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Math 104: Calculus14 chapters | 115 lessons | 11 flashcard sets
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Free 5-day trialThe 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?
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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Math 104: Calculus14 chapters | 115 lessons | 11 flashcard sets