f(x)=sqrt x/6x-3 at the point (1, f(1)) find the equation of the tangetn line. 



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Find the equation of the line tangent to `f(x)=sqrt(x)/(6x-3)` when x=1:

`f(1)=sqrt(1)/(6-3)=1/3` so the point of tangency is `(1,1/3)` . To find the slope of the tangent line we evaluate `f'(1)` :

We can use the quotient rule to find `f'(x)` --


It is unnecessary to simplify this; we only need to evaluate when x=1.


So the point is `(1,1/3)` and the slope is `m=-1/2` so the equation of the tangent line is:



The equation of the tangent line is `y=-1/2x+5/6`


The graph of f(x) and the tangent line (in red):

` `

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