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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: USA Posts: 2,533 Thanks: 1390 $f(x)=\dfrac 1 x$ $f'(x) = -\dfrac{1}{x^2}$ $f(8) = \dfrac 1 8$ $f'(8) = -\dfrac{1}{64}$ $f(x) \approx \dfrac 1 8 - \dfrac {1}{64}\left(x - 8\right) = -\dfrac{1}{64}x +\dfrac 1 4$ you're trying to produce a function of $x$, not a number. Tags approximation, find, line, tangent Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post noobinmath Pre-Calculus 4 March 23rd, 2016 10:31 AM DerGiLLster Calculus 1 October 20th, 2015 07:04 PM Shamieh Calculus 2 October 7th, 2013 05:12 AM Shamieh Calculus 3 September 23rd, 2013 02:16 PM SSmokinCamaro Calculus 2 March 1st, 2009 12:54 PM

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