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Math Help - calculus linear approximations

  1. #1
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    calculus linear approximations

    (a)Find the linear approximation L1(x) of the function f(x)=sqrt(x+3) at a=1
    (b)use linear approximate that you founf in part(a) to approximate sqrt(3.98)
    (c)scetch the graph of both f(x) and L1(x) and use these to determine if estimate in part (b) is greater or less than the true value of sqrt(3.98)
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  2. #2
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    Quote Originally Posted by gracy View Post
    (a)Find the linear approximation L1(x) of the function f(x)=sqrt(x+3) at a=1
    (b)use linear approximate that you founf in part(a) to approximate sqrt(3.98)
    (c)scetch the graph of both f(x) and L1(x) and use these to determine if estimate in part (b) is greater or less than the true value of sqrt(3.98)
    For part a) you mean x = 1, yes?

    The linear approximation is the line that approximates the function near a specified point. In this case you want the line, y = mx + b, that approximates the function f(x) = \sqrt{x+3} at x = 1.

    Well, the slope of the line is easy, it's just the first derivative at x = 1:
    f'(1) = \frac{1}{2 \sqrt{x+3}}|(x = 1) = \frac{1}{2 \sqrt{1+3}} = \frac{1}{4}.

    Now we need the y-intercept for the line. We know a point on this line: It is (1, f(1)) = (1, 2).

    So
    y = \frac{1}{4}x + b

    2 = \frac{1}{4} \cdot 1 + b

    Thus b = 2 - \frac{1}{4} = \frac{7}{4}

    Thus your linear approximation to y = f(x) at x = 1 is y = \frac{x}{4} + \frac{7}{4}.

    See if you can take it from here.

    -Dan
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  3. #3
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    Quote Originally Posted by gracy View Post
    (a)Find the linear approximation L1(x) of the function f(x)=sqrt(x+3) at a=1
    (b)use linear approximate that you founf in part(a) to approximate sqrt(3.98)
    (c)scetch the graph of both f(x) and L1(x) and use these to determine if estimate in part (b) is greater or less than the true value of sqrt(3.98)
    Hi,

    to a) There are a lot of possible approximations. Because f increases rather slowly the tangent will do here:

    Slope of the tangent: f'(x)=\frac{1}{2\cdot \sqrt{x+3}}
    f'(1) = 1/4
    f(1) = 2
    Use the point-slope-formula and you'll get:
    y=\frac{1}{4}x+\frac{7}{4}

    to b) \sqrt{3.98}=\sqrt{3+0.98}

    So x = 0.98. Plug in into the equation of the straight line and you'll get:

    y=1/4*0.98+7/4 = 1.995

    to c) I've attached a diagram. As you can see the tangent is always above the graph of f. Thus the estimated value is slightly too large.

    EB
    Attached Thumbnails Attached Thumbnails calculus linear approximations-sqrt_approx.gif  
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  4. #4
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    You can also find an approximation using a Taylor Polynomial, and subsequently find an error bound for how good that approximation is.
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