F(x) = x^1/3(x+3)^2/3 (A) Find all critical values of f. If...

f(x) = x^{1/3}(x+3)^{2/3} (A) Find all critical values of f. If there are no critical values, enter None . If there are more than one, enter them separated by commas. Critical value(s) = -3,-1,0 (B) Use interval notation to indicate where f(x) is increasing. Note: When using interval notation in WeBWorK, you use I for ?, -I for -?, and U for the union symbol. If there are no values that satisfy the required condition, then enter "{}" without the quotation marks. Increasing: (-I,-1)U(0,I) (C) Use interval notation to indicate where f(x) is decreasing. Decreasing: (-1,0) (D) Find the x-coordinates of all local maxima of f. If there are no local maxima, enter None . If there are more than one, enter them separated by commas. Local maxima at x = (E) Find the x-coordinates of all local minima of f. If there are no local minima, enter None . If there are more than one, enter them separated by commas. Local minima at x =

Transcript

00:02 Okay, so we have this problem here where we're given a function f of x, pretty ugly function, and we want to use our first derivative rule to find the critical values, increasing and decreasing intervals, and then local minimum and maximum.

00:14 So you have already found those critical values of f, so i'm kind of assuming you've already done this first derivative, you've solved for our values of x and so forth and so on.

00:28 And so what i've done next is i've taken your critical values, and i've created what i call a wiggle graph for this.

00:34 It's just a number line.

00:36 My number line right here represents my x values.

00:38 And i put all of those critical values we found as points on my number line, negative three, negative one, and zero.

00:45 And i've kind of defined these intervals, these dotted lines, to kind of break up my number line based on that.

00:51 I'm still looking at my first derivative here because that tells me when my function is increasing and decreasing.

00:58 And so i want to see what's happening in all of these intervals.

01:00 We know at those specific critical points the first derivative is zero.

01:06 We don't know what's going on between them.

01:08 So i went through and i found a number that exists in each one of these intervals.

01:12 It doesn't really matter which number.

01:14 You just choose one that works for you.

01:16 So i chose some pretty nice whole numbers where i could.

01:20 So something over here to the left of negative 3, i chose x equals negative 4.

01:24 And if you try to plug that into your first derivative, you get an error because you're trying to take the square root of a negative number.

01:31 So that's telling me my graph has an asymptote right here, and it just doesn't exist at that point.

01:37 It's not doing anything anymore.

01:41 Then in between negative 3 and negative 1, i have x equals negative 2.

01:46 I just chose that.

01:46 Again, plugged it into my first derivative, and i get this answer.

01:49 It's a really ugly decimal.

01:51 We don't really care about the number of the first derivative.

01:54 We care about the sign.

01:55 So i copied down that it was negative.

01:58 Similar thing for my next interval, i used x equals negative 1⁄2, and i don't.

02:02 I got a negative plugging it into my first derivative.

02:04 And then last interval is onto the right of zero...

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