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 Bobbyjoe March 19th, 2017 06:43 PM

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?

 romsek March 19th, 2017 07:15 PM

$f(x)=\dfrac 1 x$

$f'(x) = -\dfrac{1}{x^2}$

$f(8) = \dfrac 1 8$

$f'(8) = -\dfrac{1}{64}$

$f(x) \approx \dfrac 1 8 - \dfrac {1}{64}\left(x - 8\right) = -\dfrac{1}{64}x +\dfrac 1 4$

you're trying to produce a function of $x$, not a number.

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