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[MUSIC] So, I get a number like the 
square root of two. 

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And square root of two is irrational. 
So, how am I going to figure out what 

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it's approximately equal to? 
Well, it turns out that it's 

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approximately, you know, 1.414 and a bit 
more. 

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But how would ever figure that out? 
How would I know that that's 

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approximately the value of the square 
root of two? 

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Let's use the intermediate value theorem 
to try to pull it. 

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I want to use the intermediate value 
theorem to approximate the square root of 

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two. 
To do that, I'm going to use this 

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function, 
f(x) = x^2 - 2. 

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Notice something about this function. 
This function is equal to zero when x is 

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the square root of 2. 
And how is that? 

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f of the square root of 2 is the squared 
of two squared, which is two, minus two, 

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which is zero. 
So, if I'm trying to approximate the 

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square root of 2, what I'm going to think 
about this, is that I'm going to look for 

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a positive value, that I can plug into 
this function to make it a square root of 

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zero. 
How am I going to find such a value? 

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You know, the other good thing about this 
function is that it's continuous. 

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And then, I can plug in some values, like 
f(1). 

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Now, what's f(1)? 
It's one squared minus two and that's 

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minus one. 
And I can plot that on my graph right 

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here, and it's this point right here. 
And take a look at, say, f(2). 

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f(2) would be two squared minus two, 
which is two. 

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And I can plot that on my graph, right 
here. 

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It's 2, 
2. 

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Now, f is a continuous function and at 
one, it's negative. 

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And at two, it's positive. 
So, it's a continuous function. 

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It's negative at one and it's positive at 
two. 

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By the intermediate value theorem, there 
must be a point in between when the 

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functions value is equal to zero. 
And you can see that on the graph. 

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We're looking for that point. 
That is the square root of two where this 

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graph crosses the x-axis, 
the positive x-axis. 

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And that's the point that we're looking 
for, 

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right? 
The intermediate value theorem promises 

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me there's a value in between where the 
functions value equals zero. 

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And I can see that there must be such a 
point on the graph. 

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And now, I know that that value is 
between one and two. 

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And I can do better. 
I can cut this interval between one and 

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two in half and I get that f(1.5). 
Well, 1.5 squared is 2.25 - 2 is 0.25. 

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[SOUND] Alright. 
So, that's, say, this point on my graph. 

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Now, what do I know? 
I know that the squared of two, which is 

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where this graph crosses the x-axis, is 
between one, where the function is 

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negative and 1.5, where the function is 
positive. 

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And I can do better, 
alright? 

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I can pick some other point. 
Let's be brave and pick 1.4. 

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And if I plug in 1.4, well, 14 squared is 
196. 

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So, 1.4 squared is 1.96 - 2, 
that makes this -0.4, 

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alright? 
This is barely below the x-axis. 

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And I can plot that point here. 
And now, I know that the square root of 

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two, the positive input to this function, 
which is equal to zero, 

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well, it must be between 1.4 and 1.5. 
Because the function's value at 1.4 is 

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negative and the function's value at 1.5 
is positive. 

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So, this is not so bad. 
I mean, we, we could do this with a few 

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calculations on the blackboard. 
Let's, let's bring out some sort of 

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computation device so we can try to get 
an even better approximation to the 

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square root of two. 
So here, I've got a function. 

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The function is f(x) = x^2 - 2. 
And you can see that if I plug zero into 

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the function, I get out -2 because 0^2 - 
2 is -2. 

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Now, this is a physical function. 
I've got this knob here. And as I spin 

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this knob, the function's input value 
changes. 

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So, for example, if I spin it up to 
f(05.), f(1/2), that's negative 1.75 and 

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that's right because if I take a half 
squared, I get a quarter, and a quarter 

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minus two is -1.75. 
Now, I could spin the input up to one. 

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f(1) is -1 = because 1^2 - 2 is -1 or 
f(1.5), well, 1.5 squared is 2.25, 2.25 - 

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1 is 0.25. 
Now, this is positive or I could wheel it 

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back down a little bit. 
And I see that when I get to 1.41, the 

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function's output is negative again. 
Then, I could increase the function's 

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input a bit. 
Oh, now the function's output is positive 

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again, at f(1.145). 
And then, I could decrease the value a 

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little bit. 
Oh, 

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now, the function's output value is 
negative again and that would increase 

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the output value a little bit. 
Oh, 

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now, the function's output is positive 
and I could decrease the input a bit. 

