#### Question

f(x) = (x \[-\] 1)^{2} + 3 in [ \[-\] 3,1] ?

#### Solution

\[\text { Given }: \hspace{0.167em} f\left( x \right) = \left( x - 1 \right)^2 + 3\]

\[ \Rightarrow f'\left( x \right) = 2\left( x - 1 \right)\]

\[\text { For a local maximum or a local minimum, we must have }\]

\[ f'\left( x \right) = 0\]

\[ \Rightarrow 2\left( x - 1 \right) = 0\]

\[ \Rightarrow x = 1\]

\[\text { Thus, the critical points of f are - 3 and }1 . \]

\[\text { Now, }\]

\[f\left( - 3 \right) = \left( - 3 - 1 \right)^2 + 3 = 16 + 3 = 19\]

\[f\left( 1 \right) = \left( 1 - 1 \right)^2 + 3 = 3\]

\[\text { Hence, the absolute maximum value when x = - 3 is 19 and the absolute minimum value when x = 1 is }3 . \]

Is there an error in this question or solution?

Solution F(X) = (X − 1)2 + 3 in [ − 3,1] ? Concept: Graph of Maxima and Minima.