Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

JenniferSmart1

I found the inverse of \[f(x)=-2cos(3x)\] \[f^{-1}(x)=\frac 13 cos^{-1}(-\frac x 2)\] how do I find the domain? Is it all real numbers?

  • one year ago
  • one year ago

  • This Question is Closed
  1. JenniferSmart1
    Best Response
    You've already chosen the best response.
    Medals 0

    @satellite73

    • one year ago
  2. JenniferSmart1
    Best Response
    You've already chosen the best response.
    Medals 0

    @Callisto

    • one year ago
  3. Callisto
    Best Response
    You've already chosen the best response.
    Medals 1

    Domain of... f(x) or f^(-1) (x)?

    • one year ago
  4. JenniferSmart1
    Best Response
    You've already chosen the best response.
    Medals 0

    of \[f'(x)\]

    • one year ago
  5. JenniferSmart1
    Best Response
    You've already chosen the best response.
    Medals 0

    i Mean \[f^{-1}\]

    • one year ago
  6. Callisto
    Best Response
    You've already chosen the best response.
    Medals 1

    Try \(f^{-1} (100)\), what would you get?

    • one year ago
  7. JenniferSmart1
    Best Response
    You've already chosen the best response.
    Medals 0

    it's an imaginary number according wolfram

    • one year ago
  8. Callisto
    Best Response
    You've already chosen the best response.
    Medals 1

    I suppose the function should give real output if you input a real number. Okay, when you put x= 100 into f^(-1) (x), it doesn't give you real output. So, it's not in the domain of y. Consider cosine function, range of cosine function is -1 ≤ cos x ≤ 1. Consider the inverse of cosine function, domain of the inverse is the range of the cosine function, so the domain of cos^(-1)x is [-1, 1]. Now, you have cos^(-1) (x/2) in your inverse function, hmm.. how would you get the domain of the inverse function?

    • one year ago
  9. JenniferSmart1
    Best Response
    You've already chosen the best response.
    Medals 0

    Let's see if I understand this... \[f^{-1}(f(x))=sin^{-1}(sinx)=x \;\;\;\;where \;\;\;\;-\frac{\pi}{2}\le x\le\frac{\pi}{2}\] \[f(f^{-1}(x))=sin(sin^{-1}x)=x \;\;\;\;where\;\;\;\;-1\le x\le1\]

    • one year ago
  10. JenniferSmart1
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1360683713662:dw|

    • one year ago
  11. JenniferSmart1
    Best Response
    You've already chosen the best response.
    Medals 0

    I"m trying to picture this. What's bothering me is that fact the domain and range of one function change to the range and domain of the next function....I'm trying to keep them in order somehow

    • one year ago
  12. phi
    Best Response
    You've already chosen the best response.
    Medals 0

    the cosine gives numbers between -1 and 1 there is no real number x, where cos(x)= 2 (for example) so x = acos(2) has no real solution. You would want to restrict the domain to -1 to +1 for the case of acos(x/2) -1 ≤ x/2 ≤ 1 or -2 ≤ x ≤ 2

    • one year ago
  13. JenniferSmart1
    Best Response
    You've already chosen the best response.
    Medals 0

    but the restriction of the inverse of cosine is \[0\le x \le \pi\] but since it's x/2 do I multiply or divided that by two? or am i totally on the wrong path...

    • one year ago
  14. phi
    Best Response
    You've already chosen the best response.
    Medals 0

    you mean the restriction on the cosine is 0 to pi the restriction on the domain of the inverse cos(x) is -1 ≤ x ≤ 1

    • one year ago
  15. phi
    Best Response
    You've already chosen the best response.
    Medals 0

    the restriction on the domain of the inverse cos(x) is -1 ≤ x ≤ 1 if you are given acos(x/2) then rename the variable to y= x/2 you know the domain is restricted to -1 ≤ y ≤ 1 but with y= x/2 -1 ≤ x/2 ≤ 1 solve for x, (2 separate relations), to get -2 ≤ x ≤ 2 that is the domain in terms of x for acos(x/2)

    • one year ago
  16. JenniferSmart1
    Best Response
    You've already chosen the best response.
    Medals 0

    I find gauss's law easier to understand than this for whatever reason. I'm very visual and I can't visualize this....I would much rather draw unit circles and sine graphs, and modify that to represent the current problem that we're working on.

    • one year ago
  17. JenniferSmart1
    Best Response
    You've already chosen the best response.
    Medals 0

    I'll look at it again later...too tired to comprehend.

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.