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Math Help - equation of tangent line

  1. #1
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    equation of tangent line

    define f(x) = x^3+2x+1 for all x. Find the equation of the tangent line to the graph of f:R \rightarrow R at the point (2,13).

    I know that for a function f:I \rightarrow R (I is a neighborhood of the point x_0) that is differentiable at x_0, the tangent to the graph of the point ({x_0},(f(x_0)) is given by the equation:

    y = f(x_0)+f'(x_0)(x-{x_0})

    To find the slope of the line, wehave

    \lim_{x->{x_0}} \frac{f(x)-f(x_0)}{x-{x_0}}

    = \lim_{x->{x_0}} \frac{(x^3+2x+1)-({x_0}^3+2{x_0}+1)}{x-{x_0}}

    = \lim_{x->{x_0}} (x^3-{x_0}^3)+2 = 2

    is the slope of this function is 2 then? I cannot figure out if this is correct and how to find the equation of the tangent line
    Last edited by serious331; April 28th 2010 at 12:12 AM.
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by serious331 View Post
    define f(x) = x^3+2x+1 for all x. Find the equation of the tangent line to the graph of f:R \rightarrow R at the point (2,13).

    I know that for a function f:I \rightarrow R (I is a neighborhood of the point x_0) that is differentiable at x_0, the tangent to the graph of the point ({x_0},(f(x_0)) is given by the equation:

    y = f(x_0)+f'(x_0)(x-{x_0})

    To find the slope of the line, wehave

    \lim_{x->{x_0}} \frac{f(x)-f(x_0)}{x-{x_0}}

    = \lim_{x->{x_0}} \frac{(x^3+2x+1)-({x_0}^3+2{x_0}+1)}{x-{x_0}}

    = \lim_{x->{x_0}} (x^3-{x_0}^3)+2 = 2

    is the slope of this function is 2 then? I cannot figure out if this is correct and how to find the equation of the tangent line
    The slope of your line is found as:

    \lim_{x->2} \frac{f(x)-f(2)}{x-2} = \lim_{x->2} \frac{(x^3+2x+1)-(8+4+1)}{x-2} = \lim_{x->2} \frac{x^3+2x-12}{x-2} =14

    so f'(x_0)=14

    now find the equation of the tangent using the formula you have!
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  3. #3
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    Quote Originally Posted by serious331 View Post
    define f(x) = x^3+2x+1 for all x. Find the equation of the tangent line to the graph of f:R \rightarrow R at the point (2,13).

    I know that for a function f:I \rightarrow R (I is a neighborhood of the point x_0) that is differentiable at x_0, the tangent to the graph of the point ({x_0},(f(x_0)) is given by the equation:

    y = f(x_0)+f'(x_0)(x-{x_0})

    To find the slope of the line, wehave

    \lim_{x->{x_0}} \frac{f(x)-f(x_0)}{x-{x_0}}

    = \lim_{x->{x_0}} \frac{(x^3+2x+1)-({x_0}^3+2{x_0}+1)}{x-{x_0}}
    \frac{(x^2- x_0^3)+ 2(x- x_0)}{x- x_0}= \frac{(x- x_0)(x^2+ xx_0+ x_0^2)- 2(x- x_0)}{x- x_0}= x^2+ xx_0+ x_0^2+ 2 and, as x goes to x_0, that becomes 3x_0+ 2. Although, as harish21 says, it is simpler to put the [tex]x_0= 2[tex], f(x_0)= 13 from the start.

    = \lim_{x->{x_0}} (x^3-{x_0}^3)+2 = 2

    is the slope of this function is 2 then? I cannot figure out if this is correct and how to find the equation of the tangent line
    You appear to be saying that \frac{(x^3- x_0^2)+ 2(x- x_0)}{x- x_0}= x^3- x_0^3+ 2. That is, that you are canceling the " x- x_0 only from second term. You can't do that, of course, \frac{a+ b}{b} is NOT equal to a+ 1.

    Once you have that, use the formula you gave: y= f(x_0)+ f'(x_0)(x- x_0) with x_0= 2, f(x_0)= 13, and f'(x_0)= 14.
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