Question

Use the limit definition to find an equation of the tangent line to the graph of \(f\) at the given point. Then verify your results by using a graphing utility to graph the function and its tangent line at the point. $$ f(x)=\frac{1}{x-1} ;(2,1) $$

Short answer

The equation of the tangent line to the graph of the function \(f(x)=\frac{1}{x-1}\) at the point (2,1) is \(y = x - 1\).

Step-by-step solution

Calculating the derivative of the function

Using the definition of the derivative, we have: \[ f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}\]. Now, insert the given function: \[ f'(x) = \lim_{h \rightarrow 0} \frac{\frac{1}{x+h-1} - \frac{1}{x-1}}{h}\]. This is a complex fraction, so combine the fractions in the numerator to simplify it. After simplifying it, the derivative is \[ f'(x) = \frac{1}{(x-1)^2}\].

Finding the slope of the tangent line at the given point

Solve the derivative at the x-coordinate of the given point (x=2) to know the slope: \[ f'(2) = \frac{1}{(2-1)^2} = 1 \]. Hence, the slope (m) of the tangent line is 1.

Derive the equation of the tangent line

Using point-slope form equation \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is the point on the line and m is the slope. Therefore, the equation of the tangent line is \(y - 1 = 1(x - 2)\) or simplified to \( y = x - 1\).

Verifying results with graphing utility

To verify your result, use a graphing utility. Make sure the function \(y = \frac{1}{x-1}\) and the equation of the tangent line derived \(y = x - 1\) both share the point \((2,1)\) and that the line is indeed tangent to the graph at that point.