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