Explain how to apply the First Derivative Test.

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Hi, everyone, let's take a look at this practice problem. This problem says, given the function G of X is equal to 2 x 4 minus 8 X2 + 4, identify the X values at which G of X has a local extremum. We're given 4 possible choices as our answers. For choice A, we have X is equal to 0, and X is equal to the square root of 2. For choice B, we have X is equal to 0, X is equal to the square root of 2, and X is equal to minus the square of 2. For choice C, we have X is equal to the square root of 2, and X is equal to minus the square root of 2. And for choice D, we have X's equal to 0, and X is equal to minus the square root of 2. Now we're asked to identify the X values of which G of X has a local extremum. So in order to identify the X values for our local extremum, we first need to identify our critical points. To recall that our critical points are found by setting our first derivative equal to 0 and then solving for X. So we need to calculate G of X. Which is going to be equal to the derivative with respect to X. Of the quantity of 2 X minus 8 X2 plus 4 in quantity. And so to take this derivative, we're going to need to use our constant multiple and power rule. And so using those rules, this is going to be equal to 8 X cubed. Minus 16 X. Plus 0, that when we take the derivative of a constant, it is equal to 0. So, to find our critical points, we need to set this expression equal to 0. So we will have 8 X cubed minus 16 X is equal to 0, and then we need to solve for X. So I noticed on the left hand side I can factor out an 8 X out of both terms. And so we'll have 8 X multiplied by the quantity of X2 minus 2 in quantity is equal to 0. So to solve this, I'll set each factor here equal to 0 and solve for X. So, we'll have 8 X equal to 0, and then solving for X, we'll have X is equal to 0. And then our other factor will have X2 minus 2 is equal to 0, and solving for X we'll actually get two solutions. We'll have X is equal to the square root of 2, and X is equal to minus the square root of 2. So, we have 3 points that we need to check. And so, we're going to use our first derivative test to determine what type of critical point we have, whether we have a local maximum or minimum, or neither. So, we're gonna start off with X equal to 0. As you recall for the first derivative test, I need to um calculate the value of our first derivative, using points to the left and right of our critical point. So we're going to choose X equal to -1 as our first point. So we'll look at G of -1, and that is going to be equal to 8 multiplied by -1 cubed, minus 16 multiplied by -1, and this here is going to be equal to 8 when we evaluate that expression. The other point that we're going to look at is going to be G of 1, so we'll look at X equal to 1, which is to the right of X equal to 0. And so this is going to be equal to 8 multiplied by 1 cubed, minus 16 multiplied by 1, and this is equal to -8. So we see that to the left of our critical point at X equal to 0, our first derivative is positive, meaning that our function is increasing, and to the right of our critical point, our function is decreasing since our first derivative here is negative. So if the function is increasing, then decreasing later, that means that we have a maximum. So this here is a local maximum, and so X equal to 0. is a local extremum. So next we'll look at X is equal to wave root of 2. And so again, we'll have to choose 2 points to the left and right of this critical point, and we'll choose X is equal to 1 and X is equal to 2. And so for G of 1, we've already calculated it, and so that is equal to -8. And so we just need to calculate G of 2. Which is equal to 8, multiplied by 2 cute. -16 multiplied by 2. And when we evaluate this expression, this is equal to 32. So, to the left of our critical point, our first derivative is negative, which means our function is decreasing. And to the right of our critical point, our first derivative is positive, so it is increasing. So, since the function is decreasing and then increasing, that means we have a local minimum. So, X equal to square root of 2 is a local extremum. And finally, the last point that we need to check is X is equal to minus the square root of 2. Again, we'll choose two points, one to the left and one to the right of our critical point. And so the point to the left that we're going to look at is going to be X is equal to -2, so we'll calculate G of -2, and this is going to be equal to 8 multiplied by -2 cubed minus 16 multiplied by -2. And this is going to be equal to minus 32. And the other point that we'll choose is X equal to -1, so we'll have G of -1, which we have already calculated, and this was equal to 8. So again, here we see that to the left of our critical point, our first derivative is negative, which means our function is decreasing, and to the right of our critical point, our first derivative is increasing since our, our, our function is increasing since our first derivative is positive. And therefore, that means we have a local minimum. So X equal to minus 2 is a local extremum. So that means that actually all three of our points are local extremum. So we have X equal to 0, X equal to square rooted 2, and X equal to minus 2 rooted 2 as the answer to this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

Explain how to apply the First Derivative Test.

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