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Question
f(x) = x3 \[-\] 6x2 + 9x + 15 .
Solution
\[\text { Given: } f\left( x \right) = x^3 - 6 x^2 + 9x + 15\]
\[ \Rightarrow f'\left( x \right) = 3 x^2 - 12x + 9\]
\[\text { For a local maximum or a local minimum, we must have }\]
\[f'\left( x \right) = 0\]
\[ \Rightarrow 3 x^2 - 12x + 9 = 0\]
\[ \Rightarrow x^2 - 4x + 3 = 0\]
\[ \Rightarrow \left( x - 1 \right)\left( x - 3 \right) = 0\]
\[ \Rightarrow x = 1 \text { or } 3\]
Sincef '(x) changes from negative to positive when x increases through 3, x = 3 is the point of local minima.
The local minimum value of f (x) at x = 3 is given by \[\left( 3 \right)^3 - 6 \left( 3 \right)^2 + 9\left( 3 \right) + 15 = 27 - 54 + 27 + 15 = 15\]
Since f '(x) changes from positive to negative when x increases through 1, x = 1 is the point of local maxima.
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