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Find the absolute maximum and absolute minimum values of $f$ on the given interval: $\;f(x)=x^{2}-2x+2 , [0 , 3 ]$

Note: This is part 2 of a 7 part question, split as 7 separate questions here.

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Toolbox:
  • To find the absolute maximum and minimum values of a continuous function f on a closed interval $[a,b]$
  • (i) Find the values of the critical numbers of f in $(a,b)$
  • (ii) Find the value of $f(a)$ and $f(b)$
  • (iii) The largest of the values from (i) and (ii) is the absolute maximun value, the smallest of these values is the absolute minimum value.
$f(x)=x^2-2x+2$ on $[0,3]$ is continous on $[0,3]$
Step 1:
$f'(x)=2x-2$
$f'(x)=0 =>x=1$
Step 2:
The critical value is
$f(1)=1-2+2=1$
The value of f at the end points are $f(0)=2$ and $f(3)=9-6+2=5$
Step 3:
Comparing the 3 values, the absolute maximum is $f(3)=5$ and the absolute minimum is $f(1)=1$
answered Jul 31, 2013 by meena.p
 

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