Verified answer

let y=ln(x^2 + 5x + 8), then differentiate to find any maxima/minima

dy/dx= (1/(x^2 + 5x + 8)) * (2x +5) = (2x +5) / (x^2 + 5x + 8)

for a max or min we have dy/dx=0, therefore at a max/min we have

(2x +5) / (x^2 + 5x + 8) =0,

it is clear that in [-3,3], that (x^2 + 5x + 8) is always positive, ie >0, therfore

when dy/dx=0, we must have 2x+5 =0, ie x = -5/2,

there fore a max or a min occurs when x= -5/5, ie when y=ln(25/4 - 25/2 +8)=ln(7/4)=ln7-ln4

If you need to know whether this is a maximum or minimum you should calculate the second differential to see what type of maxima/mininma.

If the sign of the second deifferetial is negative when you make x= -5/2, it is a maximum, otherwise a minimum

f'(x) = (2x + 5)/(x^2 + 5x + 8) = 0 iff x = -5/2.

f(-5/2) = 0.5596...

f(-3) = 0.6931...

f(3) = 3.4657...

Absolute max ≈ 3.466

Absolute min ≈ 0.5596