Lösung 2.3:6c
Aus Online Mathematik Brückenkurs 1
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If we complete the square of the expression, we have that | If we complete the square of the expression, we have that | ||
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x^{2} - 5x + 7 &= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \Bigl(\frac{5}{2}\Bigr)^{2} + 7\\[5pt] | x^{2} - 5x + 7 &= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \Bigl(\frac{5}{2}\Bigr)^{2} + 7\\[5pt] | ||
&= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \frac{25}{4} + \frac{28}{4}\\[5pt] | &= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \frac{25}{4} + \frac{28}{4}\\[5pt] | ||
Version vom 08:34, 22. Okt. 2008
If we complete the square of the expression, we have that
| \displaystyle \begin{align}
x^{2} - 5x + 7 &= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \Bigl(\frac{5}{2}\Bigr)^{2} + 7\\[5pt] &= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \frac{25}{4} + \frac{28}{4}\\[5pt] &= \Bigl(x-\frac{5}{2}\Bigr)^{2} + \frac{3}{4} \end{align} |
and because \displaystyle \bigl(x-\tfrac{5}{2}\bigr)^{2} is a quadratic, this term assumes a minimal value zero when \displaystyle x=5/2\,. This shows that the polynomial's smallest value is \displaystyle \tfrac{3}{4}.
