Find all absolute extrema of the given function on each indicated interval. f(x) = x^3 - 3x + 1 on a) [0,2] b) [-3,2]
ok so i substitute them in?
You want to take the derivative and find critical points. Are you able to do that portion onegirl? c:
yes i know how to find the critical points
or not guess this is something else
Since we're given an interval, we have to also check the `end points`. So if you've found the critical points correctly. Plug them into the function separately, then also plug in the end points (0 and 2 for part a). Then simply compare all of your values that they produce. The largest will be your max, smallest your min.
f(0) = ? f(-1) = ? f(1) = ? f(2) = ?
okay so the critical points are +/- 1
ok sounds good :)
so now i plug them into the problem right?
Yes, into the original function (what it looked like before you took a derivative).
okay hold on
(1)^3 - 3(1) + 1 = -1
so -1 and -3
f(1) = -1 f(-1) = -3 Ok good. We also need to check the end points now. The ends of our interval. Plug 0 and 2 into the function. f(0) = ? f(2) = ?
Woops I think f(-1) should be giving us +3, not -3.
yes you're right its positive 3, so for f(0) = -2 and f(2) its 3
f(0) = 0^3 - 3(0) + 1 f(0) = 1
ok
so -1, 3, 1 and 3
Ok so we now have the function value at our critical points and at the ends of our interval. Our interval was from 0 to 2. Hmmm -1 is NOT in that interval is it?
no
So we'll throw that point out. These are the points we care about. f(0) = 1 f(1) = -1 f(2) = 3 Which one is the largest? That will be our maximum.
ok 3 is the largest
Ok good! We could write that like this perhaps, as a coordinate pair. Max: (2,3) How about your minimum? :)
the minimum is -1
Min : (1,-1)
Yes very good. When they ask for `critical points` they simply want the X value. When they ask for `max` or `min points` they want it as an ordered pair usually.
ok thanks
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