Lösung 2.3:10b
Aus Online Mathematik Brückenkurs 1
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|align="center"|[[Image:2_3_10_b1-1.gif|center]] | |align="center"|[[Image:2_3_10_b1-1.gif|center]] | ||
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|align="center"|<small>The region y ≤ 1 - ''x''²</small> | |align="center"|<small>The region y ≤ 1 - ''x''²</small> | ||
Version vom 14:45, 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.
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| 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.
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| 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
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
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.
