Lengths of segment AD and the length of segment BD. So we've already been able to figure out several interesting things We've been able to figure We're just going to have square root of two times The square root of two, and the denominator Which would be, let's see the numerator will have So we could rationalize the denominator by multiplying by the square root of two over the square root of two, Having a radical denominator, they don't like having a rational Root of both sides, we get X is equal to one Is equal to one over two or taking the principal Two X squared is equal to one or that X squared ![]() Squared plus this X squared is equal to the hypotenuse This side has length X, that means that this side has length X. So how does that help us figure out the lengths of the sides? Well if you say that So if your base anglesĪre the same, then you also know that the corresponding When your two base angles are the same, you know you're dealing Now this is pi over four radians and this is also pi over four radians. So the triangle, weĬould make it look like, a little bit more like this. Maybe not completely fit it over but if we were to make it look like this. I can reorient it in a way that might make it a little easier to realize. That means that this side is going to be congruent to that side. That means that the corresponding sides are also Lengths of segment AD and the length of segment DB? Well sure, because we have two base angles that have the same measure, Know about this triangle? Can we figure out the The radius of the circle is length one, what else do we Of segment AB, which is a radius of the circle or is ![]() Radians, and once again we know that this is a unit circle. Now what does that help us with? So if we know that this is pi over four and that is pi over four ![]() So the measure of angleĪBD is actually the same as the measure of angleīAD, it is pi over four. So this is going to beĮqual to, you can put a common denominator ofįour, then this is four pi over four, this is minus two pi over four, and this, of course, Measure of angle ABD is equal to pi minus pi over two, minus pi over four. Them, the interior angles of this triangle, they're What would that be in radians? Well a right angle in radians, a 90 degree angle in radians, is pi over two radians. Of angle ABD in radians, plus pi over four, plus, This right over here, lets just say measure So this angle plus that angleĪre going to add up to pi. Instead of saying 180 degrees, 180 degrees is the So we could say that the sum of the angles of a triangle add up to, To 180 degrees, but now we're thinking in terms of radians. Of the interior angles of a triangle add up Now what might be a littleīit unfamiliar, we're use to saying that the sum We know that two of theĪngles of this triangle, so if you know two of theĪngles of this triangle, you should be able toįigure out the third. So I assumed you've paused the video and tried to do this on your own. To figure out is what's the radian measure of angle ABD? Actually let's just do thatįirst and then I'll talk about the other things. Our knowledge of triangles in order to figure out several things. Video, is use our knowledge of trigonometry and use So they form this triangleĪBD, and they tell us at angle BAD, angle bad, it has a ![]() And then we dropped a perpendicular from point B to point D, point D Step 2.Here is a unit circle centered at point A and the The derivative is \(f'(x)=3x^2−6x−9.\) To find the critical points, we need to find where \(f'(x)=0.\) Factoring the polynomial, we conclude that the critical points must satisfy Use the first derivative test to find the location of all local extrema for \(f(x)=x^3−3x^2−9x−1.\) Use a graphing utility to confirm your results. \): Using the First Derivative Test to Find Local Extrema
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