4. Find and identify local extrema of the following...

4. Find and identify local extrema of the following functions. Are each of these local extrema absolute extrema? Make sure to fully justify your work. (a) $f(x,y) = 4x^2 + 3y^2 - 24x + 6y + 39$ (b) $f(x,y) = (y(x - 1))^{\frac{2}{3}}$

Transcript

00:01 So i want to find the absolute extrema on the interval 0 to 2.

00:11 So in order to do that, i want to start by first finding my critical numbers.

00:16 And then i want to compare those critical numbers to the end points, which are 0 and 2.

00:22 So to find my critical numbers, i need to find the first derivative, 3x squared minus 3.

00:28 And then i want to let that equal to 0 and solve.

00:31 So if i were to solve this, i would add the 3 over.

00:34 Divide by three, and then square root both sides to get a positive and a negative one.

00:40 So that means i have two critical numbers, negative one and positive one.

00:45 Well, for this first part of the question, i'm looking from zero to two, so negative one is not between zero and two.

00:57 So i'm only going to compare zero, one, and two.

01:02 You always want to be sure to include the endpoints as we're looking for the absolute extrema, as well as any critical numbers on that interval.

01:12 As i am looking at these endpoints and the critical number, i'm filling them into the original f of x.

01:21 So when i fill in zero to the original f of x, i get one.

01:24 When i fill one into the original f of x, i get one minus three, which is negative two, plus one is negative one.

01:31 And when i fill in two to the original f of x, i get three.

01:36 With that in mind, now i can compare those values and decide which of these is the highest and which of these is the lowest.

01:44 So on the interval 0 to 2, the absolute minimum is 1 negative 1...

Question

4. Find and identify local extrema of the following functions. Are each of these local extrema absolute extrema? Make sure to fully justify your work. (a) $f(x,y) = 4x^2 + 3y^2 - 24x + 6y + 39$ (b) $f(x,y) = (y(x - 1))^{\frac{2}{3}}$

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