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Thread: local max and min

  1. #1
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    local max and min

    hi

    have function $\displaystyle g(x)=cosxexp(\frac{2}{3}sinx)$

    has derivative $\displaystyle g'(x)=(2-3sinx-2sin^2x)exp(\frac{2}{3}sinx)$

    i factor out 1/3 and ignored $\displaystyle exp(\frac{2}{3}sinx)$ as it always positive and got $\displaystyle (1-2sinx)(2+sinx)$ then set it to zero and got x= $\displaystyle \frac{\pi}{6}$ and $\displaystyle \frac{5\pi}{6}$ used these to get coordinates, which i am happy with but when i try to find the max and min locals of g(x) i get the oposite result to the graph of g(x). no matter how i try to look at it i get $\displaystyle \frac{\pi}{6}$,$\displaystyle \frac{\sqrt{3}}{2}$ as minimum point and $\displaystyle \frac{5\pi}{6}$,$\displaystyle \frac{-\sqrt{3}}{2}$ as a maximum point. have been using first derivitive test as i am unable to work out how to get g''(x)

    any help would be great
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  2. #2
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    I dont know how you conducted the first devaritive test, but observe that g'(x)>0 for 0<x<(pi/4)
    and g'(x)<0 for pi/4<x<5pi/4. Cos(x) decides the sign because exp(2/3sin(x)) is always > 0.

    That means that on the interval 0<x<(pi/4) there is a local maxima, and on the interval pi/4<x<5pi/4, there is a local minima.

    Looks like you just turned that one around.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by smartcar29 View Post
    hi

    have function $\displaystyle g(x)=cosxexp(\frac{2}{3}sinx)$

    has derivative $\displaystyle g'(x)=(2-3sinx-2sin^2x)exp(\frac{2}{3}sinx)$

    i factor out 1/3 and ignored $\displaystyle exp(\frac{2}{3}sinx)$ as it always positive and got $\displaystyle (1-2sinx)(2+sinx)$ then set it to zero and got x= $\displaystyle \frac{\pi}{6}$ and $\displaystyle \frac{5\pi}{6}$ used these to get coordinates, which i am happy with but when i try to find the max and min locals of g(x) i get the oposite result to the graph of g(x). no matter how i try to look at it i get $\displaystyle \frac{\pi}{6}$,$\displaystyle \frac{\sqrt{3}}{2}$ as minimum point and $\displaystyle \frac{5\pi}{6}$,$\displaystyle \frac{-\sqrt{3}}{2}$ as a maximum point. have been using first derivitive test as i am unable to work out how to get g''(x)

    any help would be great
    1. The roots of the equation

    $\displaystyle 2-3 \sin(x)-2\sin^2(x)=0$

    corresponding to your first factor are $\displaystyle x=\pi/6 + 2 \pi n$ and $\displaystyle x=2 \pi/3+2 \pi m$, for all $\displaystyle n, \ m \in \mathbb{Z}$, and similarly there are multiple roots corresponding to the other factor.

    2. How are you determining if a root corresponds to a maximum or minimum?

    3. How are you evaluating $\displaystyle g(x) $?

    CB
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