Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

DeoxNA

  • 2 years ago

f(x)=2x/(x^2+6) for x>=b (b is a real number) 1. Find f'(x) 2. Hence find the smallest exact value of b for which the inverse function f^(-1) exists. My question is: isn't b arbitrary? Couldn't I just pick any real number and then the domain of the inverse would just be x>f(b) since the domain and range switch between inverses. Except I couldn't pick an imaginary number, which curiously is also the domain restriction of f(x) and f'(x) i.e. sqrt(-6). Also, what is the purpose of finding the derivative?

  • This Question is Closed
  1. mark_o.
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    integral of 2x/(x^2+6) let u=x^2+6 du=2x dx therefore 1ntegral of du/u=ln u +C integral of 2x/(x^2+6)dx=ln( ? ) +C

  2. ZeHanz
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    b is not arbitrary. The question is a little unclear, imo. See attached Geogebra graph. As you can see, if b becomes too small, f doesn't have an inverse anymore. The minimum value for b is where f has its max. That's the reason why you have to differentiate f: to calculate where it has its max.

    1 Attachment
  3. ZeHanz
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    The derivative is:\[f'(x)=\frac{ 2(x^2+6)-2x \cdot 2x }{ (x^2+6)^2 }=\frac{ 2x^2+12-4x^2 }{ (x^2+6)^2 }=\frac{ 12-2x^2 }{ (x^2+6)^2 }\]To find the extremes of f we have to solve f'(x)=0. Looking at the denominator of f', we see that it is greater than 0 for every x, so we don't bother with it:\[f'(x)=0 \Leftrightarrow 12-2x^2=0 \Leftrightarrow 6-x^2=0 \Leftrightarrow (\sqrt{6} -x)(\sqrt{6} + x)=0\]Because x > b and b >0 (see graph) we only have one solution: x = √6. For this value, f has its maximum. So taking b = √6 ensures we have an inverse of f on the interval\[[\sqrt 6, \infty)\]

  4. DeoxNA
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thanks! What was confusing me was that (and I know that it sound obvious) f^(-1) HAD to be a function and hence pass the vertical line test. I was just thinking about drawing the inverse of the function and not realizing that the inverse had to be a function and not a relation.

  5. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

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.