Hmmm soo. from the comment i got from miss hwang... I am going to add to the second question...
on the interval [a,b] when a≤c≤b... it is guaranteed that at one point,
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiLztosyIEO2u05FxjNNCHaxMB6UwJtUBt5hm24McoC-Tg5RPr8bSJJd7Osz68340SpKyG3a2pqy7f1bI8WTqcKdW-d1fPCxQxVa1Ie67gQUE2EDQd5bah2osVHevsIIVEP157pzlJwZwLd/s320/mean+value+theorem.bmp)
Example
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgALBeSmQvYi1fHX4BkzTX2kfnD00Wf5LU4IgINTZJyUXaTpjvL-NzwgLb2yGGoV1iwm6zXsOpLi12waZYEBTbK4_8Ih4ONX48HRDtyfw7N6eTYGkaWKpzHOHEjWOsRBC2GdeRQR8jN7hqV/s320/graph1.bmp)
the goal is we must find where point c lies on the interval...
by using this formula
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiLztosyIEO2u05FxjNNCHaxMB6UwJtUBt5hm24McoC-Tg5RPr8bSJJd7Osz68340SpKyG3a2pqy7f1bI8WTqcKdW-d1fPCxQxVa1Ie67gQUE2EDQd5bah2osVHevsIIVEP157pzlJwZwLd/s320/mean+value+theorem.bmp)
where a=2
and b= 0
First finding the secant line
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdzQugSELO92tfryoywHJp9MTmSKCmODGmXBKb_sAc5e_8Q5ki9LJwudAWQ-0jvvsYnD4XL31uurDG3UpKMvNfLodxQNyeNL-fmJipTzAUjBHuZsAm8O2iAO7VCWVIkdZwC6JM44FpW86A/s320/graph2.bmp)
y-y1= m (x-x1)
so from f(2)=(2)^2 = (2,4)
y-4=2(x-2)
y=2x >>>> is our secant line equation
Now we can find where point c lies
the derivative of f(x) = x^2 is 2x , therefore f'(c) = 2c
2c= m
2c= 2
c=1
Now that we have found where c lies (which is at x=1)
Now we can find the equation of the tangent line
f(c) = c^2
we have found that c=1
therefore, f(1)= 1^2 =1
the tangent line equation has to pass through (1,1)
We know that the tangent line passes through point c,
and it is paralell to the secant line (which is y=2x)
the slope of y=2x is 2
y-1=2(x-1)
y-1= 2x-2
y= 2x - 1
YAY!!!!
This only works for continuosus and differentiable functions...
for example: (square root (x^2)) +1
from [-2, 2]
this function is not differentiable at x=0
the mean value theorem FAILS here!!!
there is no tangent an x=0 because corners and cusps do not have a tangent line
so from the interval [a,b] in this case is [-2,2]
m= (f(2)-f(-2)) /2--2
= [(|2|+1)-(|-2|+1)]/4
=0/4
m=0
f(2)=3
y-3=0(x-2)
y=3 >>> this is suppose to be your secant line equation
but... sadly :( there's no secant line here.
because of the corner at x=0
but if.. you use the same function f(x) but on the interval [0,2] then the secant line is y=x+1
f'(c)= 1
the tangent line is also y=x+1
YAY!!!!!
i hope that this time my explanation is clearer
:)
now i gotta go comment more blogs
Excellent job on the example for the Mean Value Theorem! I think you mightve switched what you defined as a and b, but you totally nailed what it was I was looking for.
ReplyDeleteYour counterexamples need boundaries though. Youre right in that it fails, but be specific and show why it does.
LOL They are the same!
ReplyDeleteGood explanation !
ReplyDeleteI like how you take it step by step and actually say what step you're on.
For part 2, how come? Can you be more specific ?
Wendy im almost sure that there is a secant line on that last graph, but there isn't a tangent line. And I don't think we can use x^2,. Ms. Hwang says it on one of her latest posts
ReplyDeleteBut I love the step by step explanations!, They're very helful, and allow me to see exactly why the mean value theorem works on that equation! Great Job!
..... still aint fininshed
ReplyDeleteyour blog is very organize. really good explanations.
ReplyDeleteI like.
ReplyDeleteomg..
ReplyDeletegood. clear.
wooow
haha
good job!