Find the domain of arccsc(e^2x)

[Vital]

Homework Statement

Hello!
Seems I have forgotten how to work with such values. Please, help me to find the correct approach to this type of exercises.

Homework Equations

The task is to find the domain of the function:
f(x) = arccsc(e^2x)

The Attempt at a Solution

The answer in the book is [0, ∞).

The domain of arccsc(x) is (-∞, -1] U [1, ∞)

(1) In the given function x is an exponent of epsilon, and also it has coefficient 2. So, I see that I need to divide by 2.

(2) So, if x = -1, then the exponent will be -½, and thus it would mean that I need to take "a negative square root" of e. This means that I have to discard this option, and thus discard all negative values, and concentrate only on [1, ∞) part.

(3) If x = 1, then exponent will be ½, meaning I need to take the square root of e.

I don't see my next steps. Logically it does look as if all values starting from 0 and till ∞ are correct, because all can be e's exponent, but how to get there mathematically?

Thank you very much!

 

[haruspex]

x is an exponent of epsilon

Euler's constant is e, not epsilon

if x = -1, then the exponent will be -½

Try that again.

For this question, it might be simpler to think in terms of what values of x produce a disallowed argument to arccsc, i.e. (-1,1).

 

[carpenoctem]

e^2x is always positive. Therefore, its range is (0,+∞). However, as you figured out, arccsc(x) does not accept (0,1) as input.
What you need to find is the range of the values x such that e^2x falls in [1,+∞) because if not, then arccsc(e^2x) is not defined.

Can you solve the inequality e^2x ≥ 1?

 

[Vital]

Euler's constant is e, not epsilon

Try that again.

Do I understand you correctly that -½ is not the right approach here? I divided by 2 following "graph transformation" rule, which state that we divide x by its coefficient.
For example:
The graph y = x
If I want to graph y = 2x, then I can divide all values of x in y = x by 2 to get the desired graph.

 

[haruspex]

Do I understand you correctly that -½ is not the right approach here? I divided by 2 following "graph transformation" rule, which state that we divide x by its coefficient.
For example:
The graph y = x
If I want to graph y = 2x, then I can divide all values of x in y = x by 2 to get the desired graph.

You have e^2x. If x=1, what is the exponent on e?

Anyway, I think this is already headed in the wrong direction. You have f(x)=arccsc(y) where y=e^2x. You have established that the range for y is (-∞, -1] U [1, ∞). If those are values of e^2x, how do you find the values of 2x?

 

[Vital]

e^2x is always positive. Therefore, its range is (0,+∞). However, as you figured out, arccsc(x) does not accept (0,1) as input.
What you need to find is the range of the values x such that e^2x falls in [1,+∞) because if not, then arccsc(e^2x) is not defined.

Can you solve the inequality e^2x ≥ 1?

I see ) Thank you very much.
Yes, I can solve e^2x ≥ 1
I set e^2x = 1
then x = ln(1)/2 = 0
I see that I tried to tread a wrong path.

 

[Vital]

You have e^2x. If x=1, what is the exponent on e?

Anyway, I think this is already headed in the wrong direction. You have f(x)=arccsc(y) where y=e^2x. You have established that the range for y is (-∞, -1] U [1, ∞). If those are values of e^2x, how do you find the values of 2x?

Yes, I see that I indeed took a completely wrong direction - I unnecessarily complicated the matter. I have posted the result of required computation.
Thank you very much!

 

[haruspex]

Yes, I see that I indeed took a completely wrong direction - I unnecessarily complicated the matter. I have posted the result of required computation.
Thank you very much!

Ok.