Page 1 of 3 123 LastLast
Results 1 to 10 of 23

Thread: Trouble solving a question: lim[x->-infty]f(x) * lim[x->+infty] < 0; find max of g(x)

  1. #1
    New Member
    Join Date
    Jan 2018
    Location
    Lisbon
    Posts
    12

    Trouble solving a question: lim[x->-infty]f(x) * lim[x->+infty] < 0; find max of g(x)

    Hello!
    I'm new to the forum, so please excuse any mistakes I might be making.

    I'm having trouble solving an exercise, I feel like there's something obvious I'm missing but can't see where...

    There's a function which is continuous everywhere and \[\lim_{x\to-\infty}f(x)\times\lim_{x\to+\infty}f(x)<0\] (both limits exist and are finite. I'm supposed to find the maximum of \[g(x)=\dfrac{1}{1+\left[f(x)\right]^2}\]

    Since those limits have different signs, exist and are finite, I believe f has maximum and minimum (given by those two asymptotes)...

    I got to the derivative of g easily, but I'm not sure what it gets me. \[g'(x)=\dfrac{-2f(x)f'(x)}{\left(1+\left(f(x)\right)^2\right)}\] but since I don't know much about f, I can't say anything about the sign of g'... I've tried to find the second derivative, but in one attempt it was always positive, so it couldn't have a maximum...

    Can anyone help me see what I'm missing? Thanks in advance!

  2. #2
    Elite Member
    Join Date
    Sep 2012
    Posts
    2,729
    Quote Originally Posted by Eohyn View Post
    Hello!
    I'm new to the forum, so please excuse any mistakes I might be making.

    I'm having trouble solving an exercise, I feel like there's something obvious I'm missing but can't see where...

    There's a function which is continuous everywhere and \[\lim_{x\to-\infty}f(x)\times\lim_{x\to+\infty}f(x)<0\] (both limits exist and are finite. I'm supposed to find the maximum of \[g(x)=\dfrac{1}{1+\left[f(x)\right]^2}\]

    Since those limits have different signs, exist and are finite, I believe f has maximum and minimum (given by those two asymptotes)...

    I got to the derivative of g easily, but I'm not sure what it gets me. \[g'(x)=\dfrac{-2f(x)f'(x)}{\left(1+\left(f(x)\right)^2\right)}\] but since I don't know much about f, I can't say anything about the sign of g'... I've tried to find the second derivative, but in one attempt it was always positive, so it couldn't have a maximum...

    Can anyone help me see what I'm missing? Thanks in advance!
    You are on the wrong track altogether. You cannot use derivatives because you do not know that f(x) is everywhere differentiable. (Assuming it is differentiable, you would still need to ensure that g(x) is everywhere differentiable and calculate the derivative correctly. Using the power rule might suggest squaring the denominator of g(x) when taking the first derivative.) Moreover, you cannot assume that the asymptotes are reached as x increases or decreases without bound. Nor can you assume that f(x) does not have absolute values greater than the absolute value of the asymptotes at some finite values of x.

    What you do know is that the asymptotes have opposite sign so f(x) has at least one zero. (Intermediate Value Theorem)

    [tex]f(a) = 0 \implies \{f(a)\}^2 = 0 \implies g(a) = 1.[/tex]

    [tex]f(b) \ne 0 \implies \{f(b)\}^2 > 0 \implies g(b) < 1.[/tex]

    Now what do you think the maximum of g(x) is?

    Differential calculus is a very powerful tool, but not all problems can take advantage of it.

  3. #3
    New Member
    Join Date
    Jan 2018
    Location
    Lisbon
    Posts
    12
    Quote Originally Posted by JeffM View Post
    You are on the wrong track altogether. You cannot use derivatives because you do not know that f(x) is everywhere differentiable. (Assuming it is differentiable, you would still need to ensure that g(x) is everywhere differentiable and calculate the derivative correctly. Using the power rule might suggest squaring the denominator of g(x) when taking the first derivative.) Moreover, you cannot assume that the asymptotes are reached as x increases or decreases without bound. Nor can you assume that f(x) does not have absolute values greater than the absolute value of the asymptotes at some finite values of x.

