Mean Value Theorem for Derivatives
If f (x) is a differentiable function over [a,b], then
at some point between a and b:
f b  f  a 
ba
 f c
Mean Value Theorem for Derivatives
If f (x) is a differentiable function over [a,b], then
at some point between a and b:
f b  f  a 
ba
 f c
Differentiable implies that the function is also continuous.
Mean Value Theorem for Derivatives
If f (x) is a differentiable function over [a,b], then
at some point between a and b:
f b  f  a 
ba
 f c
Differentiable implies that the function is also continuous.
The Mean Value Theorem only applies over a closed interval.

Mean Value Theorem for Derivatives
If f (x) is a differentiable function over [a,b], then
at some point between a and b:
f b  f  a 
ba
 f c
The Mean Value Theorem says that at some point
in the closed interval, the actual slope equals the
average slope.

Tangent parallel
to chord.
y
Slope of tangent:
f  c
B
Slope of chord:
f b   f  a 
ba
A
0
y  f  x
a
c
x
b

A couple of somewhat obvious definitions:
A function is increasing over an interval if the derivative
is always positive.
f   0  f(x) is increasing
A function is decreasing over an interval if the derivative
is always negative.
f   0  f(x) is decreasing

f(x) = x3 - 4x2 + 5
f‘(x) = 3x2 – 8x
What relationship is there between
this graph of a function and its
derivative?
Pg. 192, #1
Find any maximums or minimums. And the
intervals on which the function is increasing
and decreasing.
f(x) = 5x – x2
#15
Find each value of c in (a, b) that satisfies the
Mean Value Theorem for Derivatives.
f(x) = x2 + 2x – 1 , D:[0, 1]
#15, f(x) =
2
x
+ 2x – 1 , D:[0,1]
y
C
These two functions have the
same slope at any value of x.
Functions with the same derivative
differ by a constant.
y  g  x
x
0
y  f  x

Find the function f  x  whose derivative is sin  x  and
whose graph passes through  0, 2  .
d
cos  x    sin  x 
dx
d
so:
 cos  x   sin  x 
dx
 f  x    cos  x   C
2   cos  0  C
f  x  could be  cos  x  or could vary by some constant C .
Find the function f  x  whose derivative is sin  x  and
whose graph passes through  0, 2  .
d
cos  x    sin  x 
dx
d
so:
 cos  x   sin  x 
dx
Notice that we had to have
initial values to determine
the value of C.
 f  x    cos  x   C
2   cos  0  C
2  1  C
3C
f  x    cos  x   3

The process of finding the original function from the
derivative is so important that it has a name:
Antiderivative
A function F  x  is an antiderivative of a function f  x 
if F   x   f  x  for all x in the domain of f. The process
of finding an antiderivative is antidifferentiation.
You will hear much more about antiderivatives in the future.
This section is just an introduction.

Antiderivative
• The power rule in reverse
– 1. Increase the exponent by 1
– 2. Multiply by the reciprocal of the new
exponent
f(x) =
f‘(x) = 3x2 - 4
Example
Find the equation of a function whose
derivative is f’(x) = x2 + 2x + 1 and passes
through the point (1, 3).
Example (2)
Find the equation of the function whose
derivative is f’(x) = 3x3 – 2x + 1 and passes
through the point (2, -1).
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.
a  t   9.8
(We let down be positive.)
vSince
9.8t  C is the derivative of velocity,
t  acceleration
velocity must be the antiderivative of acceleration.
1  9.8  0  C
1 C
v  t   9.8t  1
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.
a  t   9.8
v t   9.8t  C
1  9.8  0  C
1 C
v  t   9.8t  1
9.8 2
s t  
t t C
2
The power rule in reverse:
Increase the exponent by one and
multiply by the reciprocal of the
new exponent.
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.
a  t   9.8
v t   9.8t  C
1  9.8  0  C
1 C
v  t   9.8t  1
9.8 2
s t  
t t C
2
s  t   4.9t  t  C
2
The initial position is zero at time zero.
0  4.9  0   0  C
2
0C
2
s  t   4.9t  t