Differentiation

Mean Value Theorem for

Derivatives

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.

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

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:

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

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

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.

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.

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.)

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.

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

Differentiation

Mean Value TheoremASSESSMENT

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:

Differentiation

Solution 2:1

c 23

Solution 3:

Solution 4:

1c

3

ln (e - 1)

SOLUTIONS

Differentiation

Multiple Choice Question

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?

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).

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.

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.

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.]

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)

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.

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.