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Now, the function's output is negative, 
could increase the input a little bit. 

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Now, it's positive, could decrease the 
input a little bit. 

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Now, it's negative again. 
Now, Look at what happened, 

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right? 
What's going on? 

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I'm looking at the function f(x) = x^2 - 
2 and I'm walking back and forth across 

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the point where this function is equal to 
zero. 

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And as I walk to the left and to the 
right, the function is positive and the 

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function is negative. 
And I'm getting closer and closer to a, 

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to the actual place where the function is 
equal to zero and what is 1.41421356? 

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It's awfully close to the square root of 
two. 

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And if you think the square root of two 
and square it and subtract two, you get 

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zero. 
What we just saw with the knob that I was 

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turning is actually an incredibly general 
technique for figuring out where a 

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continuous function crosses the x-axis. 
And we can do it very efficiently. 

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Let's take a look at how we might do this 
a little bit more carefully. 

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Here, I've drawn a graph with some random 
looking continuous function and I've 

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labeled some points along the x-axis. 
Imagine that we couldn't see this whole 

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graph. 
Just imagine that the only thing you can 

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do is interrogate this function. 
You can ask this function at some input 

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point, are you positive or negative? 
We're trying to figure out where, 

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approximately, this continuous function 
crosses the x-axis. 

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We're looking for a zero of this 
continuous function. 

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How might the game begin? 
I'll let it begin by asking the function, 

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are you positive or negative at zero? 
And the function is positive at zero. 

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And then, maybe I'd ask at sixteen. 
At sixteen, is this function positive or 

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negative? 
And I'd see the function is negative. 

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The function's value is negative at 
sixteen. 

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Because it's a continuous function and 
the function's value at zero is positive 

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and the function's value at sixteen is 
negative, I know that there's a zero in 

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between. 
There's some point in between that if I 

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plug it into the function, I get zero 
out. 

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Now, I could cut that interval in half. 
I could plug in eight and I could ask the 

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function at f(8), are you positive or 
negative? 

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And the function f(8), the value of the 
function at eight is positive. 

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Now, look. 
I know the function's value at eight is 

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positive, the function's value at sixteen 
is negative. 

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So, by the Intermediate Value Theorem, 
there must be a value in between, right 

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there, where the function's value is 
zero. 

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And now, I could take that interval and 
cut it in half again. 

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I could take a look at twelve, which is 
exactly between eight and sixteen. 

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And I could ask the function, at 
f(12),12), are you positive or negative? 

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And the function's value at twelve is 
negative. 

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[SOUND] So now, I know by the 
intermediate Value Theorem, the function 

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zero zero of this function must land 
between eight and twelve and I could cut 

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that interval in half again. 
I could look at ten, and I could ask the 

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function, 
well, if I plug in ten, are you positive 

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or negative? 
And the function's value with ten is 

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negative. 
And now, I know that the zero must land 

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between eight and ten, because the 
function's value at eight is positive and 

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the function's value at ten is negative, 
so there must be a zero in between, a 

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point where the function's value is equal 
to zero. 

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Now, I could cut that interval between 
eight and ten in half again. 

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I could look at nine. And I could ask the 
function, at nine, are you positive or 

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negative? And if I plug in nine, it looks 
like the function is positive. 

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And that means that a value that I can 
plug into the function to make the output 

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zero lies between nine and ten. 
And this is really quite remarkable. 

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I, I, I could cut the interval between 
nine and ten and half again and see if it 

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was between nine and 9.5 or 9.5 and ten. 
And I could keep doing this every time 

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I'm getting a little bit more precision 
as to the exact value of the input that 

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would make the function zero. 
The other really neat thing about this is 

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that I could have just tried all of the 
values between zero and sixteen. 

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I could have plugged all seventeen of 
those numbers into the function to figure 

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this out. But by doing this clever 
cutting in half method, this bisection 

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method, I managed to only ask six 
questions, 

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right? 
I asked the function's value at zero and 

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sixteen, at eight and twelve, at ten and 
at nine, and by only asking six 

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questions, I was able to narrow down a 
region where a zero resided. 

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Now, that's pretty quite efficient, much 
better than asking seventeen questions. 

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So, this is a very general technique for 
figuring out where zeroes of continuous 

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functions lie. 
And I encourage you to pick your favorite 

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function and see if you can approximate a 
root, a zero of that function. 

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[MUSIC]