, 21 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	The inequality
-	~~<center> [[Image:2_3_10b-1(3).gif]]~~ </~~center~~>	+	<math>y\le \text{1}-x^{\text{2}}</math>
-	{~~{NAVCONTENT_STOP~~}}	+	defines the area under and on the curve
-	{{~~NAVCONTENT_START~~}}	+	<math>y=\text{1}-x^{\text{2}}</math>, which is a parabola with a maximum at
-	<~~center~~> ~~[[Image:2_3_10b-2~~(3).~~gif]]~~ </~~center~~>	+	<math>\left( 0 \right.,\left. 1 \right)</math>. We can rewrite the other inequality
-	{{~~NAVCONTENT_STOP}~~}	+	<math>x\ge \text{ 2}y-\text{3}</math>
-	{{~~NAVCONTENT_START~~}}	+	as
-	~~<center> [[Image:2_3_10b-~~3~~(3).gif]]~~ </~~center~~>	+	<math>y\le \text{ }{x}/{2}\;+{3}/{2}\;</math>
-	{{~~NAVCONTENT_STOP~~}}	+	and it defines the area under and on the straight line
		+	<math>y=\text{ }{x}/{2}\;+{3}/{2}\;</math>.
		+
		+
	[[Image:2_3_10_b1.gif\|center]]		[[Image:2_3_10_b1.gif\|center]]
		+	Of the figures above, it seems that the region associated with the parabola lies completely under the line
		+	<math>y=\text{ }{x}/{2}\;+{3}/{2}\;</math>
		+	and this means that the area under the parabola satisfies both inequalities.
		+
	[[Image:2_3_10_b2.gif\|center]]		[[Image:2_3_10_b2.gif\|center]]
		+
		+	NOTE: If you feel unsure about whether the parabola really does lie under the line, i.e. that it just happens to look as though it does, we can investigate if the
		+	<math>y</math>
		+	-values on the line
		+	<math>y_{\text{line}}=\text{ }{x}/{2}\;+{3}/{2}\;</math>
		+	is always larger than the corresponding
		+	<math>y</math>
		+	-value on the parabola
		+	<math>y_{\text{parabola}}=\text{1}-x^{\text{2}}</math>
		+	by studying the difference between them:
		+
		+
		+	<math>y_{\text{line}}-y_{\text{parabola}}=\text{ }\frac{x}{2}+\frac{3}{2}-\left( \text{1}-x^{\text{2}} \right)</math>
		+
		+
		+	If this difference is positive regardless of how
		+	<math>x</math>
		+	is chosen, then we know that the line's
		+	<math>y</math>
		+	-value is always greater than the parabola's
		+	<math>y</math>
		+	-value. After a little simplification and completing the square, we have
		+
		+
		+	<math>\begin{align}
		+	& y_{\text{line}}-y_{\text{parabola}}=\text{ }\frac{x}{2}+\frac{3}{2}-\left( \text{1}-x^{\text{2}} \right)=x^{2}+\frac{1}{2}x+\frac{1}{2} \\
		+	& =\left( x+\frac{1}{4} \right)^{2}-\left( \frac{1}{4} \right)^{2}+\frac{1}{2}=\left( x+\frac{1}{4} \right)^{2}+\frac{7}{16} \\
		+	\end{align}</math>
		+
		+
		+	and this expression is always positive because
		+	<math>\frac{7}{16}</math>
		+	is a positive number and
		+	<math>\left( x+\frac{1}{4} \right)^{2}</math>
		+	is a quadratic which is never negative. In other words, the parabola is completely under the line.

Tekbot: Lösning 2.3:10b moved to Solution 2.3:10b: Robot: moved page

2008-09-09T08:53:43Z

Lösning 2.3:10b moved to Solution 2.3:10b: Robot: moved page

←Older revision

Revision as of 08:53, 9 September 2008

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

2008-08-21T06:43:59Z

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

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Madde at 11:53, 15 August 2008

2008-08-15T11:53:17Z

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Madde at 11:50, 15 August 2008

2008-08-15T11:50:23Z

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2008-04-01T10:54:48Z

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	\|align="center"\|[[Image:2_3_10_b1-1.gif\|center]]		\|align="center"\|[[Image:2_3_10_b1-1.gif\|center]]
	\|width="10px"\|		\|width="10px"\|
-	\|align="center"\|[[Image:2_3_10_b1-1.gif\|center]]	+	\|align="center"\|[[Image:2_3_10_b1-2.gif\|center]]
	\|-		\|-
	\|align="center"\|<small>The region y ≤ 1 - ''x''²</small>		\|align="center"\|<small>The region y ≤ 1 - ''x''²</small>