1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
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: Vertical Tangents vs Vertical Cusps

  1. Nov 6, 2013 #1


    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data

    I have an equation f'(x) = (2x)/(3(x^2-4)^(2/3))

    2. Relevant equations

    Vertical cusps are where the one sided limits of the derivative at a point are infinities of opposite signs.

    Vertical tangent lines are where the one sided limits of the derivative at a point are infinities of the same sign.

    3. The attempt at a solution

    I know that the derivative blows up (goes to infinity) at both x = 2 and x = -2 since that will zero the denominator (and nothing cancels with the denominator in the numerator.)

    However, it is not possible for the limit of the derivative at x = 2 or -2 to change signs because the denominator will always be positive; in the denominator there is a constant multiplied by a squared term. Whether x is approaching 2 from the right or left hand side and whether the term becomes a very small positive number close to 0 or a very small negative number close to zero is irrelevant as the term will be squared, making it positive. Basically, the numerator will control the sign of the limit.

    Therefore, it is impossible for the graph of f(x) to have vertical cusps at x = 2 or x = -2. It's impossible for the one sided limits at x = 2 or x = -2 to change signs.
    Is my reasoning correct?
  2. jcsd
  3. Nov 6, 2013 #2


    Staff: Mentor

    They don't have to be the same sign. For example, y = 1/x has a vertical tangent at x = 0, and has one-sided limits of the derivative as you say above. However, y = 1/x2 has vertical tangents at x = 0 with opposite signs.

    A function can be continuous everywhere and have a cusp where the derivative is undefined. Some examples are y = |x| and y = ##\sqrt[3]{|x|}##.
  4. Nov 7, 2013 #3


    User Avatar
    Gold Member

    I think I grasp the distinction now. Vertical cusps exist where the function is defined at some point c, and the function is going to opposite infinities.

    Vertical tangents are the same as cusps except the function is not defined at the point of the vertical tangent. Also for a vertical tangent the sign can change, or it may not.

    So cusp - c for continuity at point c?


    ETA: I'm also going to conclude that this study guide I paid for is incorrect:

  5. Nov 7, 2013 #4


    Staff: Mentor

    I don't think it's useful to think of vertical tangents and cusps as nearly the same.
    This is unclear.

    From the image of your study guide:
    Vert. tangents -
    "It has a cusp if the function changes signs across the first derivative point."
    "It is just a vertical tangent if it does not change signs across the first <image is cut off> at that point."

    Both bullet points are wrong. In the first, the function cannot possibly change signs from one side of the cusp to the other. The derivative does change signs, though.

    In the second, the functions f(x) = 1/x and g(x) = 1/x2 have vertical tangents at x = 0. For the first, f'(x) < 0 everywhere except at x = 0. For the second, g'(x) > 0 for x < 0 and g'(x) < 0 for x > 0.

    The second bullet point is an excellent example of how NOT to write clearly. For one thing, the first "it" doesn't refer to the same thing as the second "it."

    A better way to categorize the places where a function's derivative is undefined, IMO, is to make a distinction between cusps on the graph and vertical asymptotes. At a cusp, the function is defined, but its derivative is undefined. Necessarily the derivative will have to change sign from one side to the other. A function f has a vertical asymptote at x = c if f is not defined at c but is defined for points near c. Rational functions exhibit this behavior due to the denominator becoming zero.

    I used WA to graph something close to what you asked about in post #1 - http://www.wolframalpha.com/input/?i=y=x/(x^2+-+4)^(2/3)&a=^_Real. For simplicity I graphed y = x/[x2 - 4]^(2/3). The basic features such as intercepts and vertical and horizontal asymptotes are the same as what you posted.

    From this graph you can see that the lines x = -2 and x = 2 are vertical asymptotes. Around x = -2, the function heads off to -∞. Around x = 2, the function heads off to +∞.

    I hope you didn't pay a lot for this study guide...
  6. Nov 7, 2013 #5


    User Avatar
    Gold Member

    Alright, I think we can agree on the point that at a cusp, the function is defined while the derivative is not. And at a vertical tangent, the function is undefined, and the derivative is also undefined.

    I would probably rewrite the guide as something like this then:

    1) There exists a cusp at point c if the function is

    a) Defined at point c


    b) The derivative of the function goes to infinities of opposite signs around point c.

    2) There exists a vertical asymptote if

    a) The function is undefined at point c.


    b) The derivative of the function around point c goes to infinity (the sign can either change or stay the same; it doesn't matter).

    And for an example of a vertical tangent I would probably mention the graph of 1/x, which is undefined at x = 0, and whose limits go to infinities of opposite signs as one approaches 0 from the left and right-hand sides.

    For a cusp I might mention x^(2/3). This function is defined at x = 0; one can take the third root of any real number (positive or negative or 0). However, its derivative goes to negative infinity as one approaches x = 0 from the left, and positive infinity as one approaches x = 0 from the right.

    3) There exists a vertical tangent if f'(x) = infinity (of either sign), and only if x exists on f(x)'s domain.

    And I think now that I got it. The original function I posted about cannot have a cusp at either x = -2 or x = 2 since the function simply isn't defined at either point. A cusp necessitates that a function be defined at the point.
    Last edited: Nov 7, 2013
  7. Nov 7, 2013 #6


    User Avatar
    Gold Member

    I actually didn't; I photocopied a friend's.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted