The Mean Value Theorem states the following:
- Given that f(x) is a continuous and differentiable function, there is a guaranteed point where the slope of the tangent line at point c, is equal to the slope of the secant line passing through the two end points of an interval. If the interval is [a,b], the secant line passes through the points a & b.
- The secant line and tangent line are parallel, for lines with the same slope are always parallel. They will never intersect!

The graph of f(x)= sin(x) + 2 on the interval [0,3], proves the Mean Value Theorem. The function is continuous and differentiable on the given interval, a requirement of the Mean Value Theorem. Also, the slope of the function's tangent line (y=3) is parallel to its secant line (y=2). Both of these lines have a slope of 0. The tangent line touches the function sin(x) at point c (3). The secant line passes through points x=0 (point a) and x=3 (point b), the two endpoints on the interval (0,3) [a,b0].
2. Why does the Mean Value Theorem ONLY work for continuous and differentiable functions ?
Okay, let's get down to the basics.
A continuous function is defined as a function for which small changes in the input result in small changes in the output.
Continuous functions have the following properties:
- For all the points on a continuous graph, the limit from the right equals the limit from the left. Therefore, the limit at each point exists.
[The limit
as x approaches a number is the actual output (the y value) at that x].
- Also, on a closed interval, every continuous has an absolute maximum and absolute minimum.
- A function may be continuous, but have a point where it is not differentiable [a point where the slope is undefined due to a corner, cusp, jump discontinuity, oscillating discontinuity, or infinite discontinuity (asymptote)].
Differentiable functions have the following qualities:
- The slope must be defined at a particular point in the function
- If f is differentiable at a point c, then f is continuous at that point c. (but NOT vice versa).
Consider the function f(x) = x^(7/9) + 1 on the interval [-1,1]. It is continuous at every point in this interval and also differentiable at every point except x=0, where there is a cusp. At x=0, the slope is undefined. This function fails the Mean Value Theorem first and foremost because there is no tangent line parallel to the secant line y=2. Technically, the line y=1 does not exist, since as i said before, x=0 is undifferentiable. However, if the slope at x=0 existed, it would be 0 and be parallel to the secant line x=2.This is why a graph must be differentiable and continuous to have a guaranteed point c.
Differentiable but not CONTINUOUS:
Consider the graph of f(x)=(sin(x)/x on the interval [-2,2]. This graph is differentiable at every point except x=0, where it is also discontinuous. This graph fails the Mean Value Theorem because it does not have a guaranteed point c. Point c in this case would the tangent line at y=0, but since the function is neither differentiable nor continuous there, the "tangent line" does not exist. Only the secant line that goes through both -2 and 2 exists.

i thought a continuous function also must have the right and left side limits equal to f(x), but I'm not sure. the first part is great but the second part u have more info than is needed like when u mention the maximum and minimum.
ReplyDeletefor your graph of f(x) = x^(7/9) + 1, it is also not differentiable because the derivative from each side are different
ReplyDelete