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Answer the following questions for the function f(x)=sin^2(x/5)

defined on the interval .(-15.507923,3.82699075)

Rememer that you can enter "pi" for as part of your answer.

a.what is f(x) concave down on the region

B. A global minimum for this function occurs at
C. A local maximum for this function which is not a global maximum occurs at
D. The function is increasing on what to what and what to what

please help me

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li

  1. First, let's get the derivative of f(x), using the chain rule and the identity for sin(2x)

    f' = 2 sin x/5 * cos x/5 * 1/5
    = 1/5 sin(2x/5)

    OK, a nice simple sine function.

    To find where f is concave downward, we need to find where f'' < 0

    f'' = 1/5 cos(2x/5) * 2/5
    = 2/25 cos(2x/5)

    Now, cos(u) < 0 when pi/2 < u < 3pi/2
    That is, when
    pi/2 < 2x/5 < 3pi/2
    or
    5pi/4 < x < 15pi/4
    or
    3.927 < x < 11.781
    Hmmm. That's not in our interval. So, subtracting multiples of 2pi*5/2 = 5pi = 15.708 we get
    -11.781 < x < -3.297

    All the local max/min are also global max/min, since sin^2 oscillates between 0 and 1.

    minima are where sin(x/5) = 0, or where x/5 = 0,pi,-pi,...
    x = 0, 5pi, -5pi

    maxima are where x/5 = pi/2, -pi/2, ...
    x = 5pi/2, -5pi/2, ...

    f is increasing where f' > 0. That is, where sin(2x/5) > 0
    sin(u) > 0 when 0 < u < pi
    0 < 2x/5 < pi
    0 < x < 5pi/2
    -5pi < x < -5pi/2

    Plot the graph of f(x) and you can double-check my values...

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    Steve

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