What this means is that from [a,b] , it must be continuous and differentiable, the secant lines between them is Parallel to the tangent line of c, the midpoint or average of [a,b], which is c.
For example in the equation y=sin(2x)+3, there is a secant line on the interval [0,1.56], that secant lines slope,green, will be the same as c tangent lines slope,blue, the midpoint of that interval, which makes them parallel.
2. it doesn't work for non differentiable and/or continuous functions because for non differentiable and non continuous functions is the equation y=2/x^2
since the secant line would be at 4, the midpoint of that interval,c,would be at 0,due to the discontinuity, there is not midpoint and is not differentiable, meaning c has no tangent line because of the discontinuity, and since we cant find the slope of tangent of c, then the mean value theorem is not applicable to functions that are not differentiable and not continuous.
a function that is continuous but not differentiable would be abs(-x/2)+1
It is continuous , but at x=0, there is a corner. As x approaches 0 from -1 from the negative side, its slope is not the same as x approaches 0 from the positive side from 1, they do not have the same tangent line. Since they do not have the same tangent line, it is not differentiable at point c=0, because the slopes are different and the tangent line from the negative and the positive side are not the same.
Whoa whoa whoa. Why does the slope at c have to be 0? And what tangent line and what secant line are you talking about in your "non working" graph (which is not discontinuous btw, only not differentiable.)
ReplyDeleteYou have about 40 other explanations that you can look at to get a good idea of what this means and then to re-do it for this week's blog.
Thought this one is not graded based on correctness, the next one will be.
hey how did you make the graphs.. what program???
ReplyDeletebtw... im confused about the second graph. you said that its not true that the theorem doesnt work for discontinuous functions.?? i think what you have to do is set boundaries say on[-.1,.1] .. then the function is not differentiable at x=0
therefore there will be no secant or tangent line at x=0. btw the function is continuous. its just not differentiable at x=0.
:)
yea look at other people, not mines because it sucks. Jesus and Dianna have good explanations.
ReplyDeleteAll of the above . I think its good though
ReplyDeletegood job in explaining that the fact why is f(x)=|-x/2|+1 not differentiable becase the derivates from each side are different
ReplyDeletecoool!