Lösung 2.3:10b

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The inequality
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The inequality <math>y\le 1-x^{2}</math> defines the area under and on the curve
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<math>y\le \text{1}-x^{\text{2}}</math>
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<math>y=1-x^{2}</math>, which is a parabola with a maximum at (0,1). We can rewrite the other inequality <math>x\ge 2y-3</math> as <math>y\le x/2+3/2</math> and it defines the area under and on the straight line <math>y=x/2+3/2</math>.
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defines the area under and on the curve
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<math>y=\text{1}-x^{\text{2}}</math>, which is a parabola with a maximum at
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<math>\left( 0 \right.,\left. 1 \right)</math>. We can rewrite the other inequality
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<math>x\ge \text{ 2}y-\text{3}</math>
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as
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<math>y\le \text{ }{x}/{2}\;+{3}/{2}\;</math>
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and it defines the area under and on the straight line
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<math>y=\text{ }{x}/{2}\;+{3}/{2}\;</math>.
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[[Image:2_3_10_b1.gif|center]]
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{| align="center"
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Of the figures above, it seems that the region associated with the parabola lies completely under the line
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|align="center"|[[Image:2_3_10_b1-1.gif|center]]
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<math>y=\text{ }{x}/{2}\;+{3}/{2}\;</math>
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|width="10px"|&nbsp;
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and this means that the area under the parabola satisfies both inequalities.
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|align="center"|[[Image:2_3_10_b1-1.gif|center]]
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|-
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|align="center"|<small>The region y&nbsp;≤&nbsp;1&nbsp;-&nbsp;''x''²</small>
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|align="center"|<small>The region x&nbsp;≥&nbsp;2''y''&nbsp;-&nbsp;3</small>
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|}
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[[Image:2_3_10_b2.gif|center]]
 
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NOTE: If you feel unsure about whether the parabola really does lie under the line, i.e. that it just happens to look as though it does, we can investigate if the
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Of the figures above, it seems that the region associated with the parabola lies completely under the line <math>y=x/2+3/2</math> and this means that the area under the parabola satisfies both inequalities.
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<math>y</math>
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-values on the line
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<math>y_{\text{line}}=\text{ }{x}/{2}\;+{3}/{2}\;</math>
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is always larger than the corresponding
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<math>y</math>
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-value on the parabola
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<math>y_{\text{parabola}}=\text{1}-x^{\text{2}}</math>
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by studying the difference between them:
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<math>y_{\text{line}}-y_{\text{parabola}}=\text{ }\frac{x}{2}+\frac{3}{2}-\left( \text{1}-x^{\text{2}} \right)</math>
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{| align="center"
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|align="center"|[[Image:2_3_10_b2.gif|center]]
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|-
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|align="center"|<small>The region y&nbsp;≤&nbsp;1&nbsp;-&nbsp;''x''² and x&nbsp;≥&nbsp;2''y''&nbsp;-&nbsp;3</small>
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|}
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If this difference is positive regardless of how
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Note: If you feel unsure about whether the parabola really does lie under the line, i.e. that it just happens to look as though it does, we can investigate if the ''y''-values on the line <math>y_{\scriptstyle\text{line}} = x/2+3/2</math> is always larger than the corresponding ''y''-value on the parabola <math>y_{\scriptstyle\text{parabola}} = 1-x^{2}</math> by studying the difference between them
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<math>x</math>
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is chosen, then we know that the line's
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<math>y</math>
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-value is always greater than the parabola's
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<math>y</math>
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-value. After a little simplification and completing the square, we have
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{{Displayed math||
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<math>y_{\scriptstyle\text{line}} - y_{\scriptstyle\text{parabola}} = \frac{x}{2} + \frac{3}{2} - (1-x^{2})\,\textrm{.}</math>}}
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<math>\begin{align}
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If this difference is positive regardless of how ''x'' is chosen, then we know that the line's ''y''-value is always greater than the parabola's ''y''-value. After a little simplification and completing the square, we have
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& y_{\text{line}}-y_{\text{parabola}}=\text{ }\frac{x}{2}+\frac{3}{2}-\left( \text{1}-x^{\text{2}} \right)=x^{2}+\frac{1}{2}x+\frac{1}{2} \\
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& =\left( x+\frac{1}{4} \right)^{2}-\left( \frac{1}{4} \right)^{2}+\frac{1}{2}=\left( x+\frac{1}{4} \right)^{2}+\frac{7}{16} \\
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\end{align}</math>
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{{Displayed math||<math>\begin{align}
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y_{\scriptstyle\text{line}} - y_{\scriptstyle\text{parabola}}
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&= \frac{x}{2} + \frac{3}{2} - (1-x^{2})\\[5pt]
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&= x^{2}+\frac{1}{2}x+\frac{1}{2}\\[5pt]
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&= \Bigl(x+\frac{1}{4}\Bigr)^{2} - \Bigl(\frac{1}{4}\Bigr)^{2} + \frac{1}{2}\\[5pt]
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&= \Bigl(x+\frac{1}{4}\Bigr)^{2} + \frac{7}{16}
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\end{align}</math>}}
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and this expression is always positive because
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and this expression is always positive because <math>\tfrac{7}{16}</math> is a positive number and <math>\bigl(x+\tfrac{1}{4}\bigr)^{2}</math> is a quadratic which is never negative. In other words, the parabola is completely under the line.
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<math>\frac{7}{16}</math>
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is a positive number and
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<math>\left( x+\frac{1}{4} \right)^{2}</math>
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is a quadratic which is never negative. In other words, the parabola is completely under the line.
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Version vom 14:44, 29. Sep. 2008

The inequality \displaystyle y\le 1-x^{2} defines the area under and on the curve \displaystyle y=1-x^{2}, which is a parabola with a maximum at (0,1). We can rewrite the other inequality \displaystyle x\ge 2y-3 as \displaystyle y\le x/2+3/2 and it defines the area under and on the straight line \displaystyle y=x/2+3/2.


 
The region y ≤ 1 - x² The region x ≥ 2y - 3


Of the figures above, it seems that the region associated with the parabola lies completely under the line \displaystyle y=x/2+3/2 and this means that the area under the parabola satisfies both inequalities.


The region y ≤ 1 - x² and x ≥ 2y - 3


Note: If you feel unsure about whether the parabola really does lie under the line, i.e. that it just happens to look as though it does, we can investigate if the y-values on the line \displaystyle y_{\scriptstyle\text{line}} = x/2+3/2 is always larger than the corresponding y-value on the parabola \displaystyle y_{\scriptstyle\text{parabola}} = 1-x^{2} by studying the difference between them

Vorlage:Displayed math

If this difference is positive regardless of how x is chosen, then we know that the line's y-value is always greater than the parabola's y-value. After a little simplification and completing the square, we have

Vorlage:Displayed math

and this expression is always positive because \displaystyle \tfrac{7}{16} is a positive number and \displaystyle \bigl(x+\tfrac{1}{4}\bigr)^{2} is a quadratic which is never negative. In other words, the parabola is completely under the line.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.3:10b“