mean value theorem for derivatives
DESCRIPTION
Mean Value Theorem for Derivatives. If f ( x ) is continuous over [ a , b ] and differentiable over ( a , b ), then at some point c between a and b :. The Mean Value Theorem says that at some point in the closed interval, the actual slope equals the average slope. - PowerPoint PPT PresentationTRANSCRIPT
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Differentiation
Mean Value Theorem for
Derivatives
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Differentiation
If f (x) is continuous over [a,b] and differentiable over (a,b),
then at some point c between a and b:
f b f af c
b a
Mean Value Theorem for Derivatives
The Mean Value Theorem only applies over a closed interval.
The Mean Value Theorem says that at some point in the closed interval, the actual slope equals the average slope.
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Differentiation
y
x0
A
B
a b
Slope of chord:
f b f a
b a
Slope of tangent:
f c
y f x
Tangent parallel to chord.
c
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Differentiation
A function is increasing over an interval if the derivative is always positive.
A function is decreasing over an interval if the derivative is always negative.
A couple of somewhat obvious definitions:
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Differentiationy
x0
y f x
y g x
These two functions have the
same slope at any value of x.
Functions with the same derivative differ by a constant.
C
CC
C
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Differentiation
Example 6:
Find the function whose derivative is and whose graph passes through .
f x sin x 0,2
cos sind
x xdx
cos sind
x xdx
so:
f x could be cos x or could vary by some constant .C
cosf x x C
2 cos 0 C
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DifferentiationExample 6:
Find the function whose derivative is and whose graph passes through .
f x sin x 0,2
cos sind
x xdx
cos sind
x xdx
so:
cosf x x C
2 cos 0 C
2 1 C 3 C
cos 3f x x Notice that we had to have initial values to determine the value of
C.
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Differentiation
The process of finding the original function from the derivative is so important that it has a name:
Antiderivative
A function is an anti-derivative of a function
if for all x in the domain of f. The process of
finding an anti-derivative is anti-differentiation.
F x f x
F x f x
You will hear much more about antiderivatives in the future.
This section is just an introduction.
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Differentiation
Example 7b: Find the velocity and position equations for a downward acceleration of 9.8 m/sec2 and an initial velocity of 1 m/sec downward.
9.8a t
9.8 1v t t
1 9.8 0 C
1 C
9.8v t t C
(We let down be positive.)
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Differentiation
Since velocity is the derivative of position, position must be the anti-derivative of velocity.
Example 7b: Find the velocity and position equations for a downward acceleration of 9.8 m/sec2 and an initial velocity of 1 m/sec downward.
9.8a t
9.8 1v t t
1 9.8 0 C
1 C
9.8v t t C 29.8
2s t t t C
The power rule in reverse: Increase the exponent by one and multiply by the reciprocal of the new exponent.
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Differentiation
Example 7b: Find the velocity and position equations for a downward acceleration of 9.8 m/sec2 and an initial velocity of 1 m/sec downward.
9.8a t
9.8 1v t t
1 9.8 0 C
1 C
9.8v t t C 29.8
2s t t t C
24.9s t t t C The initial position is zero at time zero.
20 4.9 0 0 C 0 C
24.9s t t t p
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Differentiation
Mean Value TheoremASSESSMENT
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Differentiation
State and proov Mean Value Theorem.Que 1:
Verify Mean Value Theorem for the function f(x) = -2x 3 + 6x – 2 in the interval [– 2 , 2] and find the value of c .Find the value of c that satisfies the conclusion
of the mean value theorem for f (x) = x3 -
2x2 - x + 3 on[0, 1].Find the value of c that satisfies the conclusion
of the mean value theorem for the function f (x)
= ln x in[1, e].
Que 2:
Que 3:
Que 4:
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Differentiation
Solution 2:1
c 23
Solution 3:
Solution 4:
1c
3
ln (e - 1)
SOLUTIONS
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Differentiation
Multiple Choice Question
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DifferentiationQue 1: We verify the "Mean Value theorem" for a
function f (x):
Que 2: State whether the function f (x) = 3x2 - 2 on [2, 3]satisfies the mean value theorem.
Que 3: Which of the following functions satisfies the mean value theorem?
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DifferentiationQue 4. Determine the point on the parabola f (x) = (x - 2)2, at which the tangent is parallel to the chord joining the points (2,
0) & (3, 1).
