#### Question

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

(a) 3,4

(b) 0,6

(c) 0,3

(d) 3,6

#### 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.

Disclaimer: 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.