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Many problems in math don't require an exact solution. Some problems exist simply to find out if any solution exists. In this lesson, we'll learn how to use the intermediate value theorem to answer an age-old question.
The intermediate value theorem says that if you have some function f(x) and that function is a continuous function, then if you're going from a to b along that function, you're going to hit every value somewhere in that region (a to b). Well, why is this useful? This helps you learn a lot about functions without having to graph them.
For example, if you have the function f(x)= x^3 + x^2, you can start to take a look at what f(x) equals for various values of x, like if x=0, then f(x)= 0 ^3 + 0^2, or just zero. When x=1, f(x) = 1^3 + 1^2, or 2. When x=2, f(x) = 2^3 + 2^2, which is 12.
x | f(x) | |
---|---|---|
0 | 0 | |
1 | 2 | |
2 | 12 |
Okay, great, so you've got this table here of x values and f(x) values. Well, we know that f(x) is a continuous function, so we can use this data to determine that f(x) is going to equal 1 somewhere between 0 and 1. How do we know this? Well, f(0)=0, and f(1)=2, so some value between 0 and 1 will give me f(x)=1. Another example is f(x)=10. Well, for what value of x does f(x)=10? I don't know, it's not actually on my chart, but I know that f(1)=2, and f(2)=12, so some value between 1 and 2 will give me f(x)=10. I can graph this to verify that f(x)=1 between 0 and 1, and f(x)=10 between 1 and 2.
Let's look at another example. In this example we're going to be finding roots of an equation. So let's say we have f(x)= 4x - x^2 - 3. We want to know when f(x)=0. This is called finding the roots of f(x). Similar to the last example, we're going to make a table with x and f(x). When x=0, f(x)=-3, because we have (4)(0) - 0^2 - 3. When x=2, f(x)=1. When x=4, f(x)=-3.
x | f(x) | |
---|---|---|
0 | -3 | |
2 | 1 | |
4 | -3 |
So does this tell us when f(x) will equal zero? Well let's take a look at the three values we calculated and put them on a graph. So I've got f(x) and x. When x=0, f(x)=-3. When x=2, f(x)=1. When x=4, f(x)=-3. Now because I know 4x - x^2 - 3 is a continuous function, I know that to get from -3 to 1, my function has to travel through f(x)=0. Similarly, to go from 1 to -3, f(x) has to pass through zero. So at some point, between 0 and 2, I have a root - there is some place where f(x)=0 for an x value between 0 and 2. Similarly, between 2 and 4, I know that f(x) will equal zero at some point, so there's an x value between 2 and 4. I know for this particular equation, I have at least two roots: one between 0 and 2, and one between 2 and 4. We still could have solved that problem by factoring, so what about another example?
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What about trying to find a solution to x^2=cos(x)? There's no f(x) in here, so where do we start? Well let's first subtract x^2 from both sides, so we get 0=cos(x) - x^2. If instead of saying 0, I called this f(x), then I'm just trying to find the roots of the equation cos(x) - x^2. And this I know how to do, so let's make a table. When x=0, f(x)= cos(0) - 0^2, or 1. When x=pi, f(x)=cos(pi), which is -1, - pi^2. So I'm going to leave this as -1 - pi^2. Let's plot these two points. First, I have f(x)=1 when x=0, then I have at x=pi, f(x)= -1 - pi^2. Again this is a continuous function, so somewhere between 0 and pi, this has to have at least one solution. The answer to x^2 = cos(x) is going to be some value of x between 0 and pi.
Let's review. The intermediate value theorem says that if you're going between a and b along some continuous function f(x), then for every value of f(x) between f(a) and f(b), there is some solution. If I'm going between a and b, I'm going to hit every value between f(a) and f(b).
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