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: Single Variable function, local max/min is absolute max/min

  1. Oct 10, 2009 #1
    1. The problem statement, all variables and given/known data
    Consider a function of one variable, f(x), which is continuous on an interval I. If f has a single critical point, x = a, in I, then if that point is a local maximum (or minimum) it must also be an absolute maximum. Explain why this is.

    2. Relevant equations
    Intermediate Value Theorem
    Rolle's Theorem

    3. The attempt at a solution
    Critical point x=a, in I, is a local maximum or minimum, thus f'(x)=0 at x=a.

    The function f thus must either concave up (for a to be a local min) or concave down (for a to be a local max).

    Since both cases are similar, let consider x=a is a local max.

    For a local max, f'(x) is decreasing to 0 as x approaches a from the left and f'(x) is increasing negative away from 0 as x deviate away from a on the right.

    Since there is only one single critical point at x=a, the concavity of the graph will not change. f(x) on the left of is an increasing function to f(a) and f(x) on the right of f is a decreasing function away from f(a). Thus x=a must also be the absolute maximum.

    Is my reasoning correct? Or did I unconsciously assumed something that I shouldn't have?
  2. jcsd
  3. Oct 10, 2009 #2
    Why don't you use the theorems you mentioned?

    Assume there exists [tex]x_0[/tex] such that [tex]f(x_o)>f(a)[/tex].

    What does the intermediate value theorem tell you?
  4. Oct 10, 2009 #3
    If f(x_o) > f(a), then according to the intermediate value theorem, there should be another value in the interval I where f(x) = f(a).

    Then applying Rolle's theorem, there must be a point between x and a such that the derivative of this point is 0, hence there must be a second critical point, which contradicts with single critical point assumption.

    Thanks for the help, I didn't consider proving it using proof of contradiction. Even if I did, I would not know where to begin :d. Thanks again.
    Last edited: Oct 10, 2009
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook