00:01
In this problem, we want to use the intermediate value theorem to show that there is a root of the given equation in the specified interval.
00:12
So let's start by stating the intermediate value theorem, or ivt for short.
00:19
So according to this theorem, if we're given a function f of x that is continuous over some interval a, b, and we're given some point c that is comprised between f of a and f of b, then there exists a value of x that is comprised between a, b, such that f of x is equal to c.
01:25
So what is this theorem saying? it's saying that if a function is continuous, then the function takes every value of x within its domain.
01:50
Let's draw an example to understand this a bit better.
01:56
So let's say we have some function here, f of x, that is continuous over some interval a and b.
02:17
Let's roughly sketch where f of a and f of b will be.
02:24
So f of b will be located here in our sketch, and f of a right here.
02:42
And now what happens is that any value of c comprised between f of a and f of b will correspond to a value of x that is within a, b.
02:54
Why? because our function is continuous and has no holes.
02:58
So in our case, we want to use this theorem to the following function, f of x equal to the sine of x minus x cubed plus 1, which what we've done, we've just reworked our equation...