Lösung 2.3:6b
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
K (Lösning 2.3:6b moved to Solution 2.3:6b: Robot: moved page) |
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- | {{ | + | By completing the square, the second degree polynomial can be rewritten as a quadratic plus a constant, and then it is relatively straightforward to read off the expression's minimum value, |
- | < | + | |
- | {{ | + | |
+ | <math>x^{2}-4x+2=\left( x-2 \right)^{2}-2^{2}+2=\left( x-2 \right)^{2}-2</math> | ||
+ | |||
+ | |||
+ | Because | ||
+ | <math>\left( x-2 \right)^{2}</math> | ||
+ | is a quadratic, this term is always larger than or equal to | ||
+ | <math>0</math> | ||
+ | and the whole expression is therefore at least equal to | ||
+ | <math>-\text{2}</math>, which occurs when | ||
+ | <math>x-\text{2}=0\text{ }</math> | ||
+ | and the quadratic is zero, i.e. | ||
+ | <math>x=\text{2}</math>. |
Version vom 10:56, 21. Sep. 2008
By completing the square, the second degree polynomial can be rewritten as a quadratic plus a constant, and then it is relatively straightforward to read off the expression's minimum value,
\displaystyle x^{2}-4x+2=\left( x-2 \right)^{2}-2^{2}+2=\left( x-2 \right)^{2}-2
Because
\displaystyle \left( x-2 \right)^{2}
is a quadratic, this term is always larger than or equal to
\displaystyle 0
and the whole expression is therefore at least equal to
\displaystyle -\text{2}, which occurs when
\displaystyle x-\text{2}=0\text{ }
and the quadratic is zero, i.e.
\displaystyle x=\text{2}.