Mean Value Theorem for Derivatives4.2 Teddy Roosevelt National Park, North Dakota Photo by Vickie...

Mean Value Theorem for Derivatives4.2

Teddy Roosevelt National Park, North Dakota

Photo by Vickie Kelly, 2002Created by Greg Kelly, Hanford High School, Richland, WashingtonRevised by Terry Luskin, Dover-Sherborn HS, Dover, Massachusetts

Mean Value Theorem for DerivativesIf f (x) is a continuous function over [a, b] and

differentiable over (a, b),

then at some point c between a and b:

f b f af

bc

a

Mean Value Theorem for Derivatives

The Mean Value Theorem only applies toa continuous function

over a closed interval…

If f (x) is a continuous function over [a, b] and

differentiable over (a, b),

then at some point c between a and b:

f b f af

bc

a

Mean Value Theorem for Derivatives

The Mean Value Theorem only applies toa well-behaved function that is also

differentiable in the interior of the interval.

If f (x) is a continuous function over [a, b] and

differentiable over (a, b),

then at some point c between a and b:

f b f af

bc

a

Mean Value Theorem for Derivatives

The Mean Value Theorem says that the average slope across an interval equals

the instantaneous slope at a point somewhere on the closed interval (where x=c).

y

x0

( )f a

( )f b

a b

Average slope of chord

from a to b:

f b f a

b a

Slope of tangent AT c: f c

y f x

Tangent parallel to chord.

c

If the derivative of a function is always positive over an interval, then the function is increasing there.

If the derivative of a function is always negative over an interval, then the function is decreasing there.

A couple of definitions:

Note the IF and the THEN…. they are NOT interchangeable!!!

Trying to swap these is a typical misunderstanding!!!

y

x0

y f x

y g x

These two functions have the same slope at each value of x, but they are vertical translations of each other.

Functions with the same derivative differ from each other by a constant.

vertical

translation

Example 6:

Find the function whose derivative is sin(x),and whose graph passes through .

f x 0,2

cos sind

x xdx

cos( )d

xdx

So f x could be cos x , or could vary by a constant .C

cosf x x C

cos 02 C 2 1 C

Recognize that we need to have an initial value to determine the precise value of C!

cos 3f x x

3 C

( sin( ))

sin( )

x

x

The process of finding the original function by working backward from the derivative is so pivotal that it has a name:

Antiderivative

A function is an antiderivative of a function

if for all x in the domain of f. The process

of finding an antiderivative is called antidifferentiation.

F x f x

F x f x

You will hear much more about antiderivatives in the future;

this section is just an introduction!

Since acceleration is the derivative of velocity, velocity must be the antiderivative of acceleration.

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; s(0) = -3.

9.8a t

9.8 1v t t

1 9.8 0 C 1 C

19.8

1

tv t C

(Let “down” mean a negative value.)

The power rule is now running in reverse: increase the exponent by one, then divide by the new exponent. 9.8v t t C

Since velocity is the derivative of position, position must be the antiderivative 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; s(0) = -3.

9.8a t

9.8 1v t t

2 19.8 1

2 1

t ts t C

The power rule running in reverse: increase the exponent by one, then divide by the new exponent.

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; s(0) = -3.

9.8a t

9.8 1v t t

1 9.8 0 C 1 C

9.8v t t C 29.8

12

s t t t C

24.9 1s t t t C The initial position is -3 at time zero. 2

3 4.9 0 1(0) C

3 C

24.9 1 3s t t t