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Second Derivative Test (trig problem)

  1. Jul 2, 2011 #1
    1. The problem statement, all variables and given/known data
    I am doing the various ins and outs of curve sketching and the mean value theorem and all that jazz with this function:
    [tex]f(x)=sec(x)+tan(x)[/tex]

    2. Relevant equations

    3. The attempt at a solution
    I took the first derivative to be:
    [tex]f'(x)=sec(x)tan(x)+sec^{2}(x)[/tex]

    I am having trouble finding the inflection points with the second derivative.
    I took the second derivative to be:
    [tex](sec(x)tan(x))tan(x)+sec(x)(sec^{2}(x))+2sec(x)(sec(x)tan(x))[/tex]
    I simplified to: (left out the x's for brevity)
    [tex]sectan^{2}+sec^{3}+2sec^{2}tan[/tex]
    [tex]sec(sec^{2}-1)+sec^{3}+2sec^{2}tan[/tex]
    [tex]sec^{3}-sec+sec^{3}+2sec^{2}tan[/tex]
    [tex]2sec^{3}-sec+2sec^{2}tan=0[/tex]

    Trying to solve this set to zero for the second derivative test and I have no idea how to go about it without using a calculator so far. I have not been able to simplify it into something that I can solve?

    I was thinking that maybe I could square both sides (including the zero) and then use the pythag identity to convert that last tangent into a secant, and then maybe I would have a polynomial that I could use PQ test to solve or something (but idk if that is ok to do, seems reasonable to me). Any pointers?
     
  2. jcsd
  3. Jul 2, 2011 #2

    micromass

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    Hi QuarkCharmer! :smile:

    Your first and second derivative are correct. Note however, that we can write your second derivative in a nicer form:

    [tex]sec(x)(tan^2(x)+2sec(x)tan(x)+sec^2(x))[/tex]

    Now, isn't this a nice form?? No? Don't you recognize a nice formula now??
     
  4. Jul 2, 2011 #3
    Instead of simplifying the second derivative the way you did, try another way. Put it into terms of sin and cos right away. Then you have three fractions whose denominators are all the same. Put them together, and the top half easily reveals a zero value.
     
  5. Jul 2, 2011 #4
    Okay, I think I get what you are saying.

    From here, I did this:
    [tex]sectan^{2}+sec^{3}+2sec^{2}tan[/tex]
    [tex]\frac{sin^{2}+2sin+1}{cos^{3}}[/tex]
    Now, I see that the top results in 3pi/2 +/- 2kpi, but I am confused as to whether I should concern myself with solving for when the denominator (cos^3) is equal to zero to find the undefined points?
     
  6. Jul 2, 2011 #5

    SammyS

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    Usually, that's a good idea. The concavity certainly can be different on either side of the locations at which the second derivative is undefined.

    For this function, f(x), you will also want to determine such things as the domain & sign(s) of the first derivative, etc.
     
  7. Jul 2, 2011 #6
    Okay, I have everything worked out but I am having trouble determining the Absolute/Local Max and Mins. The function sec + tan is restricted to the domain where 0>x>pi/2, and it is undefined at pi/2, either way, that is not part of the range as the domain restriction is not greater than or equal to. So uh, how would I figure out the max/mins? I get that the max is infinity, but the min has me stumped.
    I'm thinking the max/min does not exist?

    It looks like this:
    2kir9h.jpg
     
    Last edited: Jul 3, 2011
  8. Jul 3, 2011 #7

    SammyS

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    No, it's not restricted to the interval (0, π/2).
    You should probably graph at least one period of this function. Yes, it's periodic.

    Yes, it's undefined at π/2, but also at many other places.

    What are the domains of sec(x) & tan(x) ?
     
  9. Jul 3, 2011 #8
    I mean the problem states that it is restricted on that interval. I understand how the full function should look. So it looks like the Max/Min is undefined for that interval right?
     
  10. Jul 3, 2011 #9

    SammyS

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    Right, as long as x ≠ 0.
     
  11. Jul 3, 2011 #10
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