# critical values of f(r)= |r^2-3r+2|?

and then Find the absolute maximum and minimum of f(r) on the interval [0, 3].

### 1 Answer

- No MythologyLv 77 years agoFavorite Answer
f(r) = |(r - 2)(r - 1)|. As a piecewise defined function

f(r) = (r - 2)(r - 1) if r ≤ 1 or r ≥ 2

f(r) = -(r - 2)(r - 1) if 1 < r < 2.

d/dr r² - 3r + 2 = 2r - 3, so f '(r) = 0 if 2r - 3 = 0, i.e. r = 3/2.

Even a poorly drawn graph will make it obvious that f '(r) is undefined at r = 1 and r = 2. You could also use the piecewise definition above to conclude that

lim f '(r) = (2 - 3) = -1 and lim f '(r) = -(2 - 3) = 1.

r->1- . . . . . . . . . . . . . . . . . r->1+

This shows that f ' isn't defined at r = 1.

f has three critical values on the interval [0, 3], these are 1, 3/2, and 2.

f(0) = 2,

f(1) = 0,

f(3/2) = 1/4,

f(2) = 0, and

f(3) = 2

The absolute maximum is 2 taken at r = 0 and at r = 3. The absolute minimum is zero taken at r = 1 and r = 2.

- Login to reply the answers