Que 5. Find the point on the parabola y = (x + 3)2, at which the tangent is parallel to the chord of the parabola joining the points (- 3, 0) & (- 4, 1).
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Differentiation
Ques 7. State whether the function f (x) = sin x - sin
2x, x ∈ [0, π] satisfies the mean value theorem.
Que 6. State whether the function f(x) = ln x on[1,
2] satisfies the mean value theorem.
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Differentiation
SolutionsAns 1.d. If f(x) is continuous on [a, b] and is differentiable on
(a, b), then only it follows the hypotheses of mean value theorem for which the mean value theorem can be verified
Ans 2. b. f (x) = 3x2 - 2, x ∈ [2, 3][Given function.]The given function is continuous on [2, 3] and differentiable in (2, 3). So, there exists c ∈ (2, 3) such that f ′ (c)
=f(3) - f(2)3 - 2. [By mean value Theorem.] f ′(c) = 15.[f ′(c) = 6c. ]
6c = 15 ⇒ c = 5 / 2 ∈ (2, 3).Hence, the given function satisfies the mean value theorem.
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Differentiation
Ans 3.c. f (x) = x2 - 6, the function f (x) is continuous on [7, 8] and differentiable on (7, 8) [Consider the choice A.]
f ′ (c ) = f (8) - f (7)8 - 7 = 15
[Try to get c ∈ (7, 8) such that f ′ (c ) =f (b) - f (a)b - a]
2c = 15[f ′ (x) = 2x]
c = 15 / 2 ∈ [7, 8]
[There exists c ∈ (7, 8) such that f ′ (c ) =f (b) - f (a)b - a]
So, the function in the choices A satisfies the mean value theorem
f (x) = log x, x ∈ [- 7, 8] is not continuous in ∈ [- 7, 8] and hence does not satisfy the mean value theorem.[Consider the choice B.]
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Differentiation
Ans 4.
f (x) = [x], x ∈ [- 7, 7] is not continuous in [- 7, 7] and hence does not satisfy the mean value thoerem
[Consider the choice C.]f (x) = |x|, x ∈ [- 8, 8] is continuous in [- 8, 8] but not differentiable in (- 8, 8) and hence does not satisfy the mean value theorem .[Consider the choice D.]a. Slope of the chord joining (2, 0) & (3, 1) = 1-0 / 3-2 = 1Slope of the tangent to the curve at any point (x, f(x)) is f ′ (x) = 2(x - 2)f ′ (c) = 1[By Mean Value Theorem.] 2(c - 2) = 1
c = 5 / 2 ∈ (2, 3)
f (c) = (c - 2)2 = (5 / 2 - 2)2 = 1 / 4Hence, the point where the tangent to the parabola is
parallel to the given chord is (c, f (c)) = (5 / 2, 1 /4)
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DifferentiationAns 5.d.
Slope of the chord joining the points (- 3, 0) & (- 4, 1) = (1 - 0)( - 4 + 3) = - 1 Slope of the tangent to the curve at any point (x, f(x)) is f ′ (x ) = 2(x + 3)f ′(c) = -1[By mean value theorem.]2(c + 3) = -1
c = - 7 / 2 ∈ (-4, - 3)
f (c) = (c + 3)2 = (- 72 + 3)2 = 14.
Hence, the point where the tangent to the parabola is parallel to
the given chord is (c, f(c)) = (- 7 / 2, 1 /4). Ans 6.
a. f(x) = ln x, x ∈ [1, 2][Given function.]f(x) is continuous on [1, 2] and differentiable on (1, 2).So, there exists c ∈ (1, 2) such that f ′ (c) =f(2) - f(1)2 - 1.[By mean value Theorem.]1c = ln 2 ⇒ c = 1ln 2 ∈ (1, 2).[f ′(x) = 1x]Hence, the given function satisfies the mean value theorem.
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Differentiation
Ans 7.a. f (x) = sin x - sin 2x, x∈ [0, π][Given function.]The given function is continuous on [0, π] and differentiable in (0, π).So, there exists c ∈ (0, π) such that f ′ (c) =f(π) - f(0)π - 0.[By mean value theorem.]cos c - 2 cos 2 c = 0[f ′ (x) = cos x - 2 cos 2x.]cos c = 4 cos2c - 2[cos 2c = 2 cos2c - 1.] 4 cos2 c - cos c - 2 = 0c = 32o.53′ , 126o.37′ ∈ (0, π) )[Solve use the calculator.]Hence, the function f (x) satisfies the mean value theorem.