`f(x) = x - 2sin(x), [-pi,pi]` 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.
f(-pi) = -pi, f(pi) = pi. The secant line goes through the points `(-pi, -pi)` and `(pi, pi)` . The equation is obviously y=x, its slope is 1.
Let's find points where f'(x)=1:
f'(x) = 1 - 2cos(x). It is =1 when cos(x)=0, there are two such points on `(-pi, pi)` , `x=pi/2` and `x=-pi/2` . `f(pi/2)=pi/2-2` , `f(-pi/2)=-pi/2+2` .The equations of tangent lines are therefore `y=pi/2-2+(x-pi/2)` and `y=-pi/2+2+(x+pi/2)`. The equivalent equations are y=x-2 and y=x+2.
Graph is here: https://www.desmos.com/calculator/5ucbk6vzji