Solution 2.3:7b - Revision history

Tek at 11:55, 29 September 2008

Tek — Mon, 29 Sep 2008 11:55:53 GMT

←Older revision		Revision as of 11:55, 29 September 2008
Line 1:		Line 1:
-	We rewrite the expression by completing the square:	+	We rewrite the expression by completing the square,

		+	{{Displayed math\|\|<math>\begin{align}
		+	-x^{2}+3x-4
		+	&= -\bigl(x^{2}-3x+4\bigr)\\[5pt]
		+	&= -\Bigl(\Bigl(x-\frac{3}{2}\Bigr)^{2}-\Bigl(\frac{3}{2}\Bigr)^{2}+4\Bigr)\\[5pt]
		+	&= -\Bigl(\Bigl(x-\frac{3}{2}\Bigr)^{2}-\frac{9}{4}+\frac{16}{4}\Bigr)\\[5pt]
		+	&= -\Bigl(\Bigl(x-\frac{3}{2}\Bigr)^{2}+\frac{7}{4}\Bigr)\\[5pt]
		+	&= -\Bigl(x-\frac{3}{2}\Bigr)^{2}-\frac{7}{4}\,\textrm{.}
		+	\end{align}</math>}}

-	~~<math>\begin{align}~~	+	Now, we see that the first term <math>-(x-\tfrac{3}{2})^{2}</math> is a quadratic with a minus sign in front, so that term is always less than or equal to zero. This means that the polynomial's largest value is <math>-7/4</math> and that occurs when <math>x-\tfrac{3}{2}=0\,</math>, i.e. <math>x=\tfrac{3}{2}\,</math>.
-	~~& -x^{2}+3x-4=-\left( x^{2}-3x+4 \right)=-\left( \left( x-\frac{3}{2} \right)^{2}-\left( \frac{3}{2} \right)^{2}+4 \right) \\~~	+
-	~~& =-\left( \left( x-\frac{3}{2} \right)^{2}-\frac{9}{4}+\frac{16}{4} \right)=-\left( \left( x-\frac{3}{2} \right)^{2}+\frac{7}{4} \right)=-\left( x-\frac{3}{2} \right)^{2}-\frac{7}{4} \\~~	+
-	~~\end{align}</math>~~	+
-		+
-		+
-	Now, we see that the first term	+
-	<math>-~~\left~~( x-\~~frac~~{3}{2} ~~\right~~)^{2}</math>	+
-	is a quadratic with a minus sign in front, so that term is always less than or equal to zero. This means that the polynomial's largest value is	+
-	<math>-{7}/{4~~}\;~~</math>	+
-	and that occurs when	+
-	<math>x-\~~frac~~{3}{2}=0</math>, i.e.	+
-	<math>x=\~~frac~~{3}{2}</math>.	+

Ian at 11:14, 21 September 2008

Ian — Sun, 21 Sep 2008 11:14:53 GMT

←Older revision		Revision as of 11:14, 21 September 2008
Line 1:		Line 1:
-	{{~~NAVCONTENT_START~~}}	+	We rewrite the expression by completing the square:
-	<~~center~~> ~~[[Image:2_3_7b~~.~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	<math>\begin{align}
		+	& -x^{2}+3x-4=-\left( x^{2}-3x+4 \right)=-\left( \left( x-\frac{3}{2} \right)^{2}-\left( \frac{3}{2} \right)^{2}+4 \right) \\
		+	& =-\left( \left( x-\frac{3}{2} \right)^{2}-\frac{9}{4}+\frac{16}{4} \right)=-\left( \left( x-\frac{3}{2} \right)^{2}+\frac{7}{4} \right)=-\left( x-\frac{3}{2} \right)^{2}-\frac{7}{4} \\
		+	\end{align}</math>
		+
		+
		+	Now, we see that the first term
		+	<math>-\left( x-\frac{3}{2} \right)^{2}</math>
		+	is a quadratic with a minus sign in front, so that term is always less than or equal to zero. This means that the polynomial's largest value is
		+	<math>-{7}/{4}\;</math>
		+	and that occurs when
		+	<math>x-\frac{3}{2}=0</math>, i.e.
		+	<math>x=\frac{3}{2}</math>.

Tekbot: Lösning 2.3:7b moved to Solution 2.3:7b: Robot: moved page

Tekbot — Tue, 09 Sep 2008 09:03:23 GMT

Lösning 2.3:7b moved to Solution 2.3:7b: Robot: moved page

←Older revision

Revision as of 09:03, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 06:48:49 GMT

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

←Older revision		Revision as of 06:48, 21 August 2008
Line 1:		Line 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:2_3_7b.gif]] </center>	+	<center> [[Image:2_3_7b.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:2_3_7b.gif {{NAVCONTENT_STOP}}

Oskar — Tue, 01 Apr 2008 10:15:48 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:2_3_7b.gif </center> {{NAVCONTENT_STOP}}

New page

{{NAVCONTENT_START}}
<center> [[Bild:2_3_7b.gif]] </center>
{{NAVCONTENT_STOP}}