`f(x) = ln(x^2 + x + 1), [-1, 1]` Find the absolute maximum and minimum values of f on the given interval
The function is defined because x^2+x+1 always >0.
Further, ln(y) is monotone increasing and therefore it has min and max where y has min and max.
x^2+x+1 has a global minimum at x=-1/2 and it is 3/4, and global maximum at x=1 (it is 3). So f has minimum ln(3/4) and maximum ln(3) .