Study Materials for 1 st Year B.Tech Students Paper Name: Mathematics Paper Code : M101 Teacher Name: Amalendu Singha Mahapatra
Lecture 5: Objective:
The Mean Value Theorem Chapter - 3
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 point c in in (a,b) (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) f(x) is alway alwayss 0, then then f(x) = 0 for all all x in [a,b] [a,b] and we we are done. done. If f(x) f(x) varies varies on [a,b], [a,b], then then there there must be be points points where f(x) f(x)0 or points where f(x)0. 0. Assume Assume first first that that there there are are points points where f(x) f(x) 0. 0 . By the Extreme Value Theorem f has a maximum maximum at some some point point c in in [a,b]. [a,b]. Then f(c)0, so c is not an endpoint. At this this maxi maximu mum, m, f(c) (c) = 0. Now assume assume that that there there are are points points where f(x) f(x) 0. 0 . Then, again by the Extreme Value Theorem, f has a minimum at some point c in [a,b]. Again, c is not an endpo endpoin intt sin since ce f(c) f(c) 0 while while f(a) f(a) = f(b) f(b) = 0. At this minimum, 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 no t 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 tangent at P(c, f(c)) is parallel 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) = 2 2 – 8 (2) + 12 = 0 f (6) = 36 - 48 + 12 = 0 ⇒ f
(2) = f (3)
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 ⇒
f ' (c ) = 2c − 8 ⇒ 2c − 8 = 0 ⇒ c = 4 ∈ (4, 6)
Rolle's theorem is verified. Example2: Verify Rolle's theorem for f ( x ) = x in [-1,1].
The function
WBUT 03
is continuous in [-1,1] and f(-1) = f(1) = 1. But f ( x ) = x has no
x
derivative at x = 0 belongs to (-1,1).Hence Rolle's theorem is not app licable 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 x →
π
2
and
π ε
2
→∞
[ 0, π ] .
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]. 2 2 (b) f ( x) = sin x cos x in [0,
π
2
].
1
(c) f ( x) = x( x + 3)c − 2 x in [-3,0]. Objective type questions:
(1) The functi function on f ( x) = x − 1 satisfies Rolle's theorem in the interval [0,2] (a) Yes
(b) no
(2) The curve represented represented by the the parabola x 2 = 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 f ( x) x) is defined and continuous on the interval [a states that if f [a,b] and differentiable on (a,b), then there is at least one number c number c in the interval (a (a,b) (that is a < c < b) such that
The special case, when f when f (a) = f = f (b) is known as Rolle's Theorem. In this case, we have f have f '(c '(c) =0. In other words, there exists a point in the interval (a ( 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 (a f , f (a)) and (b (b f , f (b)). So the conclusion of the Mean Value Theorem states that there exists a point c (a,b)such that the tangent line line is parallel to the line passing through (a (a f , f (a)) and (b (b f , f (b)). (see Picture)
Example1. Let f ( x) =
1 x
, 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) 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 x) be whether a function with derivative zero is constant. The answer is yes. Indeed, let f let f ( x) x) =0, for every xε I . Then for any a and a differentiable function on an interval I interval I , with f with f '( '( x) b in I in I , the Mean Value Theorem implies
for some c between a and b. So our assumption implies
x) is constant Thus f Thus f (b) = f = f (a) for any aand b in I in I , which means that f that f ( x) Some important Results Follows from Lagrange’s Theorem. ' Theorem1: If f(x) is differentiable in [a,b] and f ( x ) = 0 for all xε [ a, b] then f(x) is constant in [a,b].
Theorem2: If f:[a,b]→R be 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 Applying Lagrange’s Lagrange’s mean value value Theorem prove that x 1 + x
≤
log(1 + x ) < x for all x>0
(2) Verify Lagrange’s mean value Theorem theorem for the following function
(a) g(x) g(x) = x(x-1) x(x-1)(x(x-3) 3) in [1,5] [1,5] (b) (b) f(x) (x) = x − a in [a -1,a + 1] . a>0 (3) Show by M.V.T
1 4
<
4 1 log( ) < . 3 3
Objective type questions: 3 (a) Lagrange’s Lagrange’s M.V.T M.V.T is applicable applicable on the function function f ( x) = x on [2,4]
(1) Yes
(2)
no.
(b) The function f(x) = log log (x + 1) obeys L.M.V.T L.M.V.T in [1,5] (1) Yes
(2)
no.
