Question

(a) find an equation of the tangent line to the graph of \(f\) at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of a graphing utility to confirm your results. $$ f(x)=\frac{(x-1)}{(x+1)} \quad\left(2, \frac{1}{3}\right) $$

Short answer

The equation of the tangent line to the graph of the given function at the point (2, 1/3) is \(y = \frac{2}{9}x - \frac{1}{9}\).

Step-by-step solution

Find the derivative of the function

The derivative of \(f(x)=\frac{(x-1)}{(x+1)}\) is calculated using the quotient rule for derivatives; \(\frac{d}{dx} (u/v) = (v(du/dx) - u(dv/dx))/(v^2)\) Letting \(u = x - 1\), \(du/dx = 1\) and \(v = x + 1\), \(dv/dx = 1\). So, \(f'(x) = \frac{((x + 1)(1) - (x - 1)(1))/((x + 1)^2)} = \frac{2}{(x + 1)^2}\)

Substitute the given x-value into the derivative

Now substitute \(x = 2\) into the derivative to find the slope of the tangent at that point. The outcome of \(f'(2)\) is \(\frac{2}{9}\)

Find the equation of the tangent line

Knowing that the equation of the tangent line is \(y = mx + c\), the slope \(m\) is \(f'(2) = \frac{2}{9}\), and given that the tangent line goes through the point (2, 1/3), substitute these values into the equation to find \(c\): \(\frac{1}{3} = \frac{2}{9} * 2 + c \Rightarrow c = -\frac{1}{9}\). Therefore, the equation of the tangent line is \(y = \frac{2}{9}x - \frac{1}{9}\)

Confirm the results with a Graphing Utility (Optional)

This part would usually be done visually with a graphics calculator or computer software: plotting the function \(f(x) = \frac{(x - 1)}{(x + 1)}\) and the line \(y = \frac{2}{9}x - \frac{1}{9}\) on the same graph, it should be possible to visually confirm that the line is indeed tangent to the function at the point (2, 1/3). Confirming graphically whether the line is a tangent to the curve at the point (2, 1/3).