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  1. #1
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    derivatives

    if f(1)=2 and f'(1)=5, use the equation of the line tangent to the graph of f at x=1 to approximate f(1.2).
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  2. #2
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    Quote Originally Posted by asnxbbyx113 View Post
    if f(1)=2 and f'(1)=5, use the equation of the line tangent to the graph of f at x=1 to approximate f(1.2).
    Point-Slope Form?

    f(1) = 2 gives the point
    f'(1) = 5 gives the slope.

    y - 2 = 5(x-1)

    Substitute x = 1.2

    That's about it.
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  3. #3
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    Quote Originally Posted by asnxbbyx113 View Post
    if f(1)=2 and f'(1)=5, use the equation of the line tangent to the graph of f at x=1 to approximate f(1.2).
    Or you can think of it as a truncated Taylor series.
    f(1.2) \approx f(1) + \frac{1}{1!}f^{\prime}(1) \cdot 0.2

    So
    f(1.2) \approx 2 + 5 \cdot 0.2
    etc.

    -Dan
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  4. #4
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    Quote Originally Posted by topsquark View Post
    Or you can think of it as a truncated Taylor series.
    f(1.2) \approx f(1) + \frac{1}{1!}f^{\prime}(1) \cdot 0.2

    So
    f(1.2) \approx 2 + 5 \cdot 0.2
    etc.

    -Dan
    That is not the best explanation to give this poster. Because he will not hear the term "Taylor Series" for another year.
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