From the point (4,1) draw the tangent to the curve y =

From the point $(4,1)$ draw the tangent to the curve $y = \dfrac{x-1}{x}$ and find the point of tangency.

Step 1: Derivative of the curve

Write $y = 1 - x^{-1}$; differentiating, \[y' = \frac{1}{x^{2}}. \tag{1}\]

At an arbitrary point $P = (x_{0},y_{0})$ on the curve, $y_{0} = \dfrac{x_{0}-1}{x_{0}}$ and the slope of the tangent is $k = \dfrac{1}{x_{0}^{2}}.$

Step 2: Tangent at $P$

\[ y - y_{0} = \frac{1}{x_{0}^{2}}(x-x_{0}). \tag{2} \]

Step 3: Require the tangent to pass through $(4,1)$

Substituting $x=4,\ y=1$ into (2) and simplifying the left side, $1 - \dfrac{x_{0}-1}{x_{0}} = \dfrac{1}{x_{0}}$, gives

\[ \frac{1}{x_{0}} = \frac{4-x_{0}}{x_{0}^{2}} \;\Longrightarrow\; x_{0} = 4-x_{0} \;\Longrightarrow\; x_{0} = 2. \tag{3} \]

Step 4: Point of tangency and tangent equation

Then $y_{0} = \dfrac{2-1}{2} = \dfrac12$ and $k = \dfrac14$, so the tangent at $P = \left(2,\dfrac12\right)$ is

\[ y - \frac12 = \frac14(x-2) \;\Longrightarrow\; y = \frac{x}{4}. \tag{4} \]

Step 5: Verification

\[\text{On the curve: } \frac{2-1}{2} = \frac12 = y_{0}\ \checkmark\]

\[\text{Tangent passes through } (4,1)\text{: } y(4) = \frac{4}{4} = 1\ \checkmark\]

\[\text{Slope matches derivative: } y'(2) = \frac14 = k\ \checkmark\]