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Thread: draw a curve

  1. #1
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    draw a curve

    i have got this information on a assigment:

    f (x) = ln 2x , g(x) = cos(x - p / 2) , a (x) = 2x+ | x - 2 | ,

    and the question is to draw the curve of y = f (|1- x |)

    i think it will then be y = f (|1-ln2x|) right ?
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  2. #2
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    Quote Originally Posted by paulaa View Post
    i have got this information on a assigment:

    f (x) = ln 2x , g(x) = cos(x - p / 2) , a (x) = 2x+ | x - 2 | ,

    and the question is to draw the curve of y = f (|1- x |)

    i think it will then be y = f (|1-ln2x|) right ?
    $\displaystyle f(x)=ln2x$

    $\displaystyle f(|1-x|)\Rightarrow$ replace $\displaystyle x$ with $\displaystyle |1-x|$

    $\displaystyle f(|1-x|)=ln[2(|1-x|)]$
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  3. #3
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    then how will my curve look like ?
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  4. #4
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    Quote Originally Posted by paulaa View Post
    then how will my curve look like ?
    If you are unsure of the general shapes of the abs value and log functions, you can make a table of values and solve for any intercepts/asymptotes.
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  5. #5
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    okej.

    f(|1-x|)=ln[2(|1-x|)] is it the same as y=ln[2(|1-x|)]
    ?
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  6. #6
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    Quote Originally Posted by paulaa View Post
    okej.

    f(|1-x|)=ln[2(|1-x|)] is it the same as y=ln[2(|1-x|)]
    ?
    $\displaystyle y=ln(2|1-x|)\Rightarrow\ e^y=2|1-x|$

    $\displaystyle e^{-\infty}\rightarrow\ 0$

    so there is a vertical asymptote at $\displaystyle x=1$ where the graph goes to $\displaystyle -\infty$

    both to the left and right of $\displaystyle x=1.$

    Also the graph is defined for positive and negative x as the modulus is being taken.
    Therefore the graph is symmetrical about the asymptote.
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  7. #7
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    If you're familiar with ln(x) you could also do it this way.
    $\displaystyle f(x)=ln(x)$
    Has a vertical asymptote at x=0 and f(x) = 0 at x = 1.
    $\displaystyle f(x)=ln|x|$
    makes the function even and symmetric about the y-axis.
    $\displaystyle f(x)=ln|-x|$
    Reflect the graph about the y-axis.
    $\displaystyle f(x)=ln|-(x-1)|$
    Shift the graph 1 unit to the right.
    $\displaystyle f(x)=ln|-2(x-1)|=ln|-(x-1)|+ln2$
    Shift the graph up ln2 units.
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