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Math Help - relative maximum,Newton’s Method,Integrate.......

  1. #1
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    relative maximum,Newton’s Method,Integrate.......

    please try to solve these questions.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by m777 View Post
    please try to solve these questions.
    Question # 1
    The derivative of the continuous function is given. Find all critical points and determines whether a relative maximum, relative minimum or neither occurs there

    f'(x)=2\,\sin^3(x) - \sin^2(x),\ \ \ 0<x<2\,\pi

    The critical points are solutions of:

    f'(x)=2\,\sin^3(x) - \sin^2(x)=0,\ \ \ 0<x<2\,\pi

    which occur either when sin(x)=0 or 2\,\sin(x)-1=0, the roots of these are:

    \pi for the first and \pi/6 and 5 \pi/6 for the second.

    Now these correspond to local maxima when f''(x)<0, minima when f''(x)>0 and (stationary) points of inflection when f''(x)=0.

    f''(x)=2\,\sin(x)\cos(x)[3\, \sin(x)-1]

    So the first of the roots gives f''(x)=0, the second f''(x)>0 and the third f''(x)<0

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by m777 View Post
    please try to solve these questions.
    Question # 2
    Solve it with the help of Newton’s Method up to four terms

    sin(x)=x^2

    We want to find the roots of: f(x)=\sin(x)-x^2 using Newton's method. This is the ittereation:

    x_{n+1}=x_n-f(x_n)/f'(x_n)

    and here: f'(x)=\cos(x)-2x.

    This can be set up in Excel (see attachment) from which we see that there is a root at x~=0.8767
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  4. #4
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    Hello, m777!

    1) The derivative of a continuous function is given.
    Find all critical points and determine if a rel.max., rel.min. or neither occurs there.

    f'(x)\:=\:2\sin^3x - \sin^2x,\quad 0 < x <2\pi

    Solve f'(x) = 0
    . . 2\sin^3x - \sin^2x \:= \:0\quad\Rightarrow\quad \sin^2x\left(2\sin x - 1\right)\:=\:0

    We have two equations to solve:
    . . \sin^2x \,= \,0\quad \Rightarrow\quad \sin x \,= \,0\quad \Rightarrow \quad \boxed{ x \:=\:\pi}
    . . 2\sin x - 1 \:=\:0 \quad\Rightarrow\quad \sin x \:=\:\frac{1}{2} \quad\Rightarrow\quad \boxed{ x \:=\:\frac{\pi}{6},\,\frac{5\pi}{6}}


    Second derivative: . f''(x) \:=\:6\sin^2x\cos x - 2\sin x\cos x\:=\:2\sin x\cos x\left(3\sin x - 1\right)

    f''(\pi) \:=\:2\!\cdot\!\sin\pi\!\cdot\!\cos\pi(3\sin\pi - 1) \:=\:2\cdot0\cdot(\text{-}1)\,[3\cdot0 - 1] \:=\:0\quad\hdots\quad\boxed{\text{ neither at }x = \pi}

    f''\left(\frac{\pi}{6}\right) \:=\:2\!\cdot\!\sin\frac{\pi}{6}\!\cdot\cos\frac{\  pi}{6}\left(3\sin\frac{\pi}{6} - 1\right) \:=\:2\!\left(\frac{1}{2}\right)\!\left(\frac{\sqr  t{3}}{2}\right)\!\left(3\!\cdot\!\frac{1}{2} - 1\right)\:>\:0
    . . f(x) is concave up: \cup\quad\hdots\quad\boxed{\text{relative minimum at }x = \frac{\pi}{6}}

    f''(x)\left(\frac{5\pi}{6}\right) \:=\:2\cdot\sin\!\frac{5\pi}{6}\!\cdot\cos\!\frac{  5\pi}{6}\left(3\sin\!\frac{5\pi}{6} - 1\right)  \:=\:2\!\left(\frac{1}{2}\right)\!\left(\text{-}\frac{\sqrt{3}}{2}\right)\!\left(3\!\cdot\!\frac{  1}{2} - 1\right) \:<\:0
    . . f(x) is concave down: \cap\quad\hdots\quad\boxed{\text{relative maximim at }x = \frac{5\pi}{6}}

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  5. #5
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    #3 Integrate it with respect to “y”

    (i) \int (2+y^2)^2dy

    Integrate it with respect to “x”
    (ii) \int \frac{1}{x^6} dx
    Hello,

    to (i). Expand the bracket first and then integrate:

    \int (2+y^2)^2dy=\int(4+4y^2+y^4)dy=4y+\frac{4}{3}y^3+\  frac{1}{5}y^5+C

    to (ii):

    \int \frac{1}{x^6} dx=\int x^{-6}dx=-\frac{1}{5}x^{-5}+C

    EB

    Merry Christmas and a happy New Year!
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  6. #6
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    once again thanks.

    Hello,
    Merry christmas and happy new year to all of you.specially to soroban, earboth and captain black.
    from
    m777
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