A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

rational

  • one year ago

Consider a quadratic function \(f(x)=x^2 + mx + n\). Find the real numbers \(m\) and \(n\) such that the maximum distance between \((x,0)\) and \((x, f(x))\) is minimum in the interval \([-1,~1]\)

  • This Question is Closed
  1. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    .

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

    hey @rational so we want the max D to be an element of [-1,1]?

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

    sorry distance

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

    not only that but we want it to be a minimum of [-1,1]?

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

    sorry \( x~\in ~[-1, 1] \) we want to find \(m,~n\) such that the deviation of \(f(x)\) from \(x\) axis is minimum

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

    oh okay

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

    |dw:1432795011006:dw|

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

    we get different max deviations for different values of m,n need to find m,n that give minimum value for the max deviation

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

    I think I might be interpreting things wrong Take f(x)=x^2+n for example 0 is in [-1,1] and the distance between (0,f(0)) and (0,0) is largest but the thing is there is no largest real number n

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

    yes there is no largest "maximum deviation" but there certainly exists a least "maximum deviation"

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

    for \(f(x,n) = x^2+n\), it seems the least "maximum deviation" will be less than \(1\) https://www.desmos.com/calculator/rtanhulgw9

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

    My intuition tells me the correct answer is going to be \[f(x) = x^2-\frac{1}{2}\] The maximum deviation of f(x) from the line y=0 is the absolute value of f(x), and |f(x)| at the end points to be 1/2 and vertex to be 1/2. Any other value of m will increase these values. Look at this graph of the absolute value. https://www.desmos.com/calculator/9cozemnmgj when m>0 the endpoints are larger than 1/2 and when m<0 the vertex is larger than 1/2, maknig m=0 the local min.

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

    that looks very convincing, so if i understand correctly the least "max deviation" for \(f(x)=x^2 + mx + n\) is \(\frac{1}{2}\) and is achieved when \((m,n)=(0,-\frac{1}{2})\)

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

    Yeah, that's what I'm thinking, I wish I could come up with a better argument but it seems kinda obviously right to me. In my mind, it has to be at m=0 since it minimizes the derivative in a sense on the interval [-1,1] and the reason that matters is because the derivative is what causes |f(x)| to increase.

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

    Interestingly when I first did this I calculated the average minimum deviation and got m=0 and n=-1/3 which is slightly different, but also useful and definitely practical in physics. =D

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

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

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.