Solution 2.3:7c

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Current revision (11:58, 29 September 2008) (edit) (undo)
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If we complete the square
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If we complete the square,
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{{Displayed math||<math>x^{2}+x+1=\Bigl(x+\frac{1}{2}\Bigr)^{2}-\Bigl(\frac{1}{2} \Bigr)^{2}+1 = \Bigl(x+\frac{1}{2}\Bigr)^{2} + \frac{3}{4}\,,</math>}}
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<math>x^{2}+x+1=\left( x+\frac{1}{2} \right)^{2}-\left( \frac{1}{2} \right)^{2}+1=\left( x+\frac{1}{2} \right)^{2}+\frac{3}{4}</math>
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we see on the right-hand side that we can make the expression arbitrarily large simply by choosing <math>x+\tfrac{1}{2}</math> sufficiently large. Hence, there is no maximum value.
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we see on the right-hand side that we can make the expression arbitrarily large simply by choosing
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<math>x+\frac{1}{2}</math>
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sufficiently large. Hence, there is no maximum value.
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Current revision

If we complete the square,

\displaystyle x^{2}+x+1=\Bigl(x+\frac{1}{2}\Bigr)^{2}-\Bigl(\frac{1}{2} \Bigr)^{2}+1 = \Bigl(x+\frac{1}{2}\Bigr)^{2} + \frac{3}{4}\,,

we see on the right-hand side that we can make the expression arbitrarily large simply by choosing \displaystyle x+\tfrac{1}{2} sufficiently large. Hence, there is no maximum value.

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