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Math Help - Limits

  1. #1
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    Limits

    Find:
    a) the slope of the tangent
    b) the slope at x = 7
    c) where the curve is differentiable


    equation:
    f(x) = \sqrt (x - 3)

    a)

    m = \lim_{h \to \-0} \frac{f(x + h) - f(x)}{h}

    m = \lim_{h \to \-0} \frac{\frac {1}{x + h + 3} - \frac {1}{x + 3}}{h}

    m = \lim_{h \to \-0} \frac{\frac {x + 3 - x - h - 3}{(x + h + 3)(x + 3)}}{h}

    m = \lim_{h \to \-0} \frac{-h}{(x + h + 3)(x + 3)} \times \frac {1}{h}

    m = \lim_{h \to \-0} \frac{-1}{(x + h + 3)(x + 3)}

    m = \frac{-1}{(x + 0 + 3)(x + 3)}

    m = \frac{-1}{(x + 3)^2}


    Textbook Answer: m = \frac{1}{(x + 3)^2}

    I get a negative answer instead of a positive answer, I can't seem to find my mistake, please help?


    Are questions b and c correct?

    b)

    y = \frac{-1}{(x + 3)^2}x + b
    y = \frac{-1}{(7 + 3)^2} \times 7 + b
    y = \frac{-1}{10^2} \times 7 + b
    y = \frac{-7}{100} + b

    m = \frac{-7}{100}

    c)

    x ER (x at all real numbers)
    Last edited by Macleef; April 11th 2008 at 04:35 PM.
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  2. #2
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    Quote Originally Posted by Macleef View Post
    Find:
    a) the slope of the tangent
    b) the slope at x = 7
    c) where the curve is differentiable


    a)

    m = \lim_{h \to \-0} \frac{f(x + h) - f(x)}{h}

    m = \lim_{h \to \-0} \frac{\frac {1}{x + h + 3} - \frac {1}{x + 3}}{h}

    m = \lim_{h \to \-0} \frac{\frac {x + 3 - x - h - 3}{(x + h + 3)(x + 3)}}{h}

    m = \lim_{h \to \-0} \frac{-h}{(x + h + 3)(x + 3)} \times \frac {1}{h}

    m = \lim_{h \to \-0} \frac{-1}{(x + h + 3)(x + 3)}

    m = \frac{-1}{(x + 0 + 3)(x + 3)}

    m = \frac{-1}{(x + 3)^2}


    Textbook Answer: m = \frac{1}{(x + 3)^2}

    I get a negative answer instead of a positive answer, I can't seem to find my mistake, please help?


    Are questions b and c correct?

    b)

    y = \frac{-1}{(x + 3)^2}x + b
    y = \frac{-1}{(7 + 3)^2} \times 7 + b
    y = \frac{-1}{10^2} \times 7 + b
    y = \frac{-7}{100} + b

    m = \frac{-7}{100}

    c)

    x ER (x at all real numbers)
    For a) your problem is that you are correct. The book is wrong.

    b) is also correct.

    c) What happens when x = -3?

    -Dan
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  3. #3
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    Quote Originally Posted by Macleef View Post
    Find:
    a) the slope of the tangent
    b) the slope at x = 7
    c) where the curve is differentiable

    a)

    m = \lim_{h \to \-0} \frac{f(x + h) - f(x)}{h}

    m = \lim_{h \to \-0} \frac{\frac {1}{x + h + 3} - \frac {1}{x + 3}}{h}

    m = \lim_{h \to \-0} \frac{\frac {x + 3 - x - h - 3}{(x + h + 3)(x + 3)}}{h}

    m = \lim_{h \to \-0} \frac{-h}{(x + h + 3)(x + 3)} \times \frac {1}{h}

    m = \lim_{h \to \-0} \frac{-1}{(x + h + 3)(x + 3)}

    m = \frac{-1}{(x + 0 + 3)(x + 3)}

    m = \frac{-1}{(x + 3)^2} Mr F says: CORRECT.


    Textbook Answer: m = \frac{1}{(x + 3)^2} Mr F says: INCORRECT.

    I get a negative answer instead of a positive answer, I can't seem to find my mistake, please help?


    Are questions b and c correct? Mr F says: b) is wrong. You're meant to substitute x = 7 into {\color{red}m = \frac{-1}{(x + 3)^2}} .

    And c) is wrong. You're meant to consider when {\color{red}m = \frac{-1}{(x + 3)^2}} is defined. In this case, it's easier to consider when it's NOT defined. It doesn't look to healthy at the real number x = - 3 .....

    b)
    y = \frac{-1}{(x + 3)^2}x + b
    y = \frac{-1}{(7 + 3)^2} \times 7 + b
    y = \frac{-1}{10^2} \times 7 + b
    y = \frac{-7}{100} + b

    m = \frac{-7}{100}

    c)

    x ER (x at all real numbers)
    ..
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  4. #4
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    So 3 is the limit ?
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  5. #5
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    Quote Originally Posted by Macleef View Post
    So 3 is the limit ?
    What do you mean? Limit of what?
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    What do you mean? Limit of what?
    of the curve
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  7. #7
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    Quote Originally Posted by Macleef View Post
    of the curve
    I have no idea what you're asking ....

    "So 3 is the limit of the curve" makes absolutely no sense. State the complete question that you think 3 is the answer to.
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  8. #8
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    is the curve differentiable at x = 3 because the restriction is at this x value? or is f(x) not differentiable because it's a discontinuity curve?
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  9. #9
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    Quote Originally Posted by Macleef View Post
    is the curve differentiable at x = 3 because the restriction is at this x value? or is f(x) not differentiable because it's a discontinuity curve?
    Of course it is. m = \frac{-1}{(x + 3)^2} has no trouble with x = 3: m = -1/36, a perfectly fine slope.

    As both I and topsquark have already said, m = \frac{-1}{(x + 3)^2} has trouble when x = -3. Because the denominator is equal to zero and hence the expression becomes undefined.

    So clearly the curve is differentiable for all real values of x except x = -3.
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  10. #10
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    Quote Originally Posted by topsquark View Post
    For a) your problem is that you are correct. The book is wrong.

    b) is also correct. Mr F begs to differ....

    c) What happens when x = -3?

    -Dan
    Nope. I nearly thought so too at first glance .....
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