Solution 2.3:6c

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m (Lösning 2.3:6c moved to Solution 2.3:6c: Robot: moved page)
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If we complete the square of the expression, we have that
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<center> [[Image:2_3_6c.gif]] </center>
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{{NAVCONTENT_STOP}}
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<math>\begin{align}
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& x^{2}-5x+7=\left( x-\frac{5}{2} \right)^{2}-\left( \frac{5}{2} \right)^{2}+7 \\
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& =\left( x-\frac{5}{2} \right)^{2}-\frac{25}{4}+\frac{28}{4}=\left( x-\frac{5}{2} \right)^{2}+\frac{3}{4} \\
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\end{align}</math>
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and because
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<math>\left( x-\frac{5}{2} \right)^{2}</math>
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is a quadratic, this term is at least equal to zero when
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<math>x={5}/{2}\;</math>. This shows that the polynomial's smallest value is
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<math>\frac{3}{4}</math>.

Revision as of 11:03, 21 September 2008

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


\displaystyle \begin{align} & x^{2}-5x+7=\left( x-\frac{5}{2} \right)^{2}-\left( \frac{5}{2} \right)^{2}+7 \\ & =\left( x-\frac{5}{2} \right)^{2}-\frac{25}{4}+\frac{28}{4}=\left( x-\frac{5}{2} \right)^{2}+\frac{3}{4} \\ \end{align}


and because \displaystyle \left( x-\frac{5}{2} \right)^{2} is a quadratic, this term is at least equal to zero when \displaystyle x={5}/{2}\;. This shows that the polynomial's smallest value is \displaystyle \frac{3}{4}.

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