Solution 2.3:7b

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m (Lösning 2.3:7b moved to Solution 2.3:7b: Robot: moved page)
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We rewrite the expression by completing the square:
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<center> [[Image:2_3_7b.gif]] </center>
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{{NAVCONTENT_STOP}}
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<math>\begin{align}
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& -x^{2}+3x-4=-\left( x^{2}-3x+4 \right)=-\left( \left( x-\frac{3}{2} \right)^{2}-\left( \frac{3}{2} \right)^{2}+4 \right) \\
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& =-\left( \left( x-\frac{3}{2} \right)^{2}-\frac{9}{4}+\frac{16}{4} \right)=-\left( \left( x-\frac{3}{2} \right)^{2}+\frac{7}{4} \right)=-\left( x-\frac{3}{2} \right)^{2}-\frac{7}{4} \\
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\end{align}</math>
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Now, we see that the first term
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<math>-\left( x-\frac{3}{2} \right)^{2}</math>
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is a quadratic with a minus sign in front, so that term is always less than or equal to zero. This means that the polynomial's largest value is
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<math>-{7}/{4}\;</math>
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and that occurs when
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<math>x-\frac{3}{2}=0</math>, i.e.
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<math>x=\frac{3}{2}</math>.

Revision as of 11:14, 21 September 2008

We rewrite the expression by completing the square:


\displaystyle \begin{align} & -x^{2}+3x-4=-\left( x^{2}-3x+4 \right)=-\left( \left( x-\frac{3}{2} \right)^{2}-\left( \frac{3}{2} \right)^{2}+4 \right) \\ & =-\left( \left( x-\frac{3}{2} \right)^{2}-\frac{9}{4}+\frac{16}{4} \right)=-\left( \left( x-\frac{3}{2} \right)^{2}+\frac{7}{4} \right)=-\left( x-\frac{3}{2} \right)^{2}-\frac{7}{4} \\ \end{align}


Now, we see that the first term \displaystyle -\left( x-\frac{3}{2} \right)^{2} is a quadratic with a minus sign in front, so that term is always less than or equal to zero. This means that the polynomial's largest value is \displaystyle -{7}/{4}\; and that occurs when \displaystyle x-\frac{3}{2}=0, i.e. \displaystyle x=\frac{3}{2}.

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