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# Find the critical numbers of the function.$h(t) = 3t - \arcsin t$

## $t=\pm \frac{2 \sqrt{2}}{3}$ and $t=\pm 1$ are the four Critical numbers of $h(t)$

Derivatives

Differentiation

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##### Kristen K.

University of Michigan - Ann Arbor

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we're going to find the critical numbers of the function H f D equals three times t minus arc sine of T. So um first we talk about the domain of this function. The domain of age is values of T. For which uh three T N R. Sine of T R defined three D. Is defined for every real number T. But the art sign of tea is to find only four numbers Between 81 and one. Including those values. That is because we look at Axinn means the angle at which the sign and equal the number T. So then everything can only be A value between 81 and one and can be either of these two values 81 or one. So it's a close interval. Now the 1 1 that's the main of age. And now we calculate the first serve a tive of age We get 3 -1 over and over the root of one minus t square. This is expression here for the relative of the art sign of T. The inverse of the sine function. And we remember definition of critical number which we're going to use here. So see into the main of the function I have is a critical number of the function. Yes. The derivative at sea the radio of the function at the value. See what you see here, wow see does not exist or it exists and Its value is equal to zero. So this is definition of critical uh number of a function we are going to use. So the first of the first thing we have to note here is that the values that can be critical numbers of functions are in the domain of the function first of all. And that's the first condition. The second one is either that the derivative at that point does not exist or derivative at that point. It's you. So the only Possible values that can be critical numbers are the values between 81 and one including native one and 1. Now let's see the conservative here and we have a square root of an expression here. This is fresh in yes, positive four T Square less than one. And that for absolute value of teal is the one that is T between 81 and one. This means that the derivative of the art sign of the inverse of the same function is the interval negative 11 without including the In Points 91 and one. So the derivative is defined for all values of T in this domain of the function except for the importance. And at that file that those values one and 91, the derivative does not exist because we have an expression that can not be eve elated at all. So the values that we can do is is that equal Plus one under negative one is in the domain of the function because they are the endpoints and they are included in the domain and derivative of age does not exist at these points. And so we used the definition and we must say that one and negative one. Art critical numbers of the function age this is because again I say it again because These numbers 1 -1 are into the main and say hear ours better. Okay okay. To take this. But for a moment Here is Erin say are because we are talking about two points. Mhm Okay. So which again? And so uh these points negative 1-1 are in the domain into function age but or besides that there's a relative that those numbers one and everyone does not exist then by definition these two values what another one must be critical numbers of age. So we have already these two critical numbers. Now we got to see if there are more than that. That is Uh points in the domain or values in the domain where derivative is zero. So we got to stay that equation here. Have you been to equal zero Is the same as 3 -1. Over square root of one. Manistee Square is zero. Which is the same as one over square root of one minus t square is three. And that's the same as saying that square root of one minus t square is one third or Yeah, the same as one minus t square gotta be 1/9. And here to have the equivalence here it's important to notice that he got to be between native one and one. That's a requirement Already said when we calculate the derivative, we can only consider values of TV T-91 and one excluding and one is good and nearly one. And for that reason because he is in the opening of the net they want one. We can say that this equivalence here is true. So going that way we can take these squares both sides because they are positive quantities. Okay, now this is equivalent also to t square ical one minus 1/9. And this is the same as the square equal to eight nine. And so to get to be equal to more or less squares of eight overnight that is squares of a key which is two squares of 2/3. And so t equal more or less. Two skirts of 2/3 are critical numbers of age but I haven't verified exactly that they are in the internet even morning, I can't do that. There is no that To spirit of two or three. It's about if you calculate that number, it's about 0.94 to age. 28 0904 16. So the positive and negative value with this magnitude here, He's in the injured for native 1-1. And so it's true that these are critical numbers because they are in the mayor of the functions and the derivative at those points is a good zero. So this uh this is the second part of the problem if you want. So in summary f sorry. H amen function age has four critical numbers T equal more or less one and T equals more or less two squares of two or 3. And this is the final answer. And I recall that we have uh look at the domain of the function, we have calculate the derivative. And we saw that there are points where the of the domain function that For those points the relative does not exist. And so those two points in this case 91-1 get to be considered critical numbers as a function that's a part of derivative not existing and the other parties there a difficult zero. And the points we found are both in the interval Nettie 11. And so they are critical numbers of age. Also In summer we have four critical members of dysfunction. The end points of the domain -1 and one where the relative that's not exist. And the numbers more or less, two scores of two or 3 which are critical numbers because they are in the domain of the functions and the derivative at those moments is equal to zero.