`f(x) = x/(x + 1), [(-1/2),2]` Use a graphing utility to (a) graph the function `f` on the given interval, (b) find and graph the secant line through points on the graph of `f` at the endpoints of the given interval, and (c) find and graph any tangent lines to the graph of `f` that are parallel to the secant line.
(1) `f(x)=x/(x+1) ` is a rational function. It has a vertical asymptote at x=-1, and a horizontal asymptote of y=1. On the interval [-1/2,2] the graph is increasing towards the limiting value of y=1.
f(-1/2)=-1 and f(2)=2/3
(2) The slope of the secant line through the endpoints of the interval is `m=(2/3-(-1))/(2-(-1/2))=2/3 ` so the equation of the secant line is `y=2/3x-2/3 `
(3) Since the function is continuous and differentiable on the interval the Mean Value theorem guarantees the existence of a c in the interval such that f'(c) is the slope of the secant line.
` 1/(x+1)^2=2/3 ==> x=sqrt(3/2)-1 `
The corresponding point on the graph is `(sqrt(3/2)-1,1-sqrt(2/3)) `
So the tangent line parallel to the secant line is `y-1+sqrt(2/3)=2/3(x-sqrt(3/2)+1) `
The graph of the function, secant line, and tangent line: