Lösung 2.3:10b
Aus Online Mathematik Brückenkurs 1
The inequality \displaystyle y\le \text{1}-x^{\text{2}} defines the area under and on the curve \displaystyle y=\text{1}-x^{\text{2}}, which is a parabola with a maximum at \displaystyle \left( 0 \right.,\left. 1 \right). We can rewrite the other inequality \displaystyle x\ge \text{ 2}y-\text{3} as \displaystyle y\le \text{ }{x}/{2}\;+{3}/{2}\; and it defines the area under and on the straight line \displaystyle y=\text{ }{x}/{2}\;+{3}/{2}\;.
Of the figures above, it seems that the region associated with the parabola lies completely under the line \displaystyle y=\text{ }{x}/{2}\;+{3}/{2}\; and this means that the area under the parabola satisfies both inequalities.
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 \displaystyle y -values on the line \displaystyle y_{\text{line}}=\text{ }{x}/{2}\;+{3}/{2}\; is always larger than the corresponding \displaystyle y -value on the parabola \displaystyle y_{\text{parabola}}=\text{1}-x^{\text{2}} by studying the difference between them:
\displaystyle y_{\text{line}}-y_{\text{parabola}}=\text{ }\frac{x}{2}+\frac{3}{2}-\left( \text{1}-x^{\text{2}} \right)
If this difference is positive regardless of how
\displaystyle x
is chosen, then we know that the line's
\displaystyle y
-value is always greater than the parabola's
\displaystyle y
-value. After a little simplification and completing the square, we have
\displaystyle \begin{align}
& 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} \\
& =\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} \\
\end{align}
and this expression is always positive because
\displaystyle \frac{7}{16}
is a positive number and
\displaystyle \left( x+\frac{1}{4} \right)^{2}
is a quadratic which is never negative. In other words, the parabola is completely under the line.