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Math Help - How to find equation of tangent of function

  1. #1
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    How to find equation of tangent of function

    Given g(x) = √(X-1), find the equation of tangent line to the graph at the point (5,2).

    I have no idea why this was given as an assignment; this is not something I have any recollection of learning how to do.

    Can somebody please help me and explain how I to do this.

    Thank you.
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  2. #2
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    let y = mx + n be the equation of the tangent line we are looking for.

    First, you know the line passes through the point (5,2) -- so you can put x=5, y=2 and get your first equation.

    Second, you know the line y=mx+n is tangent to the curve g(x) = \sqrt{x-1} at the point (5,2) -- that means that they have the same slope at x=5, or more precisely that g'(5) = m.

    Can you follow with the rest?
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  3. #3
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    Thank you so much for your answer.

    I'm not sure I'm following, should I plug in 5 in my function?

    That would give me g(5) = 2? So that my slope would be 2.

    That would give me y= 2X - 8 which doesn't really make sense.

    There must be something I don't get.
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  4. #4
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    Quote Originally Posted by Mrs. White View Post
    Thank you so much for your answer.

    I'm not sure I'm following, should I plug in 5 in my function?

    That would give me g(5) = 2? So that my slope would be 2.

    That would give me y= 2X - 8 which doesn't really make sense.

    There must be something I don't get.
    g(5) = 2 is the point on the curve which the tangent line "touches".

    the slope of that line is g'(5) ... know what that means?
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  5. #5
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    Quote Originally Posted by skeeter View Post
    the slope of that line is g'(5) ... know what that means?
    No I don't. How am I to find the equation of this line if the slope is g'(5)?

    Thank you so much.
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  6. #6
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    Quote Originally Posted by Mrs. White View Post
    No I don't. How am I to find the equation of this line if the slope is g'(5)?

    Thank you so much.
    g'(5) = \lim_{x \to 5} \frac{g(x) - g(5)}{x-5}
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  7. #7
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    Quote Originally Posted by skeeter View Post
    g'(5) = \lim_{x \to 5} \frac{g(x) - g(5)}{x-5}


    Does that mean I would replace h by 5 in the difference quotient equation I found?

    This is what I found:
    1/√(x+h)-1 + √(x-1)

    So now I would do


    √(5+5)-1 + √(5-1) = √9 + √4 = 5, which would mean my slope is 1/5 ?

    If so my equation would be: Y= 1/5(X) +1 ??

    I hope this is correct, that would be such a relief.



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  8. #8
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    Quote Originally Posted by Mrs. White View Post
    Does that mean I would replace h by 5 in the difference quotient equation I found?

    This is what I found:
    1/√(x+h)-1 + √(x-1)

    So now I would do


    √(5+5)-1 + √(5-1) = √9 + √4 = 5, which would mean my slope is 1/5 ?

    If so my equation would be: Y= 1/5(X) +1 ??

    I hope this is correct, that would be such a relief.


    sorry ... but that is not how a limit of a difference quotient is determined.

    if you do it correctly, you will find that the the slope of g(x) at x = 5 is g'(5) = \frac{1}{4}
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  9. #9
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    Quote Originally Posted by skeeter View Post
    sorry ... but that is not how a limit of a difference quotient is determined.

    if you do it correctly, you will find that the the slope of g(x) at x = 5 is g'(5) = \frac{1}{4}
    This is what I had previously found ! But I thought I was wrong. I plugged in 0 for h and it gave me a slope of 1/4.

    My equation was then y=1/4X + 3/4.

    Was that the correct way to proceed (plug in 5 for x and 0 for h)?
    Last edited by Mrs. White; September 28th 2009 at 01:41 PM.
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