Let `f(x)=2x^3-9x^2+12 x+6.` Discuss the global maxima and minima of `f(x)in[0,2]a n d(1,3)` and, hence, find the range of `f(x)` for corresponding intervals.

Explanation

`f(x) =2x^(3)-9x^(2)+12x+6`
`f(x)=6x^(2)-18x+12=6(x^(2)-3x+2)=6(x-1)(x-2)`
Clearly the critical point of f(x) in [0,2] is x=1
Now f(0) =6, f(1)=11,f(2)=10
Thus ,x =0 is the point of global minimum of f(x) in [0,2] and x=1 is the point of global maximum.
Hence , range is [6,11].
For x `in` (1,3), clearly,x=2 is the only critical point in (1,3) `f(2)=10 underset(xrarr1)lim f(x)=11` and `underset(xrarr3)lim f(x) =15`
Thus x=2 is the point of global minimum in (1,3) and the global maximum in (1,3) does not exist
Hence range is [10,15]