# What is the equation of the line tangent to #f(x)=1/abs(x-3) # at #x=4#?

y > 0.

The graph is for the purpose. The scales are befittingly chosen.

For x > 3, y =1/x-3 and for x < 3, y = -1/(x-3). Together, the graph is a \

branch of a hyperbola and its mirror image, with respect to the

asymptote x = 3.

The first is relevant, as x = 4 > 3.

Easily, y'=-1/(x-3)^2=-1, at x = 4.

So, the equation to the tangent at (4, 1) is

y-1=-(x-4).

graph{y(y|x-3|-1)(x-3+.0001y)(x+y-5)=0 [-10, 10, -1, 5]}

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The equation of the line tangent to f(x)=1/abs(x-3) at x=4 is y = -1/4(x-4) + 1/abs(4-3).

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