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Homework Help: Investigating max and min value of a function

  1. Mar 19, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex] f(x) = (1/2)sin2x + cosx [/tex]


    [tex] f^2 min +f^2 max = ? [/tex]

    2. Relevant equations

    Differentiation not allowed... only by transformations and analysis.

    3. The attempt at a solution

    I am confused by what it means by f^2 min +f^2 max... does it imply we have to find max and min values seperately, square them and add them? Or does this formulation imply we can directly get the required value?

    Do we have to square the function before investigating it?
  2. jcsd
  3. Mar 20, 2008 #2


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    I think what they're asking you to do is find the sum of the squares of the maximum and minimum values of this function.

    I'd suggest first writing out what sin(2x) is: you'll find that f(x) can be expressed as a product of two factors. It should be reasonably straightforward to see what the greatest and least values of that product are. Can f(x) be zero?
  4. Mar 20, 2008 #3
    you need to take the derivative of that..

    [tex] f(x) = (1/2)sin2x + cosx [/tex]

    after that you solve f'(x)=0
    find the extreme points and put the y values in the asked equation

    [tex] f^2 min +f^2 max = ? [/tex]
  5. Mar 20, 2008 #4


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    The problem statement says no differentiation allowed...

    (If it were, the derivative would be f'(x)=1*cos2x - sinx .)
    Last edited: Mar 20, 2008
  6. Mar 21, 2008 #5
    i'm getting

    [tex]f(x) = cosx(1+sinx)[/tex]

    ok i got min value = 0 (the problem says 0 <= x <= pi/2)
    how do we get the max value of the product?
  7. Mar 21, 2008 #6
    you have a point in x=pi/2 +pi*k
    and for x=3/4*pi +2pi*k

    substitute them in the fuction and find their y values
    Last edited: Mar 21, 2008
  8. Mar 21, 2008 #7
    how does that work?
  9. Mar 21, 2008 #8


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    If the interval is [0, pi/2], then you don't have to worry about the other periodic values of sine and cosine. (You didn't mention the interval earlier...)

    The minimum is zero at pi/2 because of the cosine term. For the maximum, you could either look at the terms in f(x) or square your result for f(x) first. In any case, using x = 0 would give you
    f(0) = 1, but there's a place where we can do better. The problem with the endpoints is that sine is high when cosine is low and vice versa. What value of x gives both fairly high values for sine and cosine? (Consider graphs of those functions.)
  10. Mar 22, 2008 #9
    pi/4 gives equal values for sine and cosine... however, how do we know there isn't a value thats higher? i think there may be a more rigorous proof...

    e.g. on another similar problem i could obtain a quadratic equation containing f(x) in a constant term (by squaring function) and getting something like:

    [tex] ax^2 +bx + c = f^2(x) [/tex]

    [tex] ax^2 +bx + (c - f^2(x)) = 0 [/tex]

    [tex] b^2 - 4a(c - f^2(x)) >or= 0 [/tex] (for real f(x))

    thus obtaining min and max values simultaneously...

    on this example, on squaring i get a quartic equation on sin(x)... i'm unable to bring it to a simpler (quadratic) form or otherwise...
    Last edited: Mar 22, 2008
  11. Mar 22, 2008 #10


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    Maybe we shouldn't look at the quartic polynomial, but rather the factored form. The function squared is

    [tex] (cos x)^{2} (1+sin x)^{2} = (1 - [sin x]^{2})(1+sin x)^{2} = (1 - sin x)(1+sin x)^{3}

    So we make the substitution t = sin x and ask for the maximum on the interval [0,1] of

    [tex] (1 - t)(1+ t)^{3}

    [I'm still thinking about how to solve this without calculus. (The maximum turns out to occur at sin x = 1/2 , BTW, not {sqrt(2)}/2 , as I'd earlier thought.) The instruction "use transformations and analysis" isn't very descriptive, so I'm still trying out ideas...]
    Last edited: Mar 22, 2008
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