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# Find the absolute maximum and absolute minimum values of $f$ on the given interval: $\;f(x)=\sqrt{9-x^{2}} ,[-1 , 2]$

Note: This is part 4th of a 7 part question, split as 7 separate questions here.

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.
$f(x)=\sqrt {9-x^2}$ is continous in $[-1,2]$
Step 1:
$f'(x)= \large\frac{-2x}{2 \sqrt {9-x^2}}=\large\frac{-x}{\sqrt {9-x^2}}$ is defined in $[-1,2]$