very MEAN value Theorem (edited version)

Hmmm soo. from the comment i got from miss hwang... I am going to add to the second question...

The very MEAN value theorem states, that given f(x) of a smooth continuous and differentiable curve, there is at least one point on f(x) at which the derivative (slope) of the curve is parallel to the "average" derivative of the arc.

on the interval [a,b] when a≤c≤b... it is guaranteed that at one point,

Example

for the graph y=x^2 on [0,2]

the goal is we must find where point c lies on the interval...

by using this formula

where a=2

and b= 0

First finding the secant line

slope of the secant line is 2

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