btech 1st sem: maths: mean value theorem
DESCRIPTION
Study Materials for 1st Year B.Tech Students Paper Name: Mathematics Paper Code : M101 Teacher Name: Amalendu Singha Mahapatra Lecture 5: Objective: The Mean Value Theorem Chapter - 3We begin with a common-sense geometrical fact: somewhere between two zeros of a non-constant continuous function f, the function must change directionFor a differentiable function, the derivative is 0 at the point where f changes direction. Thus, we expect there to be a point c where the tangent is horizontal. TTRANSCRIPT
Study Materials for 1st Year B.Tech StudentsPaper Name: Mathematics
Paper Code : M101Teacher Name: Amalendu Singha Mahapatra
The Mean Value Theorem Lecture 5: Chapter - 3 Objective: We begin with a common-sense geometrical fact: somewhere between two zeros of a non-constant continuous function f, the function must change direction
For a differentiable function, the derivative is 0 at the point where f changes direction. Thus, we expect there to be a point c where the tangent is horizontal. These ideas are precisely stated by Rolle's Theorem:
Rolle's Theorem
Statement:Let f be differentiable on (a,b) and continuous on [a,b]. If f(a) = f(b) = 0, then there is at least one point c in (a,b) for which f(c) = 0.
Notice that both conditions on f are necessary. Without either one, the statement is false!
(1) For a discontinuous function,the conclusion of Rolle's Theorem may not hold:
(2) For a continuous, non-differentiable function,again this might not be the case:
Though the theorem seems logical, we cannot be sure that it is always true without a proof.
Proof of Rolle's Theorem
Note that either f(x) is always 0 on [a,b] or f varies on [a,b].
If f(x) is always 0, then f(x) = 0 for all x in [a,b] and we are done. If f(x) varies on [a,b], then there must be points where f(x)0 or points where
f(x)0.
Assume first that there are points where f(x)0. By the Extreme Value Theorem f has a maximum at some point c in [a,b]. Then f(c)0, so c is not an endpoint. At this maximum, f(c) = 0.
Now assume that there are points where f(x)0. Then, again by the Extreme Value Theorem, f has a minimum at some point c in [a,b]. Again, c is not an endpoint since f(c)0 while f(a) = f(b) = 0. At this minimum, f(c) = 0.
This completes the proof.
Working Rule for Verifying Rolle's Theorem
Let f (x) be a function defined on [a, b].
Step 1: Show that the function is continuous in the given interval. Some known standard functions which are continuous, can be mentioned directly.
Step 2: Differentiate f (x) and examine if f '(x) is defined at every point in the open interval (a, b).
Step 3: Check if f (a) = f (b), If all the above condition are satisfied, then Rolle's theorem is applicable else the Rolle's theorem is not applicable.
If Rolle's theorem is applicable, solve f '(c) = 0. Show that one of these roots lie in the open interval (a, b).
Geometrical meaning
Let A (a,f (a)) and B (b,f (b)) be two points on the graph of f (x) such that f(a) = f(b), then c (a, b) such that the tangent at P(c, f(c)) is parallel to x - axis
Worked out problems:
Example1: Verify Rolle's theorem for the function f (x) = x2 – 8x + 12 on (2, 6).
Since a polynomial function is continuous and differentiable everywhere f (x) is differentiable and continuous (i) and (ii) conditions of Rolle's theorem is satisfied.
f (2) = 22 – 8 (2) + 12 = 0
f (6) = 36 - 48 + 12 = 0
Therefore (iii) condition is satisfied.
Rolle's theorem is applicable for the given function f (x).
There must exist c (2, 6) such that f '(c) = 0
f '(x) = 2x – 8
Rolle's theorem is verified.
Example2: Verify Rolle's theorem for in [-1,1]. WBUT 03
The function is continuous in [-1,1] and f(-1) = f(1) = 1. But has no
derivative at x = 0 belongs to (-1,1).Hence Rolle's theorem is not applicable for this function.
Example3: Verify Rolle's theorem for the function f(x) = tanx in [0,π].
The function f(x) = tanx is not continuous throughout the interval [0,π]. Since tanx
as and .
Hence Rolle's theorem is not applicable for this function
Assignment:
(1)Verify Rolle's theorem for the following function
(a)f(x) = x tanx in [0, π].
(b) in [0, ].
(c) in [-3,0].
Objective type questions:
(1) The function satisfies Rolle's theorem in the interval [0,2]
(a) Yes (b) no
(2) The curve represented by the parabola = 4y follows Rolle's theorem in [-1,1]
(a) Yes (b) no
Lecture 6:
The Mean Value Theorem is one of the most important theoretical tools in Calculus. It states that if f(x) is defined and continuous on the interval [a,b] and differentiable on (a,b), then there is at least one number c in the interval (a,b) (that is a < c < b) such that
The special case, when f(a) = f(b) is known as Rolle's Theorem. In this case, we have f '(c) =0. In other words, there exists a point in the interval (a,b) which has a horizontal tangent. In fact, the Mean Value Theorem can be stated also in terms of slopes. Indeed, the number
is the slope of the line passing through (a,f(a)) and (b,f(b)). So the conclusion of the Mean Value Theorem states that there exists a point c (a,b)such that the tangent line is parallel to the line passing through (a,f(a)) and (b,f(b)). (see Picture)
Example1. Let , a = -1and b=1. We have
On the other hand, for any c (-1,1), not equal to 0, we have
So the equation
does not have a solution in c. This does not contradict the Mean Value Theorem, since f(x) is not even continuous on [-1,1].
Remark. It is clear that the derivative of a constant function is 0. But you may wonder whether a function with derivative zero is constant. The answer is yes. Indeed, let f(x) be a differentiable function on an interval I, with f '(x) =0, for every . Then for any a and b in I, the Mean Value Theorem implies
for some c between a and b. So our assumption implies
Thus f(b) = f(a) for any aand b in I, which means that f(x) is constant
Some important Results Follows from Lagrange’s Theorem.
Theorem1: If f(x) is differentiable in [a,b] and for all then f(x) is constant in [a,b].
Theorem2: If f:[a,b]→R be such that f(x) is continuous in [a,b] and f '(x) ≥0 in (a,b) then f(x) is non-decreasing in [a,b].
Assignment:
(1) Applying Lagrange’s mean value Theorem prove that
for all x>0
(2)Verify Lagrange’s mean value Theorem theorem for the following function
(a) g(x) = x(x-1)(x-3) in [1,5]
(b) f(x) = in [a -1,a + 1] . a>0
(3)Show by M.V.T
.
Objective type questions:
(a) Lagrange’s M.V.T is applicable on the function on [2,4]
(1) Yes (2) no.
(b) The function f(x) = log (x + 1) obeys L.M.V.T in [1,5]
(1) Yes (2) no.