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Math Help - linearization

  1. #1
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    linearization

    Find the linear approximation of the function f(x) = √(25-x) at a = 0 and use it to approximate the numbers √24.9 and √24.99.

    L(x)=

    L(x)=f(a)+f'(a)(x-a)

    f'(a)=\frac{1}{2sqrt(25-x)}
    f'(0)=\frac{1}{10}
    f(a)=5
    <br />
L(x)=5 +\frac{1}{10}(x-25)
    (I used this to get the
    √24.9 and √24.99; (4.99, 4.999 respectively), but...

    <br />
L(x)=\frac{50}{10} + \frac{1}{10}x - \frac{25}{10}

    L(x)=

    I put this in for part a, and it was marked wrong. A=0, but if I put (x-0) instead of (x-25) it becomes a number greater than 5, which won't make sense considering you plug in a number lower than 25.
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  2. #2
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    Quote Originally Posted by hazecraze View Post
    Find the linear approximation of the function f(x) = √(25-x) at a = 0 and use it to approximate the numbers √24.9 and √24.99.

    L(x)=

    L(x)=f(a)+f'(a)(x-a)

    f'(a)=\frac{1}{2sqrt(25-x)} ***
    f'(0)=\frac{1}{10}
    f(a)=5
    <br />
L(x)=5 +\frac{1}{10}(x-25)
    (I used this to get the
    √24.9 and √24.99; (4.99, 4.999 respectively), but...

    <br />
L(x)=\frac{50}{10} + \frac{1}{10}x - \frac{25}{10}

    L(x)=

    I put this in for part a, and it was marked wrong. A=0, but if I put (x-0) instead of (x-25) it becomes a number greater than 5, which won't make sense considering you plug in a number lower than 25.
    noticed a sign error ***

    f(x) = \sqrt{25-x}

    f'(x) = -\frac{1}{2\sqrt{25-x}}
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  3. #3
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    I set it up in point slope form like you would for the tangent line and got:
    (y-5)=(\frac{-1}{10}(x-0))
    y= 5 + \frac{-1}{10}x

    So x if really delta(x)? (25-24.9 or 24.99)
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  4. #4
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    f(x) = \sqrt{25-x}

    f'(x) = -\frac{1}{\sqrt{25-x}}


    f(.1) = \sqrt{24.9}

    f(.1) \approx L(.1) = f(0) + f'(0)(.1 - 0) = 5 - \frac{1}{5} \cdot (0.1) = 4.98


    f(.01) = \sqrt{24.99}

    f(.01) \approx L(.01) = f(0) + f'(0)(.01 - 0) = 5 - \frac{1}{5} \cdot (0.01) = 4.998
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