Question
Let the interval $$I = \left[ { - 1,\,4} \right]$$ and $$f:I \to R$$ be a function such that $$f\left( x \right) = {x^3} - 3x.$$ Then the range of the function is :
A.
$$\left[ {2,\,52} \right]$$
B.
$$\left[ { - 2,\,2} \right]$$
C.
$$\left[ { - 2,\,52} \right]$$
D.
none of these
Answer :
$$\left[ { - 2,\,52} \right]$$
Solution :
$$\eqalign{
& f'\left( x \right) = 3{x^2} - 3 = 3\left( {{x^2} - 1} \right) \cr
& {\text{So, }}f'\left( x \right) \leqslant 0{\text{ if }} - 1 \leqslant x \leqslant 1{\text{ and }}f'\left( x \right) > 0{\text{ if }}1 < x \leqslant 4 \cr} $$
$$\therefore f\left( x \right)$$ is decreasing in $$ - 1 \leqslant x \leqslant 1$$ and increasing in $$1 < x \leqslant 4$$
$$\therefore \min f\left( x \right) = f\left( 1 \right),\,\max f\left( x \right) = $$ the greatest among $$\left\{ {f\left( { - 1} \right),\,f\left( 4 \right)} \right\}$$
$$\eqalign{
& {\text{Now,}}\,f\left( 1 \right) = {1^3} - 3.1 = - 2 \cr
& f\left( { - 1} \right) = {\left( { - 1} \right)^3} - 3\left( { - 1} \right) = 2 \cr
& f\left( 4 \right) = 64 - 12 = 52 \cr} $$
$$\therefore $$ the range of the function $$\left[ { - 2,\,52} \right]$$