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Thread: Find the equation of the tangent line(s)

  1. #1
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    Find the equation of the tangent line(s)

    Find the equation of the tangent line(s),which is perpendicular to the line x+6y+5=0, to the graph of F(x) at the tangent point(s) (x,y)
    Hint:Find the slope first then use y-y=m(x-x)

    The equation is $\displaystyle f(x)= 2x^3-3 $

    i dont really get this equation anyone know were to start
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  2. #2
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    Find the slope of x+6y+5=0
    Find the derivative of f(x)

    what is the relationship of the slope between 2 lines that are perpendicular?

    Set that slope equal to f'(x)

    Hope this helps. Ask if you are still confused.
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  3. #3
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    Find the slope of x+6y+5=0

    $\displaystyle m=-1/6 $

    Find the derivative of f(x)

    $\displaystyle 6x^2 $

    what is the relationship of the slope between 2 lines that are perpendicular?

    I believe the slopes the same

    Set that slope equal to f'(x)

    $\displaystyle -1/6=6x^2 $

    Still confused
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  4. #4
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    Quote Originally Posted by john doe View Post
    Find the slope of x+6y+5=0

    $\displaystyle m=-1/6 $

    Find the derivative of f(x)

    $\displaystyle 6x^2 $

    what is the relationship of the slope between 2 lines that are perpendicular?

    I believe the slopes the same

    Set that slope equal to f'(x)

    $\displaystyle -1/6=6x^2 $
    Slopes are the same when they are parallel.
    When 2 lines are perpendicular, their slopes are opposite reciprocal of each other.

    So you would set $\displaystyle 6x^2=6 $

    Find x and put it back into the original equation to find y and that will be your point.

    Now just use the point-slope equation since you know the point and the slope.
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  5. #5
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    hows this look

    $\displaystyle y+1=6(x-1) $
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  6. #6
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    Quote Originally Posted by john doe View Post
    hows this look

    $\displaystyle y+1=6(x-1) $
    There should be 2 lines.
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  7. #7
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    i dont understand
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  8. #8
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    Quote Originally Posted by john doe View Post
    i dont understand
    haha sorry it's late here, I forgot to add in the explanation.

    $\displaystyle 6x^2=6 $

    x=1, -1
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  9. #9
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    i must be losing it because i dont know what that means



    my graph kinda turned out with a parabola and a tangent line
    going though at 1,-1





    its kinda 3 aclock
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  10. #10
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    Let's back up a bit.

    Here is your question: Find the equation of the tangent line(s),which is perpendicular to the line x+6y+5=0, to the graph of F(x) at the tangent point(s) (x,y)


    so we need to find at where does F(x) have slopes of 6.

    So we did $\displaystyle F'(x)=6x^2=6$
    the solution would be x=1 and -1 because x is squared

    That means the function F(x) has 2 points where the slope is 6 that is at (1, F(1)) and (-1, (F(-1)).

    This means there are 2 lines which satisfies the question.

    Hope this clears things up!
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  11. #11
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    I think i see it at
    so that would look like a x^3 graph shifted to the bottom
    theres a slope 6 that hits the graph 2 times 1 and -1
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  12. #12
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    Quote Originally Posted by john doe View Post
    I think i see it at
    so that would look like a x^3 graph shifted to the bottom
    theres a slope 6 that hits the graph 2 times 1 and -1
    Hmm, I'm not quite sure what you mean.

    The slope of x+6y+5=0 would be -1/6

    Perpendicular slope=6

    $\displaystyle f'(x)=6x^2=6 $
    $\displaystyle x= 1,-1 $

    $\displaystyle f(1)=-1 $
    $\displaystyle f(-1)=-5 $

    so the 2 lines would be

    $\displaystyle y+5=6(x+1)$
    $\displaystyle y+1=6(x-1)$

    All I was trying to tell you after your first solution is that there is more than one answer.
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  13. #13
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    alright thax for the help

    i dont fully understand it but

    i appreciate all the time your talking

    this late at night to explain to me

    (ill take a look at it tomorrow)
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  14. #14
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    In general, let $\displaystyle f:[t_{0},\infty)\to\mathbb{R}$ be a differentiable function.
    Then at the point $\displaystyle s\in(t_{0},\infty)$ tangent line of $\displaystyle f$ is $\displaystyle g(t):=f^{\prime}(s)(t-s)+f(s)$ for all $\displaystyle t\in[t_{0},\infty)$.
    And let $\displaystyle h(t):=at+b$ for some $\displaystyle a,b\in\mathbb{R}$ with $\displaystyle a\neq0$ be a line, which is supposed to be perpendicular to the line $\displaystyle g$.
    Then, we have to determinate $\displaystyle s$ in such a way that $\displaystyle g$ is perpendicular to $\displaystyle h$; that is, $\displaystyle a*f^{\prime}(s)=-1$.
    If $\displaystyle f^{\prime}$ as inverse function $\displaystyle (f^{\prime})^{-1}$ (as in this case), then
    $\displaystyle s=(f^{\prime})^{-1}(-1/a)$
    and so
    $\displaystyle g(t)=-\frac{1}{a}\big(t-(f^{\prime})^{-1}(-1/a)\big)+f\big((f^{\prime})^{-1}(-1/a)\big)$.
    The rest is just simple calculation.
    I hope you will find this clearifying.
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