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

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

$f(0)=0^2-4.0+5=5$
$f(1)=1-4+5=2$
$f(2) =4 -8+5-1$
$f(3)=9-12+5=2$
$f(0) \leq f(x)$ for all $x \in [0,3]$