    What you do know is that the asymptotes have opposite sign so f(x) has at least one zero. (Intermediate Value Theorem)

    [tex]f(a) = 0 \implies \{f(a)\}^2 = 0 \implies g(a) = 1.[/tex]

    [tex]f(b) \ne 0 \implies \{f(b)\}^2 > 0 \implies g(b) < 1.[/tex]

    Now what do you think the maximum of g(x) is?

    Differential calculus is a very powerful tool, but not all problems can take advantage of it.
    First of all, thanks for the help!

    You're right, I can't assume the function is differentiable or that the asymptotes are the maximum and minimum... Silly me!

    If I'm understanding correctly, then g has a maximum when [tex] x:f(x)=0 [/tex]. I know such x exists because of the Intermediate Value Theorem. Another question: can I guarantee there's only one such x?

  4. #4
    Elite Member
    Join Date
    Sep 2012
    Posts
    2,729
    Quote Originally Posted by Eohyn View Post
    First of all, thanks for the help!

    You're right, I can't assume the function is differentiable or that the asymptotes are the maximum and minimum... Silly me!

    If I'm understanding correctly, then g has a maximum when [tex] x:f(x)=0 [/tex]. I know such x exists because of the Intermediate Value Theorem. Another question: can I guarantee there's only one such x?
    You cannot guarantee that f(x) has a unique zero. But you do not need to do so. You are asked to find the maximum of g(x). Its value will be 1 at EVERY x such that f(x) = 0. The value of g(x) will be less than 1 but more than zero at every x such that f(x) [tex]\ne[/tex] 0. Consequently the maximum value of g(x) is 1 no matter how many zeroes f(x) may have.

  5. #5
    New Member
    Join Date
    Jan 2018
    Location
    Lisbon
    Posts
    12
    Quote Originally Posted by JeffM View Post
    You cannot guarantee that f(x) has a unique zero. But you do not need to do so. You are asked to find the maximum of g(x). Its value will be 1 at EVERY x such that f(x) = 0. The value of g(x) will be less than 1 but more than zero at every x such that f(x) [tex]\ne[/tex] 0. Consequently the maximum value of g(x) is 1 no matter how many zeroes f(x) may have.
    Thank you, you've been a great help!

  6. #6
    New Member
    Join Date
    Jan 2018
    Location
    Lisbon
    Posts
    12
    Hello,

    Just one final question. Do you think it's necessary to prove that I can find a closed and limited interval from a to b and formally prove such an interval is bounded? Personally, I don't think it's needed since the function is continuous everywhere...

    Thanks in advance!
    Last edited by Eohyn; 01-27-2018 at 01:38 PM.

  7. #7
    New Member
    Join Date
    Jan 2018
    Location
    Lisbon
    Posts
    12
    Quote Originally Posted by Eohyn View Post
    Hello,

    Just one final question. Do you think it's necessary to prove that I can find a closed and limited interval from a to b and formally prove such an interval is bounded? Personally, I don't think it's needed since the function is continuous everywhere...

    Thanks in advance!
    Any help, please?

  8. #8
    Elite Member
    Join Date
    Sep 2012
    Posts
    2,729
    Quote Originally Posted by Eohyn View Post
    Any help, please?
    I don't understand the question. The interval [a, b] is bounded; that is what the notation means.

  9. #9
    New Member
    Join Date
    Jan 2018
    Location
    Lisbon
    Posts
    12
    Quote Originally Posted by JeffM View Post
    I don't understand the question. The interval [a, b] is bounded; that is what the notation means.
    That's what I thought, but the teacher is saying unless you prove the interval is closed and bounded, the answer is very incomplete...

  10. #10
    Senior Member
    Join Date
    Nov 2017
    Location
    Rochester, NY
    Posts
    1,590
    Quote Originally Posted by Eohyn View Post
    That's what I thought, but the teacher is saying unless you prove the interval is closed and bounded, the answer is very incomplete...
    It will help if you show us the exact answer you submitted, so we can see what might have been omitted. As it is, it is not clear what a and b you are referring to.

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •