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Math Help - Help with local max/min, concavity, increasing decreasing?

  1. #1
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    Help with local max/min, concavity, increasing decreasing?

    13) f(x)=sin x + cos x   0\leq x \leq 2\pi
    f'(x)=cosx-sinx = 0
    x = \frac{\pi}{4} and \frac{5\pi}{4}
    how do I find where the function increase and decrease?
    If I put these into a sign chart, I end up with all 0's
    also how do I find the concavity?
    I get that  f''(x) = -sin x - cos x
    so x = \frac{7\pi}{4} and \frac{3\pi}{4}
    and for this also, if I plug it into f''(x) for a sign chart, the results will still be 0
    is there another method for trig to do these problems?
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  2. #2
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    Quote Originally Posted by dorkymichelle View Post
    13) f(x)=sin x + cos x   0\leq x \leq 2\pi
    f'(x)=cosx-sinx = 0
    x = \frac{\pi}{4} and \frac{5\pi}{4}
    how do I find where the function increase and decrease?
    x = \frac{\pi}{4} and \frac{5\pi}{4} are split points that split up the domain into three intervals

    0\leq x\leq  \frac{\pi}{4}

    \frac{\pi}{4}\leq x\leq  \frac{5\pi}{4} and

    \frac{5\pi}{4}\leq x\leq 2\pi

    Quote Originally Posted by dorkymichelle View Post
    If I put these into a sign chart, I end up with all 0's
    To determine the sign of f'(x) in each of these intervals use a value of x in the interior of the interval

    To determine the sign of f'(x) in the interval, 0\leq x\leq  \frac{\pi}{4}, for example, you could use x=\frac{\pi}{6}

    f'(\frac{\pi}{6})=\cos{\frac{\pi}{6}}-\sin{\frac{\pi}{6}}=\frac{\sqrt{3}}{2}-\frac{1}{2}>0

    If f'(x)>0, then f(x) is increasing in the interval

    If f'(x)<0, then f(x) is decreasing in the interval

    Determine the sign of f'(x) in the other two intervals

    Quote Originally Posted by dorkymichelle View Post
    also how do I find the concavity?
    I get that  f''(x) = -sin x - cos x
    so x = \frac{7\pi}{4} and \frac{3\pi}{4}
    x = \frac{7\pi}{4} and \frac{3\pi}{4} are split points for the second derivative

    They split up the domain into the intervals

    0\leq x\leq  \frac{3\pi}{4}

     \frac{3\pi}{4}\leq x\leq  \frac{7\pi}{4} and

     \frac{7\pi}{4}\leq x\leq 2\pi

    Determine the sign of f"(x) in each interval

    If f"(x)>0, then f(x) is concave up

    If f"(x)<0, then f(x) is concave down
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  3. #3
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    Oh I see, I guess I just had a "trig scares me" moment and forgot to find a value between those intervals and used the crit. points instead.
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