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Math Help - slope of tangent

  1. #1
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    slope of tangent

    y=x-x^5

    find the slope of the tangent line to the curve at point (1,0)

    slope =

    <br />
\frac{f(x)-f(a)}{x-a}<br />

    I'm a little confused on what 'a' equals, but going off my notes im gonna say its 1, being x value in the point

    <br />
\frac{(x-x^5)-(1-1^5)}{x-1}<br />

    =

    <br />
\frac{x-x^5}{x-1}<br />

    and thats where I get stuck
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by silencecloak View Post
    y=x-x^5

    find the slope of the tangent line to the curve at point (1,0)

    slope =

    <br />
\frac{f(x)-f(a)}{x-a}<br />

    I'm a little confused on what 'a' equals, but going off my notes im gonna say its 1, being x value in the point

    <br />
\frac{(x-x^5)-(1-1^5)}{x-1}<br />

    =

    <br />
\frac{x-x^5}{x-1}<br />

    and thats where I get stuck
    factor out the x, you get

    \lim_{x \to 1} \frac {x(1 - x^4)}{x - 1}

    now use the difference of two squares for the numerator
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    factor out the x, you get

    \lim_{x \to 1} \frac {x(1 - x^4)}{x - 1}

    now use the difference of two squares for the numerator
    \lim_{x \to 1} \frac { x(1-x)(1+x)(1+x^2)}{x - 1}

    =

    \lim_{x \to 1} \frac{x-(x-1)(1+x)(1+x^2)}{x-1}

    =

    \lim_{x \to 1} \ x-(1+x)(1+x^2)

    =

    \lim_{x \to 1} 1-(1+1)(1+1)

    =

    \lim_{x \to 1}  1-(1+1)(1+1)

    = -3

    pretty sure i goofed somewhere or im doing it wrong
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by silencecloak View Post
    \lim_{x \to 1} \frac { x(1-x)(1+x)(1+x^2)}{x - 1}

    =

    \lim_{x \to 1} \frac{x{\color{red}-}(x-1)(1+x)(1+x^2)}{x-1}
    well, yeah. obviously if you put stuff there that's not supposed to be there that will happen exactly where did that minus sign come from? do you realize that you are changing the value of the function?
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    well, yeah. obviously if you put stuff there that's not supposed to be there that will happen exactly where did that minus sign come from? do you realize that you are changing the value of the function?
    Ya, I was trying to pull a negative out so I could cancel out the denominator



    EDIT:

    durrr

    \frac{-x+1}{x-1} = -1

    so

    <br />
-x(1+x)(1+x^2) = -4<br /> <br />
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by silencecloak View Post
    Ya, I was trying to pull a negative out so I could cancel out the denominator



    EDIT:

    durrr

    \frac{-x+1}{x-1} = -1

    so

    <br />
-x(1+x)(1+x^2) = -4<br /> <br />
    yes

    of course you have to write the limit until you actually take the limit
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