# min/max with a twist

• Nov 2nd 2007, 08:25 AM
winterwyrm
min/max with a twist
Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = ln(x2 + 3x + 5)
[-2, 2]

• Nov 2nd 2007, 09:47 AM
ticbol
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.
• Nov 2nd 2007, 10:17 AM
winterwyrm
wow, I kept thinking the derivative was 2x+5 for who knows what reason, lol..... I kept checking everything and wondering why it just wouldn't work out:D