Rolle's theorem:
Statement :
Let, F(x) be a real valued
function in interval [a, b] such
that,
1. F(x) is continuous in closed interval [a, b].
2. F(x) is differentiable in
open interval (a, b)
3. F(a) = F(b).
2. Rolle's theorem:
Statement :
Let, F(x) be a real valued
function in interval [a, b] such
that,
1. F(x) is continuous in
closed interval [a, b].
2. F(x) is differentiable in
open interval (a, b)
3. 3. F(a) = F(b).
Then there exist at least one
point X = C € (a, b) such that
' (c) = 0.
Now, meaning and
geometrical interpretation of
above three conditions.
'(c) represents slope of the
line at point C.
'(c) = 0, means that slope
of line is zero i.e. Line is
tangent to the curve (in
4. graph) at point c.
Equation solve के यावर जे हा
line x-axis ला या point la
parallel येईल ते हा या point
ला function चा derrivative
'zero' येईल.
Geometrical interpretation :
( posibility of tangent parallel
to x - axis ).
5. You can see full immage here
Click on following :
SEE
Examples :
1. Verify Rolle's theorem
for () = (- )² in the
interval [ 0, 1].
6.
() = (- )².
1. Since, () is polynomial
so it is continuous in [0, 1]
i.e since () is exist for all
€ [ 0, 1] so () is
continuous in [0, 1].
2. '() = 3X² - 4X + 1 is
exist for all X € ( 0, 1) so
() is differentiable in
(0, 1).
3. And f(0) = f(1).
7. Hence () satisfy all three
conditions of Rolle's theorem
so there exist a value X = C
such that,
' (c) = 0. ---------------- (1)
Now,
'() = 3X² - 4X + 1
Put value as "C" in equation,
'(c) = 3C² - 4C + 1 -----(2)
From equation (1) and (2)
3C² - 4C + 1 = 0
Now use factorization method
8. (3C-1) (C -1) = 0
C = 1/3 and C = 1.
But only C = 1/3 belongs to
(0, 1) and
'(1/3) = 0
Hence Rolle's theorem is
verified.
S. Tupe
(8605104356/9970106236
![Rolle's theorem:
Statement :
Let, F(x) be a real valued
function in interval [a, b] such
that,
1. F(x) is continuous in
closed interval [a, b].
2. F(x) is differentiable in
open interval (a, b)](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)