Solution 2.3:7b
From Förberedande kurs i matematik 1
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- | {{ | + | We rewrite the expression by completing the square, |
- | + | ||
- | {{ | + | {{Displayed math||<math>\begin{align} |
+ | -x^{2}+3x-4 | ||
+ | &= -\bigl(x^{2}-3x+4\bigr)\\[5pt] | ||
+ | &= -\Bigl(\Bigl(x-\frac{3}{2}\Bigr)^{2}-\Bigl(\frac{3}{2}\Bigr)^{2}+4\Bigr)\\[5pt] | ||
+ | &= -\Bigl(\Bigl(x-\frac{3}{2}\Bigr)^{2}-\frac{9}{4}+\frac{16}{4}\Bigr)\\[5pt] | ||
+ | &= -\Bigl(\Bigl(x-\frac{3}{2}\Bigr)^{2}+\frac{7}{4}\Bigr)\\[5pt] | ||
+ | &= -\Bigl(x-\frac{3}{2}\Bigr)^{2}-\frac{7}{4}\,\textrm{.} | ||
+ | \end{align}</math>}} | ||
+ | |||
+ | Now, we see that the first term <math>-(x-\tfrac{3}{2})^{2}</math> 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 <math>-7/4</math> and that occurs when <math>x-\tfrac{3}{2}=0\,</math>, i.e. <math>x=\tfrac{3}{2}\,</math>. |
Current revision
We rewrite the expression by completing the square,
\displaystyle \begin{align}
-x^{2}+3x-4 &= -\bigl(x^{2}-3x+4\bigr)\\[5pt] &= -\Bigl(\Bigl(x-\frac{3}{2}\Bigr)^{2}-\Bigl(\frac{3}{2}\Bigr)^{2}+4\Bigr)\\[5pt] &= -\Bigl(\Bigl(x-\frac{3}{2}\Bigr)^{2}-\frac{9}{4}+\frac{16}{4}\Bigr)\\[5pt] &= -\Bigl(\Bigl(x-\frac{3}{2}\Bigr)^{2}+\frac{7}{4}\Bigr)\\[5pt] &= -\Bigl(x-\frac{3}{2}\Bigr)^{2}-\frac{7}{4}\,\textrm{.} \end{align} |
Now, we see that the first term \displaystyle -(x-\tfrac{3}{2})^{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-\tfrac{3}{2}=0\,, i.e. \displaystyle x=\tfrac{3}{2}\,.