Revision of Mean Value Theorem

1) we all have an idea of what the mean value theorem is and we are doing this entry because of some unclear area in this topic. the mean value theorem finds a point in the given interval that is parallel to two points a and b also in the interval there are certain condition the graph most pass before the mean value theorem can be used and give a definite answer. off the back i do not know the conditions but i believe those conditions are the function must be continuous and differentiable. if it passes that, then the mean value theorem can be used. the equation that i am going to use is √(x) in the interval [0, infinity). REMEMBER the point of the blogs is to show an understanding so I'm going to show u my understanding with this simple Graph. the graph look like this











I'm going to choose x=4, and x=9 for a and b. the out put for x=4 is 2 and for x=9 is 3. simple rite. using the mean value theorem we get a slope of 1/5 . doing the math the point x=25/4 would have an instantaneous slope of 1/5. the tangent line would be y = (x/5)+(5/4) which is going to be represented by the green line. f(a) = 4 and f(b)=9. the secant line witch is not in the graph would have the same slope as the green line y=(x/5)+(5/4).

A very MEAN Value Theorem

1) f'(c) = [f(b) - f(a)]/(b - a) means there is a point in a continuous function c that has a slope that is equal to the secant line of two points x = a and x = b. in the graph there are 3 function. the green function is x^2 and the blue line indicates the two points which are 0 and 2. the blue line shows the secant line that connects the two. and the green line is the point c in this case it is 1, where the slope is equal to the secant line as u can see. green line slope(f'(c)) is equal to the blue secant line([f(b) - f(a)]/(b - a)). note the secant(blue) line is parallel with the green line, which represents f'(c).


2) the mean value theorem only works on continues and differentiable functions because other wise the theorem can not give a guaranteed point C that exist in the function.

(C) if the function is discontinuous such as |1/x| the function does not have point c where the slope is equal to 0 and is parallel to the the secant line. this is what makes discontinuous function unable to use the mean value theorem.

(D) the function |x| is a continues function but not differentiable. in this graph the secant line crosses X=-2 , and x=3 and with the mean value theorem f'(c) should be equal to 1/5 but there are only two possible slopes which are -1 and 1, therefore c does not exist. which proves wrong the mean value theorem. that is why functions that are discontinuous or not differentiable are not guaranteed a f'(c) that is equal to the secant line.