Question

f(x) =  x³ \[-\] 6x² + 9x + 15 . 

Answer 1

\[\text { Given: } f\left( x \right) = x^3 - 6 x^2 + 9x + 15\]

\[ \Rightarrow f'\left( x \right) = 3 x^2 - 12x + 9\]

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

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

\[ \Rightarrow 3 x^2 - 12x + 9 = 0\]

\[ \Rightarrow x^2 - 4x + 3 = 0\]

\[ \Rightarrow \left( x - 1 \right)\left( x - 3 \right) = 0\]

\[ \Rightarrow x = 1 \text { or } 3\]

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.

The local maximum value of  f (x) at x = 1 is given by \[\left( 1 \right)^3 - 6 \left( 1 \right)^2 + 9\left( 1 \right) + 15 = 1 - 6 + 9 + 15 = 19\]