I don't know how to use any graphing calculator so I cannot see now the graph of f(x).

Let us see if there are maximun or minimum between [-2,2].

f(x) = ln(x^2 +3x +5)

Differentiate both sides with respect to x,

f'(x) = [1/(x^2 +3x +5)](2x +3)

Set that to zero,

0 = 2x +3

x = -1.5 ------a critical point, maximum or minimum.

Now let us just see the values of f(x) at the ends of the interval, at x = -1.5, and at x=0.

f(-2) = ln[(-2)^2 +3(-2) +5] = ln(3)

f(-1.5) = ln[(-1.5)^2 +3(-1.5) +5] = ln(2.75)

f(0) = ln[(0)^2 +3(0) +5] = ln(5)

f(2) = ln[(2)^2 +3(2) +5] = ln(15)

So, in that interval, f(-1.5) is the lowest, while f(2) is the highest.

Therefore, in that interval, the absolute maximum is ln(15), and the absolute minimum is ln(2.75). --------------answer.