Answer
$$y = -\frac{8}{25}x + \frac{8}{5}$$
Work Step by Step
In this exercise, we are given a point $(x_0,y_0)$ that is in a line tangent to $f(x)$, but that is not part of the function itself. In other words, $(x_0,y_0)$ is not the point of tangency between $f(x)$ and the line. As such, the approach is somewhat different to previous exercises. We can construct the following equation for the tangential line using the$(x_0,y_0) = (5, 0)$ provided: $$y - y_0 = m(x - x_0)$$ $$y - 0 = m(x - 5)$$ $$y = m(x - 5)$$ where m is the slope of the tangential line. We know that, at the point of tangency, both the function $f(x)$ and the equation above must have the same $(x,y)$. Therefore: $$f(x) = y$$ $$\frac{2}{x} = m(x-5)$$ for the point of tangency, where by rewriting the equation, we get: $$\frac{2}{(x)(x-5)} = m$$ We can now proceed to find the first derivative of the original function, since we know it gives us a formula for finding the slope of a tangential line to any function $f$: $$f(x) = \frac{2}{x}$$ $$f(x) = 2x^{-1}$$ $$f'(x) = -2x^{-1 -1}$$ $$f'(x) = -2x^{-2}$$ $$f'(x) = -\frac{2}{x^{2}}$$ Since we know that $f′(x)=m$, we can now proceed to do the following: $$f′(x)=m$$ $$-\frac{2}{x^{2}} = \frac{2}{(x)(x-5)}$$ $$\frac{1}{-x^{2}} = \frac{1}{(x)(x-5)}$$ $$(x)(x-5) = -x^{2}$$ $$x^{2} - 5x = -x^{2}$$ $$2x^{2} - 5x = 0$$ $$x(2x -5) = 0$$ where $x = \frac{5}{2}$ since $x=0$ (the other solution) is undefined for the function $f(x)$. With this value of $x$, we can solve for $f'(x)$: $$f'(x) = -\frac{2}{x^{2}}$$ $$f'(\frac{5}{2}) = -\frac{2}{(\frac{5}{2})^{2}}$$ $$f'(\frac{5}{2}) = -\frac{8}{25}$$ Finally, with the slope at hand, we can finish the equation of the tangential line we constructed initially: $$y = m(x-5)$$ $$y = -\frac{8}{25}(x - 5)$$ $$y = -\frac{8}{25}x + \frac{8}{5}$$
