Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: F(x)=0, 2 - (cscx)^2 = 0

  1. Nov 16, 2011 #1
    1. The problem statement, all variables and given/known data

    Analyze the function f(x) = 2x + cotx over the interval [0,pi] and state the increases/decreases and concavity.

    2. Relevant equations

    I already found the first and second derivatives. However, I'm having trouble when setting them equal to zero.

    2 - csc2x = 0

    and

    2(cscx)^2(cotx)^2 = 0

    3. The attempt at a solution

    2 - cscx^2 = 0

    cscx^2 = 2

    maybe 1/sinx^2 = 2

    maybe (1/sinx)^2 = 2

    and then i'm stuck...
     
  2. jcsd
  3. Nov 16, 2011 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Pause with 1/(sin(x)^2)=2. Use algebra to rearrange that to sin(x)^2=1/2. Do you know how to do that? I'm not sure why you are getting stuck. You are doing fine so far.
     
  4. Nov 17, 2011 #3
    Thanks for the help. Yea i was doubting whether I should isolate sinx or just x. I was able to arrange the equation and solve for the angles to get x = pi/4, 3pi/4 since the interval is [0,pi].

    However, I'm getting stuck when trying to find the inflection points. The second derivative is 2cosx/sin(x)^3. I know cosx/sinx is cotx. But the power of 3 in the denominator is throwing me off. Also, I know is easier to find the angles of sinx and cosx than cotx, but is it necessary for the equation to be isolated into one trig term, such as cotx = (stuff), or can this be arranged to where sinx and/or cosx = (stuff)?

    Ultimately, I am stuck at f"(x) = 2cosx/sin(x)^3 = 0 with little to no idea of what to do.
     
  5. Nov 17, 2011 #4
    [tex]\frac{2\cos x}{\sin^3 x}[/tex]

    You are trying to differentiate this right?

    There are a couple ways you can do this. Start off with the constant rule.

    [tex] 2 (\frac{\cos x}{\sin x^3}[/tex]

    Hint: [itex]\frac{a}{b^2} = \frac{a}{b}\frac{1}{b} or \frac{1}{x} = x^{-1}[/itex]

    There is also a simple differentiation rule that is very helpful in calculus I. Rings a bell?
     
  6. Nov 17, 2011 #5
    No sorry, that is already the second derivative of the original function I posted.

    I am trying to solve f"(x) = 0 for f"(x) = 2cosx/sin(x)^3 = 0
     
  7. Nov 18, 2011 #6

    eumyang

    User Avatar
    Homework Helper

    Take the denominator and set it equal to 0. The values of x you find will be values that CANNOT be in the solution set.

    Then take the numerator and set it equal to 0. Remove from this solution set any values of x that also make the denominator equal to 0, if there are any.
     
  8. Nov 18, 2011 #7
    So 2cosx = 0, so x should be 2(∏/2) and 2(3∏/2).

    so x = ∏ and 3∏

    Since the interval given in the problem is [0,∏],

    the only allowable x value is x = ∏

    -------------------------------------

    Now setting (sinx)^3 = 0

    I would guess sinx = 0, x = 1 and -1 and since its to the 3rd power,

    the x values are x = (1^3 and -1^3) or maybe ([itex]\sqrt[3]{1}[/itex] and[itex]\sqrt[3]{-1}[/itex])

    Are either of these correct?

    So since none of the denominator roots = the numerator roots, then it stands that the only x value to make 2cosx/(sinx)^3 = 0 within interval [0,∏] is:

    x = ∏

    And if this is the correct answer, is the way I arrived at that answer correct?

    And if its not correct, how do I solve and think about this problem?
     
  9. Nov 18, 2011 #8
    Wait a minute...

    2 should not be included in setting cosx = 0

    because 2 can be made a seperate term and set equal to zero, such as 2≠0

    So I just have to set cosx = 0, which is x = ∏/2 and 3∏/2

    And since they still don't match the denominator's roots,

    x = ∏/2 is my revised guess.

    If this is correct, is my method/reasoning correct?

    If not, why and how do I get the correct answer?
     
  10. Nov 18, 2011 #9

    eumyang

    User Avatar
    Homework Helper

    No, that's not right. sin 1 ≠ 0. Also, does (-1)3 = 1?
     
  11. Nov 18, 2011 #10
    of course, so if sinx = 0 , x = 0 and x = ∏

    would that mean that sinx^3 = 0, x = [itex]\sqrt[3]{0}[/itex] and [itex]\sqrt[3]{∏}[/itex]

    which means those two values are undefined

    and that the values of the numerator can't equal 0 or ∏,

    And since the values of the numerator equal 2/∏ and 3∏/2,

    The only x value that exists within the given interval [0,∏] is x = ∏/2

    Is this correct?
     
  12. Nov 19, 2011 #11

    eumyang

    User Avatar
    Homework Helper

    No, that's not right either. You're getting confused with regards to the cube root. From
    [itex]\sin^3 x = 0[/itex]
    take the cube root of both sides first.
    [itex]\sqrt[3]{\sin^3 x} = \sqrt[3]{0}[/itex]
    [itex]\sin x = 0[/itex]
    Then find x, which you did (0 and π).
    The rest looks okay to me.
     
  13. Nov 19, 2011 #12
    Yea, I was confused between getting the values x = (0, ∏) and x = (^3√0, ^3√ ∏)

    This was my guess that lead me to the incorrect answer

    [itex]sin^{3}x = 0[/itex]

    [itex]arcsin^{3}0 = ∏[/itex]

    [itex]^{3}√[/itex][itex]arcsin^{3}0[/itex] = [itex]^{3}√∏[/itex]

    arcsin0 = [itex]^{3}√∏[/itex]

    sin([itex]^{3}√∏) = 0[/itex]

    .....and the same process for the other x value of 0.

    But now I realize that [itex]sin^{3}√∏[/itex] can't equal 0

    Because it is sin∏ that equals 0!!

    So now I believe I understand why its correct to take the cube root of both sides first as you said.
     
    Last edited: Nov 19, 2011
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook