Question
Let $$f\left( x \right) = 6 - 12x + 9{x^2} - 2{x^3},\,1 \leqslant x \leqslant 4.$$ Then the absolute maximum value of $$f\left( x \right)$$ in the interval is :
A.
$$2$$
B.
$$1$$
C.
$$4$$
D.
none of these
Answer :
$$2$$
Solution :
$$\eqalign{
& f'\left( x \right) = - 12 + 18x - 6{x^2} \cr
& = - 6\left( {{x^2} - 3x + 2} \right) \cr
& = - 6\left( {x - 1} \right)\left( {x - 2} \right) \cr
& \therefore \,f'\left( x \right) > 0\,{\text{if }}1 < x < 2\,{\text{and }}f'\left( x \right) < 0\,{\text{if }}2 < x \leqslant 4 \cr
& \therefore \,f\left( x \right){\text{is}}\,\,{\text{m}}{\text{.i}}{\text{.}}\,\,{\text{in}}\,1 < x < 2\,{\text{and}}\,{\text{m}}{\text{.d}}{\text{.}}\,{\text{in}}\,\,2 < x \leqslant 4 \cr
& \therefore {\text{absolute}}\,{\text{maximum}} = {\text{the}}\,{\text{greatest}}\,{\text{among}}\,\left\{ {f\left( 1 \right),\,f\left( 2 \right)} \right\} \cr
& = {\text{the}}\,{\text{greatest}}\,{\text{among}}\,\left\{ {1,\,2} \right\} = 2 \cr} $$