Find the maximum and minimum values of the function `y=(log)_e(3x^4-2x^3-6x^2+6x+1)AAx in (0,2)` Given that`(3x^4-2x^3-6x^2+6x^2+6x+1)>0Ax in (0,2)`

Answer

Point of local maxima :x=`1//2`
Point of local minima:x=1

Explanation

`f(x) = log_(e) (3x^(4)-2x^(3)-6x^(2)+6x+1),x in(0,2)`
`f(x)=(12x^(3)-6x^(2)-12x+6)/(3x^(4)-2x^(3)-6x^(2)+6x+1)`
`=6(2x^(3)=x^(2)-2x+1)/((3x^(4)-2x^(3)-6x^(2)+6x+1)`
`=6(x^(2-2)(2x-1))/(3x^(4)-2x^(3-6x^(2)+6x+1)`
Sign scheme of f(X) is as follows:
Hence x =`1//2` is point of maxima and x =1 is point of minima
Hence `f_(min) =f(1)=ln2`
and `f_(max)=f(1//2)=ln(39//16)`