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Continuity Example 1
Hi, everyone!
Welcome back to integralcalc.com.
Today we’re going to be talking about continuity.
And in this one we’ve been given the function f of x equals the third root of x plus one divided by x minus one.
And we’ve been asked to find the domain of the function, in other words, where the function is defined or continuous.
So in this one, what we need to realize is that the function is going to be continuous wherever it’s defined.
So we just need to find out where it’s undefined.
And there’s a couple things you want to look for to find out where a function is undefined.
When you’ve got a rational function like this, a fraction, we’ve got x plus one divided by x minus one.
When you have something like that, you know that the function is going to be undefined
whenever the denominator of the fraction is equal to zero.
So if we set x minus one equal to zero, and we solve for x, obviously by adding one to both sides,
you know that the function is undefined at x equals one.
If we plugged in one for x, we would get zero here in the denominator. This would be equal to zero.
And we can’t divide by zero.
We can’t have zero when in the denominator of a fraction, so we know that the function is undefined there.
We would also know that a function is undefined whenever we have a negative answer or negative number inside of our square root.
So if we were able to figure out where x plus one divided by x minus one was equal to a negative number,
the function would also be undefined there.
However, in the case of this function, we’re looking at the third root of this fraction, not just the square root.
And you can take the third root of a negative number.
For example, if you take the third root of negative one, you get negative one because negative one cubed is negative one,
which is what’s inside the square root sign here.
So you can take the third root of a negative number. You can’t just take the square root of a negative number, right?
The square root of negative one would be undefined because you can’t multiply anything by itself to get a negative number.
So because we have the third root of this fraction here, we know that we can have a negative number inside the square root
so we don’t have to worry about the function being undefined for the points at which this fraction would be equal to in negative value.
So, the only thing that we do have to worry about is where the denominator is equal to zero and we already figured that out.
That’s x equals one. So we can say that the function is continuous everywhere where x is not equal to one.
So we can just say the domain is x not equal to negative one, which implies all real numbers except for one.
And we can say that the function is continuous for x not equal to one. And that’s it.
So I hope that video helped you, guys and I will see you in the next one.
Bye!