f '(c) = f(b)-f(a)/(b-a)
1. Explain what this means graphically by showing an example.


The graph shown here is f(x)= x^2+1. First of all the graph is both continuous and differentiable so it is a possible candidate for the Mean Value Theorem. Now let's apply a secant line (or an average rate of change)
The green line [y=2x+1] represents f(b)-f(a)/(b-a). The two points where the green line intersects our parabola are the points "a" and "b".
The tangent line is parallel to the secant line, meaning that instantaneous rate of change is equal to the average rate of change.2. Explain why this only works for continuous and differentiable functions.
The Mean Value Theorem only applies to continuous and differentiable functions because if a graph has a discontinuity of any kind, then it disrupts the normal flow of the graph and it would mean that the interval has stops. You cannot find instantaneous or average rates of change when there is a chunk of the graph missing. The function also has to be differentiable because there needs to be a slope in order for any graph to have a rate of change.
This graph, y= |x|, is not differentiable at x=o because it has a corner. A "corner" has an infinite amount of tangent lines.
A great start Rocio. A few questions to help you along:
ReplyDeleteWhat does a "rate of change" look like graphically?
A secant line needs to go through two points. What are the two points that you used?
The counterexamples need boundaries. What are they? Why doesn't it work?
What do you mean by "the interval has stops" (part 2)?
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ReplyDeletefor your first example you should add what are the close intervals
ReplyDeleteexample:
f(x)=x^2, [0, 4]
something like that because if you dont add the closed intervals they may believe that the function in general has to be differentiable and continuous when the truth is it has to be within the closed intervals
Yeah you should add boundaries to the |x| because there is still slopes, but the mean value theorem only doesn't apply when c=0, since there is no slope there (: I still like your explanations, Josh!
ReplyDeletehey nice way of explaining it but yeah your missing the boundaries in the function for to understand it better!
ReplyDelete