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Trig question

  1. Jul 21, 2004 #1
    For the function f(x) = 3sin bx + d where b and d are positive constants, determine an expression for the smallest positive value of x that produces the maximum value of f(x).
    :confused:
     
  2. jcsd
  3. Jul 22, 2004 #2
    oops i guess i should post what i have:

    well the smaller the period of the sin graph, the smaller the value of x

    as for the largest maximum value would be the amplitude + d where as d increases, the larger the f(x)
    3+d at 2pi/b

    So how exactly do you put that as an expression? My teacher is quite picky about these little things...
     
  4. Jul 22, 2004 #3

    AKG

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    You can see that this function has it's greatest value when sin(bx) has it's greatest value. sin(bx) has a maximum value of 1. You know that the smallest argument for the sine function that gives a value of 1 is [itex]\pi /2[/itex]. Therefore:

    [tex]bx = \pi /2[/tex]

    [tex]x = \frac{\pi}{2b}[/tex]

    I think you said something like [itex]2\pi /b[/itex] which is wrong. Anyways, the expression you're looking for is:

    [tex]\frac{\pi}{2b}[/tex]
     
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