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Determining where the function is concave up? (Given an integral)

  1. Apr 12, 2014 #1
    1. The problem statement, all variables and given/known data

    x
    Determine all intervals over which the function f(x)=∫√(1+t^2) dt is concave upward
    1


    2. Relevant equations

    I know concave up means the f"(x)>0, so you have to get the second derivative


    3. The attempt at a solution

    f'(x)=√(1+x^) by susbituting in x

    Then f"(x)=1/2*(1+x^2)^(-1/2)*2x

    This is equal to:
    2x/2√(1+x^2)

    Then, I'm really confused on what to do next because the denominator is never zero, so would the only critical point be 0 and you check to the left and right of that?
     
  2. jcsd
  3. Apr 12, 2014 #2

    LCKurtz

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    Apparently that is ##\int_1^x \frac 1 {\sqrt{1+t^2}}~dt##

    Yes. Find where it is positive. Don't be afraid to cancel the 2's.
     
  4. Apr 12, 2014 #3
    I apologize for the typo! So it would be:

    x/√(1+x^2)

    The only critical number would be 0 then, so wouldn't it be concave up on (-infinity, 0) and (0, infinity) since the denominator is always positive because of the square root?
     
  5. Apr 12, 2014 #4

    LCKurtz

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    Is the numerator always positive?
     
  6. Apr 12, 2014 #5
    The numerator would give only a critical point of 0 since that is the only place it is not defined, right? So then it would be concave up on (0, infinity)?
     
  7. Apr 12, 2014 #6

    LCKurtz

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    There is no ##x## where that fraction is not defined. Instead of asking me if ##(0,\infty)## is the answer, explain why you think it is.
     
  8. Apr 12, 2014 #7
    It would be (0, ∞) since the critical point is 0 because it makes the fraction zero. Concave up means f"(x)>0 and f"(x) is positive when the function is evaluated at any number greater than 0. If you evaluate the function at -1, for example, you would get a negative number, so it would be concave down less than 0. If that makes sense?
     
  9. Apr 12, 2014 #8

    LCKurtz

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    Yes, that makes perfect sense. Since ##f''(x)=0## when ##x=0##, that is a point where the second derivative might change sign (or maybe not). That's why you check the signs on both sides. For ##x<0## you have ##\frac - +## and for ##x>0## you have ##\frac + +## for the signs. That's why ##f''(x)>0## for ##x>0##. That's how you want to analyze problems like this.
     
  10. Apr 12, 2014 #9
    Thank you so much! I really appreciate it!
     
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