Lösung 2.3:6a

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Version vom 14:12, 22. Okt. 2008

Using the rule \displaystyle (a+b)^2=a^2+2ab+b^2, we recognize the polynomial as the expansion of \displaystyle (x-1)^{2}\,,

\displaystyle x^{2}-2x+1 = (x-1)^{2}\,\textrm{.}

This quadratic expression has its smallest value, zero, when \displaystyle x-1=0, i.e. \displaystyle x=1. All non-zero values of \displaystyle x-1 give a positive value for \displaystyle (x-1)^{2}.


Note: If we draw the curve \displaystyle y=(x-1)^{2}, we see that it has a minimum value of zero at \displaystyle x=1\,.