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If $f(x)=x^{2}-4x+5$ on $[0 ,3 ]$ then the absolute maximum value is

\[\begin{array}{1 1}(1)2&(2)3\\(3)4&(4)5\end{array}\]

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$f(0)=0^2-4.0+5=5$
$f(1)=1-4+5=2$
$f(2) =4 -8+5-1$
$f(3)=9-12+5=2$
$f(0) \leq f(x) $ for all $x \in [0,3]$
The absolute maximum is 5.
Hence 4 is the correct answer
answered May 20, 2014 by meena.p
 
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