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 March 19th, 2017, 06:43 PM #1 Member   Joined: Feb 2017 From: henderson Posts: 36 Thanks: 0 Find the tangent line approximation Find the tangent line approximation for f(x) = 1/x near x=8 so I used the f(a) +f'(a)(x-a) f = 1/x f' = -1/x^2 I end up writing y - 1/8 = -1/64 (x + 8) my answer I get is -1/64 - 1/4, but it's wrong so I am stuck. :/ Any help? Last edited by skipjack; March 19th, 2017 at 11:12 PM.
 March 19th, 2017, 07:15 PM #2 Senior Member     Joined: Sep 2015 From: CA Posts: 1,300 Thanks: 664 $f(x)=\dfrac 1 x$ $f'(x) = -\dfrac{1}{x^2}$ $f(8) = \dfrac 1 8$ $f'(8) = -\dfrac{1}{64}$ $f(x) \approx \dfrac 1 8 - \dfrac {1}{64}\left(x - 8\right) = -\dfrac{1}{64}x +\dfrac 1 4$ you're trying to produce a function of $x$, not a number.

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