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Math Help - Inc/dec/maxima/concavity

  1. #1
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    Inc/dec/maxima/concavity

    f(x) = sinx + cosx, 0 <= x <= 2pi

    a) Find the intervals on which f is increasing or decreasing.
    b) Find the local maximum and minimum values of f
    c) Find the intervals of concavity and the inflection points.

    a)For f'(x) I got my critical points to be x = pi / 4 and x = 5pi / 4 but didnt know how to put this on a sign table in order to find when it is inc/dec

    b) f(pi / 4) = (2)^1/2, but I didnt know how to work out
    f(5pi / 4) = sin(5pi / 4) + cos(5pi / 4) without a calculator

    c)For f''(x) my critical points were x = 3 pi / 4 and x = 3 pi / 2 and had the same problem as a)
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  2. #2
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    Quote Originally Posted by TsAmE View Post
    f(x) = sinx + cosx, 0 <= x <= 2pi

    a) Find the intervals on which f is increasing or decreasing.
    b) Find the local maximum and minimum values of f
    c) Find the intervals of concavity and the inflection points.

    a)For f'(x) I got my critical points to be x = pi / 4 and x = 5pi / 4 but didnt know how to put this on a sign table in order to find when it is inc/dec

    b) f(pi / 4) = (2)^1/2, but I didnt know how to work out
    f(5pi / 4) = sin(5pi / 4) + cos(5pi / 4) without a calculator

    c)For f''(x) my critical points were x = 3 pi / 4 and x = 3 pi / 2 and had the same problem as a)
    Make a graph of the function and you can easily see the answer to all these questions.

    Here is a graph of the derivative.


    If this were the function, you can clearly see that it is decreasing from 0->3pi/4, increasing from 3pi/4->7pi/4, decreasing from 7pi/4->2pi

    local max is (\frac{7\pi}{4},\sqrt{2})
    local min is (\frac{3\pi}{4},-\sqrt{2})

    inflections at (\frac{\pi}{4},0) and (\frac{5\pi}{4},0)
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  3. #3
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    How would you do this without sketching the graph e.g using f'(x), f''(x) like I have done
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  4. #4
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    Quote Originally Posted by TsAmE View Post
    How would you do this without sketching the graph e.g using f'(x), f''(x) like I have done
    Alright, consider this image:



    What relation do you see between the critical points and the zeros of f'(x) and those of f(x)? This is your answer.
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  5. #5
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    Quote Originally Posted by TsAmE View Post
    How would you do this without sketching the graph e.g using f'(x), f''(x) like I have done
    f'(x) = \cos{x} - \sin{x}

    f'(x) = 0 at x = \frac{\pi}{4} and x = \frac{5\pi}{4}

    perform the first derivative test to locate extrema ...

    in the interval \left[0, \frac{\pi}{4}\right) , f'(x) > 0 ... f(x) is increasing.

    in the interval \left(\frac{\pi}{4}, \frac{5\pi}{4}\right) , f'(x) < 0 ... f(x) is decreasing.

    in the interval \left(\frac{5\pi}{4}, 2\pi\right] , f'(x) > 0 ... f(x) is increasing.

    when f'(x) changes sign from positive to negative, f(x) has a relative maximum at that critical value.

    when f'(x) changes sign from negative to positive, f(x) has a relative minimum at that critical value.


    set f''(x) = 0 and perform the same analysis. inflection points on the graph of f(x) occur where f''(x) changes sign. f(x) will be concave up for f''(x) > 0 and concave down for f''(x) < 0.

    all this basic info is in your text. look it over.
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  6. #6
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    Oh I see makes a lot of sense, thanks for the explanation. Im just curious, you wouldnt be able to use a sign table if they didnt give you a closed interval right? e.g. for this question (without a closed interval) x < pi / 4 wouldnt be increasing (as it would soon decrease due to trig graphs oscillating infinitely)?
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