00:01
Here we have two functions.
00:04
F of x equals x squared minus x minus 2, and g of x equals x squared plus x minus 2.
00:12
If we want to find where f of x is equal to 0, that means we're just going to replace f of x here with 0, or in other words, set this function equal to 0.
00:34
Then to find the x values where it's equal to zero, we can factor it.
00:43
So that means we're breaking this into two binomials, each of which starts with x because the first term in our quadratic is x squared.
00:53
Then we just need to find two numbers that multiply to equal negative two and add up to negative one, the coefficient of x.
01:00
So that would be negative two and one, because negative two times one.
01:05
Is negative 2, negative 2 plus 1 is negative 1.
01:11
Then we apply the 0 product property and set each binomial equal to 0 to find the solutions and then solve.
01:26
So we get x equals 2 and we get x equals negative 1.
01:37
To find word g of x equals 0, we do the same thing.
01:43
Set the function equal to 0, then factor.
01:56
But this time we're looking for two numbers that multiply to equal negative 2 and add up to positive 1.
02:03
So that means we're going to have plus 2 and minus 1.
02:10
Set those equal to 0 and solve.
02:21
And we get x equals negative 2 and x equals 1.
02:32
What if we want to find where f of x equals g of x? all we have to do is to set these functions equal to each other.
02:45
So we have x squared minus x minus 2 is equal to x squared plus x minus 2.
03:00
Notice we have x squared on both sides of the equation.
03:05
They just offset each other.
03:09
Or if you want to think of it as subtracting x squared from this side to get rid of the x squared, and move it over here when we subtract x squared on the other side that also becomes zero.
03:28
So those are eliminated.
03:31
And here i like to add x to both sides because that gets rid of the negative.
03:37
So we have negative x plus x is zero minus two just leaves us with negative two.
03:46
Now if you prefer to have your x over on the other side, you can subtract x from both sides, and that will still work.
03:56
So we have negative 2 on the left is equal to x plus x, which is 2x, minus 2.
04:04
Then we need to add 2 to both sides to get x alone on one side of the equal sign, and we get 2x equals 0.
04:16
And then divide both sides by 2 to get x completely alone, and we get x equals 0 over 2 or 0.
04:27
So f of x equals g of x when x equals 0.
04:31
That means if we were to graph them both on the same coordinate grid, they would intersect or cross each other when x equals 0.
04:43
What if we wanted to find where f of x is greater than 0? we already found where f of x is equal to zero.
04:57
We just use those values to create intervals.
05:03
I'll demonstrate by putting it on a number line.
05:08
So we found that x equals 2 and x equals negative 1 are the solutions of that quadratic.
05:19
So that creates three intervals on the number line, where x is less than negative 1, where x is between negative 1 and 2, and where x is greater than 2.
05:37
So we just choose a value in each of those intervals and evaluate our function there to see if it gives us an answer or a value that is greater than 0.
05:49
I'm going to start with x equals 0 just because that's really easy.
05:56
F of 0 is equal to 0 squared minus 0 minus 2, which is negative 2.
06:09
Negative 2 is not greater than 0, so that interval is not in our solution.
06:18
If we chose x equals negative 2 to test this interval where x is less than negative 1, we would get negative 2 squared minus negative 2 minus 2, which is 4, because negative 2 times negative 2 is 4, minus negative 2, which is plus 2, so that would be 4 plus 2 is 6, and then 6 minus 2 is 4.
06:53
And 4 is greater than 0, so that interval does work, and then we could choose a value in this interval for instance, 3, and find that f of 3 is equal to 3 squared minus 3 minus 2, which is 9 minus 3 minus 2, 9 minus 3 is 6, 6 minus 2 is 4, and 4 is greater than 0.
07:27
So those intervals are the solutions to this inequality.
07:31
So you could write that x is less than negative 1 and x is greater than 2, or in interval notation, we would write it as negative infinity to negative 1 and 2 to infinity.
07:52
We use parentheses on our end points because it is a greater than inequality, not greater than or equal to.
08:06
How about g of x is less than or equal to zero? we'll scroll down to get a little more space here.
08:30
So g of x is less than or equal to zero.
08:42
So we are going to substitute this for g of x because this tells us that that's what g of x is equal to.
08:57
And we're going to say that's less than or equal to zero.
09:02
So we have x squared plus x minus 2 is less than or equal to zero.
09:11
So we have our values that we found when we set it equal to zero.
09:19
So we're going to do the same thing here on the number line.
09:24
And we found that the solutions or the roots or the zeros of g of x were at 1.
09:31
And negative 2.
09:37
So that's where our intervals will fall.
09:43
So we have x is less than negative 2, but it's actually less than or equal to because that's what our inequality is.
09:56
And then we have negative 2 is less than or equal to 0.
09:59
Oh, sorry, it's less than or equal to x.
10:07
It's less than or equal to 1.
10:10
Okay, so x is between negative 2 and 1.
10:13
And then x is greater than or equal to 1 because we're going to the right here.
10:23
Okay, now we test our intervals like we did before.
10:28
So we'll start with 0 because that's an easy one.
10:35
Okay, so g of 0 is equal to 0 squared plus 0 minus 2, which is negative 2.
10:49
Is that less than or equal to zero? yes it is.
10:55
So that interval works.
10:58
We should still go ahead and check the others just because that's good math.
11:02
So we're going to we're going to check x equals negative 3 here because that's in our interval here on the left...
