Let f(x) be a continuous function on the interval [a, b] and differentiable on
the open interval (a, b). Then there is at least one value c in the interval (a, b) such that
f'(c) = (f(b)-f(a))/(b-a)
In other words, the tangent line to the graph of f at c and the secant
through points (a,f(a)) and (b,f(b)) have equal slopes and are therefore parallel.
Interactive Tutorial Using Java Applet
1 - Click on the button "click here to start" and maximize the window obtained.
2 - Use the first slider to select a position for the secant. Use the second slider to change the position of the tangent so that the x coordinate of the point of tangence is somewhere between a and b and the tangent and secant are parallel (see figure above). You can adjust the position of the tangent such that the values of the slopes of the two lines are as close as possible. Approximate the x coordinate c of the point of tangency. Try to find the exact value by calculation and compare the two values.
3 - Try to position the secant so that you get two points of tangence and therefore two values of c.