Find domain and range of function algebraically?

[tomtisner17]

f(x) = cot(2 arccos x)

I have only had one class so far and this is due in a bit, I am not sure what to do?

So far I used identities and simplified into

f(x) = ((cos/sin)(arccos/1)-1)/(2(cos/sin)(arccos))

f(x) = (1-sinx)/2 ??

I still have no conception of if this is right or even if I'm doing the right thing. Just some slight hints or explanations of the broad concept would be fantastic

 

[Deleted member 4993]

f(x) = ((cos/sin)(arccos/1)-1)/(2(cos/sin)(arccos))

Which "identity" did you use to derive that(↑).

 

[Otis]

Hi Tom. Instead of trying to rewrite function f, we could view it as a composite function.

Work with the inside function first (arccosine).

What are its domain and range?

Use those restrictions and think about necessary adjustments to the domain and range of cot(2t), where t is the output of arccos(x).

?

 

[tkhunny]

For starters, neither of these means anything

((cos/sin)(arccos/1)

(cos/sin)(arccos)

Thus, it is safe to assume that something went seriously wrong in order for you to get here.

 

[JeffM]

f(x) = cot(2 arccos x)

I have only had one class so far and this is due in a bit, I am not sure what to do?

So far I used identities and simplified into

f(x) = ((cos/sin)(arccos/1)-1)/(2(cos/sin)(arccos))

f(x) = (1-sinx)/2 ??

I still have no conception of if this is right or even if I'm doing the right thing. Just some slight hints or explanations of the broad concept would be fantastic

First, Otis gave you the very broad brush picture for such problems.

You have a composite function [imath]f(x) = g(h(x)).[/imath]

What is the domain of h? Wherever h(x) is undefined, then f(x) will be undefined. Make sense?

What is h(x) in this problem? What is its domain? What is its range?

Now g may have a very broad domain, but not all of that domain may be relevant to the composite function. First, we are limited to the domain of h. Let h(x) = y. What is the range on y? Nothing outside that range is relevant to g(h(x)) = g(y). In other words, the range of h is going to limit the domain of g with respect to the composite function. Furthermore, any value in the range of h that is not in the domain of g is also not in the domain of the composite function. So we must ask what is the domain of g within the limits set by the range of h?

What is g(y) in this problem? What are the limits within which we must look for a domain?

Second, I presume you are working with radian measure. Is that correct?

Third, I do not see a necessity to transform the cotangent, but if you do so, you need to do so correctly.

[math]f(x) = cot(2 * arccos(x)) = \dfrac{cos\{2 * arccos(x)\}}{sin\{2 * arccos(x)\}} [/math]

 

[Deleted member 4993]

f(x) = cot(2 arccos x)

I have only had one class so far and this is due in a bit, I am not sure what to do?

So far I used identities and simplified into

f(x) = ((cos/sin)(arccos/1)-1)/(2(cos/sin)(arccos))

f(x) = (1-sinx)/2 ??

I still have no conception of if this is right or even if I'm doing the right thing. Just some slight hints or explanations of the broad concept would be fantastic

Which "identity" did you use to derive that(↑).

TomTisner - Please respond to the query above.

 

[Steven G]

Here is how I would find the domain of f(x) = cot(2 arccos x)
1) Ask yourself what you can't compute the cot of.
2) Then compute the x-values where 2arccos(x) equals those values you found in step 1
3) The domain will be all real values EXCEPT for the x-values you found in step 2.

 

[JeffM]

Here is how I would find the domain of f(x) = cot(2 arccos x)
1) Ask yourself what you can't compute the cot of.
2) Then compute the x-values where 2arccos(x) equals those values you found in step 1
3) The domain will be all real values EXCEPT for the x-values you found in step 2.

@Jomo That method ignores the limitation on the domain of the arc cosine function.

 

[Steven G]

@Jomo That method ignores the limitation on the domain of the arc cosine function.

OOPS!