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Let f(x)=x^(3)-3x^(2)+6AA x in R " and "g(x)={{:(max.f(t), x+1 le t le x+2","-3 le x le 0),(1-x " for " x ge 0):} Then find y=g(x) " for " x in [-3, 1].

Answer»

Solution :`f(X)=x^(3)-3x^(2)+6`
If `f'(x)=3x^(2)-6x=0`, then `x=0, 2` are the critical points of `f(x)`.
`x=0` is the POINT of LOCAL MAXIMA and x=2 is the point of local minima.
Clearly, `f(x)` is increasing in `(-oo,0)` and `(2, oo)` and decreasingin (0,2).

Case 1 : ` x+2 le 0 rArr x le -2`
`rArr " " g(x)=f(x+2), -3 le x le -2`
Case2: `x+1 LT 0 ` and `0 lt x+2 lt 2`
`x lt -1 ` and `-2 lt x lt 0`
i.e., `-2 lt x lt -1 " " :. g(x)=f(0)`
Case 3: `0 le x+1, x+2 le 2`
`rArr -1 le x le 0, g(x)=f(x+1)`
`rArr g(x)={{:(f(x+2)", " -3 le x lt -2),(f(0)"," -2le x lt -1),(f(x+1)"," -1 le x lt 0),(1-x ","0 lexlt1):}`


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