



 
Hi Steven, First lets try this problem with $k = 3.$ The function is $f(x) = \large \frac{1}{x} = x^{1}$ so the point $P$ has coordinates $\left(3, \large \frac13\right).$ The derivative of $f(x)$ is $f^{\prime}(x) = (1) x^{2} =  \large \frac{1}{x^2}$ and hence the slope of the tangent to $f(x)$ at $x = 3$ is $f^{\prime}(3) =  \large \frac{1}{3^2} =  \frac19.$ Thus the tangent line $l$ to $f(x)$ at $x = 3$ has equation \[\left(y  \frac13\right) =  \frac19 (x  3)\] Now you try it with $x = k.$ Penny  


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