http://www.thescienceforum.com/Calcu...ions-8994t.php

Here we are at the end of Chapter 4! I've got an easy one to start things off. My work makes perfect sense to me, but the book's answer doesn't. I am going to use AMax to mean Absolute Max, Amin to mean Absolute min, LMax to mean local max, and Lmin to mean local min.

Find the local and absoulte extreme values of the function on the given interval.

5) f(x) = x + sin 2x, [0,π]

My work:

f '(x) = 1 + 2 cos 2x

From what I understand, f(x) can only change from increasing to decreasing if f '(x) = 0 within the given interval. A point at which f '(x) = 0 is a critical point, and only a critical point may be a local min or local max value. To find the criticals point, we set f '(x) = 0 and solve for x.

f '(x) = 0 = 1 + 2 cos 2x

cos 2x = -1/2

2x = -π/3

x = -π/6

I would say that since the angle is 2x, that both the cos and sin would completely cycle every π value of x, therefore f '(x) = 0 when x = -π/6, , 2π/6, 5π/6, 8π/6 and so on. 2π/6 and 5π/6 are the only x values within the interval [0,π]

f(π/3) = Sqrt(3)/2 + π/3

f(5π/6) = -Sqrt(3)/2 + 5π/6

f(0) = 1

f(π) = π

AMax = π

LMax = Ø

Amin = 0

Lmin = Ø

(in x values)

The book answer is:

AMax = π

LMax = π/3

Amin = 0

Lmin = 2π/3

Never mind on this one. My problems with this were more or less concerning trig. I think I just need a small break.