From Förberedande kurs i matematik 1

Revision as of 10:14, 1 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:2_3_6c.gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (11:45, 29 September 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	If we complete the square of the expression, we have that
-	<center> [[Bild:2_3_6c.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\begin{align}
		+	x^{2} - 5x + 7 &= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \Bigl(\frac{5}{2}\Bigr)^{2} + 7\\[5pt]
		+	&= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \frac{25}{4} + \frac{28}{4}\\[5pt]
		+	&= \Bigl(x-\frac{5}{2}\Bigr)^{2} + \frac{3}{4}
		+	\end{align}</math>}}
		+
		+	and because <math>\bigl(x-\tfrac{5}{2}\bigr)^{2}</math> is a quadratic, this term assumes a minimal value zero when <math>x=5/2\,</math>. This shows that the polynomial's smallest value is <math>\tfrac{3}{4}</math>.

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Current revision

If we complete the square of the expression, we have that

\displaystyle \begin{align} x^{2} - 5x + 7 &= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \Bigl(\frac{5}{2}\Bigr)^{2} + 7\\[5pt] &= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \frac{25}{4} + \frac{28}{4}\\[5pt] &= \Bigl(x-\frac{5}{2}\Bigr)^{2} + \frac{3}{4} \end{align}

and because \displaystyle \bigl(x-\tfrac{5}{2}\bigr)^{2} is a quadratic, this term assumes a minimal value zero when \displaystyle x=5/2\,. This shows that the polynomial's smallest value is \displaystyle \tfrac{3}{4}.