f'(x) = -1/x^2, so at x=a, the slope is -1/a^2.
So, now you have a point and a slope, so the line is
y - 1/a = -1/a^2 (x-a)
function f(x) =1/x
assume that point has coordinates (a,1/a), where a > 0 is some fixed value.
(a) Find the equation of the tangent line. (Your answer will be a formula depending on the value a.)
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