f(x)= x, 0<and equal to x<1
= 0, x=1
is zero at x=0 and at x=1. it derivative is equal to 1 at every point between 0 and 1, so f' is never zero between 0 and 1, and the graph of f has no tangent parallel to the chord from (0,0) to (1,0).
why this does not contradict the Mean Value Theorem?
The Mean Value Theorem states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists a point c in (a, b) such that the derivative of the function at c is equal to the average rate of change of the function over the interval [a, b]. In other words, if f is continuous on [a, b], then there exists a point c in (a, b) such that f'(c) = (f(b) - f(a))/(b - a).
In the given function f(x) = x, 0 <= x <= 1, and f(0) = 0, f(1) = 0. The derivative of f(x) is equal to 1 for all x in the interval (0, 1). Since f'(c) = 1 for any value of c between 0 and 1, this means that the average rate of change of f(x) over the interval [0, 1] is also 1.
In this case, since f'(c) does equal the average rate of change of f(x) over the interval [0, 1], the Mean Value Theorem is not contradicted. The Mean Value Theorem only guarantees the existence of a point where the derivative of the function equals the average rate of change over the interval, but it does not guarantee that there are no other points where the derivative takes other values. Therefore, even though f' is never zero between 0 and 1, the Mean Value Theorem still holds because it only requires the existence of one point where the derivative matches the average rate of change.