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Math Help - [SOLVED] Calculus 1

  1. #1
    addanis
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    [SOLVED] Calculus 1

    Can someone please help me solve this problem?

    Find the absolute maximum and absolute minimum of g(x)= sin(x) - cos(x) on [0,π].

    It is on a closed interval on zero and pie.

    Thanks.
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  2. #2
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    Quote Originally Posted by addanis View Post
    Can someone please help me solve this problem?

    Find the absolute maximum and absolute minimum of g(x)= sin(x) - cos(x) on [0,π].

    It is on a closed interval on zero and pie.

    Thanks.
    f(x)=\sin x-\cos x.
    This is continous on [0,\pi].
    By Fermat's principle a relative extrema occurs when the derivative does not exist or is zero. Since the derivative is zero the only points on the interval (0,\pi) is when derivative is zero. And we need to check the boundardy points because they are not gaurennted by the theorem.
    g'(x)=\cos x+\sin x
    Thus,
    \cos x+\sin x=0
    \cos x=-\sin x, \cos x\not = 0
    1=-\tan x
    \tan x=-1
    x=\pi -\frac{\pi}{4}=\frac{3\pi}{4}
    Und,
    g(3\pi/4)=\sqrt{2}/2+\sqrt{2}/2=\sqrt{2}
    g(0)=0-1=-1
    g(\pi)=0+1=1
    Thus,
    x=3\pi/4---> Max
    x=0---> Min
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  3. #3
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    Hello, addanis!

    Find the absolute maximum and minimum of g(x) \:= \:\sin(x) - \cos(x) on [0,\,\pi]

    Set the derivative equal to 0 and solve.

    . . g'(x) \:=\:\cos x + \sin x \:=\:0\quad\Rightarrow\quad \sin x \:=\:-\cos x\quad\Rightarrow\quad \frac{\sin x}{\cos x}\:=\:-1

    . . \tan x \:=\:-1\quad\Rightarrow\quad x \:=\:\frac{3\pi}{4},\:\frac{7\pi}{4}


    Second Derivative Test: . g''(x) \:=\:-\sin(x) + \cos(x)

    g''\left(\frac{3\pi}{4}\right) \:=\:-\sin\left(\frac{3\pi}{4}\right) + \cos\left(\frac{3\pi}{4}\right) \:=\:-\sqrt{2} . . . negative: concave down \cap
    . . . . Hence: relative maximum at \left(\frac{3\pi}{4},\:\sqrt{2}\right)

    g''\left(\frac{7\pi}{4}\right)\:=\:-\sin\left(\frac{7\pi}{4}\right) + \cos\left(\frac{7\pi}{4}\right) \:=\:+\sqrt{2} . . . positive: concave up \cup
    . . . . Hence, relative minimum at \left(\frac{7\pi}{4},\:\text{-}\sqrt{2}\right)


    Test endpoints:
    . . g(0)\:=\:\sin(0) - \cos(0) \:=\:-1
    . . g(\pi)\:=\:\sin(\pi) - \cos(\pi) \:=\:+1


    Therefore . . . . \begin{array}{cc}\text{Abs. maximum: }& \left(\frac{3\pi}{4},\:\sqrt{2}\right) \\ \text{Abs. minimum: }& \left(\frac{7\pi}{4},\:\text{-}\sqrt{2}\right) \end{array}

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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, addanis!


    Set the derivative equal to 0 and solve.

    . . g'(x) \:=\:\cos x + \sin x \:=\:0\quad\Rightarrow\quad \sin x \:=\:-\cos x\quad\Rightarrow\quad \frac{\sin x}{\cos x}\:=\:-1

    . . \tan x \:=\:-1\quad\Rightarrow\quad x \:=\:\frac{3\pi}{4},\:\frac{7\pi}{4}


    Second Derivative Test: . g''(x) \:=\:-\sin(x) + \cos(x)

    g''\left(\frac{3\pi}{4}\right) \:=\:-\sin\left(\frac{3\pi}{4}\right) + \cos\left(\frac{3\pi}{4}\right) \:=\:-\sqrt{2} . . . negative: concave down \cap
    . . . . Hence: relative maximum at \left(\frac{3\pi}{4},\:\sqrt{2}\right)

    g''\left(\frac{7\pi}{4}\right)\:=\:-\sin\left(\frac{7\pi}{4}\right) + \cos\left(\frac{7\pi}{4}\right) \:=\:+\sqrt{2} . . . positive: concave up \cup
    . . . . Hence, relative minimum at \left(\frac{7\pi}{4},\:\text{-}\sqrt{2}\right)


    Test endpoints:
    . . g(0)\:=\:\sin(0) - \cos(0) \:=\:-1
    . . g(\pi)\:=\:\sin(\pi) - \cos(\pi) \:=\:+1


    Therefore . . . . \begin{array}{cc}\text{Abs. maximum: }& \left(\frac{3\pi}{4},\:\sqrt{2}\right) \\ \text{Abs. minimum: }& \left(\frac{7\pi}{4},\:\text{-}\sqrt{2}\right) \end{array}

    But the interval is [0,\pi]
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