Texts 1 8 if x 2 fx 2 x 4 6 if x 4 in this problem well use...

Texts: 1 8 If x < 2 f(x) 2 < x < 4 6 - if x > 4 In this problem, we'll use the TI-84 graphing calculator or Desmos to graph f() and its endpoints. If needed, return to a similar example in the reading or visit the Resources module (opens in a new tab). Be sure to select the appropriate page for the technology you are using (TI-84 or Desmos), and find the video pertaining to Graphing Piecewise-defined Functions. Instructions: Use your graphing calculator or Desmos to graph f(). Adjust your graphing window so that you can see the shape of the function easily. You should have three distinct graph shapes: part of a horizontal line, part of a reciprocal function, and part of a decreasing line. Your response to this question will depend on the technology you use to graph the function. Follow the instructions below corresponding to the technology you are using. Graphing Calculator: 1. Describe the left and right endpoints of the function at x = 2 and x = 4. What are the coordinates for the solid points and the holes at these x-values? (Keep in mind that we need to use the Canvas Math Editor whenever we need to type math expressions or equations.) 2. Embed a clear and easy-to-read picture of your graph (directions for embedding a picture). Desmos: 1. Describe the left and right endpoints of the function at x = 2 and x = 4 by using Desmos to place solid points and holes at the appropriate coordinates. 2. Embed a clear and easy-to-read picture of your graph (directions for embedding a picture).

Transcript

00:01 So we have the intermediate value theorem, which is to say if we have a and b in the domain of f, that is to say f is a function that goes from the domain to its...

00:13 Well, for the range, but in particular the codomain, we'll call it that.

00:18 We're going to assume in this case that d is a subset of r, and the codomain can also be r, with a not equal to b.

00:29 Then the function f takes on every value between f of a and f of b.

00:33 If we write that symbolically, that means there exists some c between a and b, such that f of c is equal to d, and this is true for all d in the interval from f of a to f of b, or f of b to f of a in the case that f of b is less than f of a.

01:00 What does this mean in our own words? well, i can't use your words, but my own words would be that if we have two points in the domain, a and b, and f is taking on some value at a, f of a, and some value at b, put that in red, f of b, that if f is continuous, which it doesn't actually say here, but f has to be a continuous function, that in order to get to b, it has to pass through at least once everything between a and b.

01:43 In particular here, that looks like this region.

01:47 Basically, it can't skip over anything in between the values of f of a and f of b.

01:54 There can't be any holes there, which means that it has to take on every value in the middle.

02:02 That's my words.

02:03 You can use your own words for the assignment here.

02:07 Now let's see.

02:08 It says, now we can apply this theorem in the special case that f is a polynomial function, which is a continuous function.

02:16 You have a point on the graph, it's at f, at x equals a, lies above the x -axis, just say f of a is greater than zero.

02:25 Another point, f of b, is less than zero.

02:29 There has to be a third point between the two where it crosses the x -axis.

02:32 That is, there has to exist some value c between a and b, or b and a, again, if b is less than a, or if a is less than b, such that f of c is equal to zero, because zero is in between any positive and negative number.

02:50 We would like to find what that is using the bisection method.

02:56 To do that, what we do is just take the midpoint and see what happens.

03:00 I'll illustrate this in this case here.

03:04 F of x is equal to x to the fourth minus 8x squared.

03:10 What i'm going to notice is that, first of all, f of zero is equal to zero, because we get f of zero is equal to zero to the fourth minus 8 times zero squared.

03:28 That's a relatively easy zero to find.

03:31 I'm also going to notice that f of one is equal to one to the fourth minus eight times one squared, which is one minus eight is negative seven.

03:44 I'm also going to notice that f of two is equal to two to the fourth, which is 16, minus eight times two squared is 16 minus 32, is equal to negative 16.

04:02 That's also a thing that is true.

04:05 F of three, let's keep trying this, three to the power of four is 81 minus eight times 72, or not 72, eight times nine, which is 72.

04:19 81 minus 72 is nine.

04:23 We see at f of two, we get negative 16.

04:28 At f of three, it's nine.

04:31 Just sketch a graph here.

04:33 That means we've got two and three.

04:36 We have to go from negative 16 to nine between two and three, which, because zero is in between those two values, means somewhere between two and three, we have to have a zero.

04:49 How can we find that? well, i'm going to make a guess that it's halfway between two and three.

04:54 It's two and a half.

04:55 This is not something we're going to calculate directly.

04:58 We say two and a half to the power of four minus, i'm going to do parentheses, eight times two and a half squared.

05:10 If we do that, we get negative 10 .9.

05:14 F of two and a half is negative 10 .93, whatever.

05:22 The point here is that it's less than zero.

05:26 Actually, if we put a point at two and a half, two and a half, it is still less than zero down here before it gets up to three, which means that this zero, which has to exist, is in between two and a half and three.

05:46 At this point, i'm going to pull up desmos because it will allow me to do a better graph.

05:51 F of x equals x to the fourth.

05:53 I'm going to hide this real quick.

05:55 Minus eight times x squared.

05:58 We'll graph the function in a minute, but first we just want to make a table, which we can do like this...

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