#### Question

The least and greatest values of f(x) = x^{3}\[-\] 6x^{2}+9x in [0,6], are ___________ .

3, 4

0, 6

0, 3

3, 6

0, 54

#### Solution

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

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

\[\text { For a local maxima or a local minima, we must have }\]

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

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

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

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

\[ \Rightarrow x = 1, 3\]

\[\text { Now,} \]

\[f\left( 0 \right) = 0^3 - 6 \left( 0 \right)^2 + 9\left( 0 \right) = 0\]

\[f\left( 1 \right) = 1^3 - 6 \left( 1 \right)^2 + 9\left( 1 \right) = 1 - 6 + 9 = 4\]

\[f\left( 3 \right) = 3^3 - 6 \left( 3 \right)^2 + 9\left( 3 \right) = 27 - 54 + 27 = 0\]

\[f\left( 6 \right) = 6^3 - 6 \left( 6 \right)^2 + 9\left( 6 \right) = 216 - 216 + 54 = 54\]

The least and greatest values of *f*(*x*) = *x*^{3}- 6*x*^{2}+9*x* in [0, 6] are 0 and 54, respectively.

#### Notes

The question in the book has some error. So, none of the options are matching with the solution. The solution here is according to the question given in the book.