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| Math Central | Quandaries & Queries |
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Question from Steven, a student: Consider the graph of the function f(x) = 1/x in the first quadrant, and a line tangent to f at a point P where x = k. Find the slop of the line tangent to f at x = k in terms of k and write an equation for the tangent line l in terms of k. |
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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