00:04
Okay, we want to determine if this equation has a solution in one of these intervals.
00:12
So if we determine that it does have a solution in this interval, we don't need to go on further to check this one.
00:20
We're just trying to determine whether this function, whether this equation has a solution in either one of these intervals.
00:28
So we just want to see if this one or this one contains a solution to this equation.
00:34
So we are going to apply the intermediate value theorem.
00:41
And the intermediate value theorem says that if you have a continuous function on a closed interval, and it takes on two values in that interval, then it takes on all the values between those two values.
00:56
Okay, let's go back and let's look at this equation.
01:00
2 to the x equals x to the third.
01:02
If i minus x q from both sides, i get 2 to the x minus x minus x.
01:07
Cubed equals zero so a solution to this equation obviously is a solution to the original equivalent equation i am going to call 2 to the x minus x cubed f of x now i want to find out what is f of 1 .25 and what is f okay remember this is our function f of x now 2 to the x minus x cubed what is f of 1 .375.
02:02
Okay, with f of x being 2 to the x minus x cubed, f of 1 .25 is a positive 0 .4253.
02:10
F of 1 .375 is a negative 0 .0059.
02:15
The intermediate value theorem says that since f of x is a continuous function on this closed interval and takes on a positive value and a negative value, then it must take on the value of zero somewhere on this interval.
02:36
I'll say it again.
02:36
The immediate value theorem works for continuous functions on closed intervals.
02:41
Since our function f of x is greater than zero on this interval and less than zero on this interval, f of x must equal zero somewhere on this interval...