From Förberedande kurs i matematik 1

Revision as of 10:15, 1 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:2_3_7b.gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (11:55, 29 September 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	We rewrite the expression by completing the square,
-	<center> [[Bild:2_3_7b.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{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>}}
		+
		+	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>.

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Current revision

We rewrite the expression by completing the square,

\displaystyle \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}

Now, we see that the first term \displaystyle -(x-\tfrac{3}{2})^{2} 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 \displaystyle -7/4 and that occurs when \displaystyle x-\tfrac{3}{2}=0\,, i.e. \displaystyle x=\tfrac{3}{2}\,.