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Question
f(x) = (x \[-\] 1)2 + 3 in [ \[-\] 3,1] ?
Solution
\[\text { Given }: \hspace{0.167em} f\left( x \right) = \left( x - 1 \right)^2 + 3\]
\[ \Rightarrow f'\left( x \right) = 2\left( x - 1 \right)\]
\[\text { For a local maximum or a local minimum, we must have }\]
\[ f'\left( x \right) = 0\]
\[ \Rightarrow 2\left( x - 1 \right) = 0\]
\[ \Rightarrow x = 1\]
\[\text { Thus, the critical points of f are - 3 and }1 . \]
\[\text { Now, }\]
\[f\left( - 3 \right) = \left( - 3 - 1 \right)^2 + 3 = 16 + 3 = 19\]
\[f\left( 1 \right) = \left( 1 - 1 \right)^2 + 3 = 3\]
\[\text { Hence, the absolute maximum value when x = - 3 is 19 and the absolute minimum value when x = 1 is }3 . \]
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