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Aphysicist uses Ihe fact that sin X = x for very small values of x to solve problem: Why was she justified in doing thal?...

Question

Aphysicist uses Ihe fact that sin X = x for very small values of x to solve problem: Why was she justified in doing thal?

Aphysicist uses Ihe fact that sin X = x for very small values of x to solve problem: Why was she justified in doing thal?



Answers

(a) Estimate $f^{\prime}(0)$ if $f(x)=\sin x,$ with $x$ in degrees. (b) In Example 3 on page $93,$ we found that the derivative of $\sin x$ at $x=0$ was $1 .$ Why do we get a different result here? (This problem shows why radians are almost always used in calculus.)

Okay, so for this question, we're gonna use the triangle ABC. So we have a B. C. We're gonna leave all this as one, this as X. So we have the sign of a is equal to X. So the side inverse of X is equal to a co sign of B is equal to x so close that the universe of X is equal to be so we can write the sine inverse of X plus. The coastline in verse of X is equal to a plus B which is equal to pi minus C which is equal to pi over pi minus pi over to which is equal to pi over too. And we know that X the left hand says is defined for X is negative 1 to 1.

Since the range of the university function is from miners pi over two to a pi over two, this statement is only true for acts in this range give a counter example if articles part sign your X, he calls zero, that's inverse sine zero, ico zero, it does not equal time.

Okay, We just need a quick explanation as to why this problem is impossible. And and, uh, I have to do my best to kind of show work. But I'm going to concentrate on this statement right here in verse. Co Sign of to Because, as you recall, if you were to do, like cosine of some data, think about what kind of answers you could get. And some people use the phrase. It's like the range of co sign of data. If you look in the unit Circle co signs the X coordinate, you know, so it can go all the way to the right is one zero and export it, or it can go all the way to the left of the exported negative one. Um, so the range of cosine is Onley between negative one and one. Um, so if you were to go, like, do the inverse cosine So that would be that the domain as I misspell that of inverse cosine. Uh, maybe I shouldn't be using data there. I should use a different variable, um co sign of inverse co sign of I don't know. I'll use X, even though that's probably wrong. Needs to be the same as the range of cosine. So it has to be negative 1 to 1. So what that means is this doesn't make any sense. Two is not within that range. Uh, so the sine inverse sign of X is equal to one possible, and that doesn't make any sense. How are you going to answer that question like X equal? Sign of no solution. The big Yikes. Big Yikes is happening. Um, I wouldn't write big Yanks on your piece of paper. Um, but this is more or less a statement, and that's how what I would go to to say, Hey, this is not true. Um, and if you want a longer explanation, could probably try and solve it, but that's why I'm sticking with

So here we have e grass. Um, sign of X this year is why people to sign X. And this here is the angry heiress. So why is he going to sign to me? Name one. So we have these two punches graph and were asked if I'd a Richard II vets with these two functions to be equal to each other. So at what will you expose? Sign effects equal to or sign of X. So once we grasp that, we can find the intersection here. The intersection What the at zero. So the value of X it was sign of X would be equal to our sign of X. Would be zero so ex would have to be, Which is you, since this is where the intersection is at and so


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