Transcript Slide 1

5.2…and...5.3
All mixed together….
Rolles Theorem
If f is continuous on [a,b] and
differentiable on (a,b) and if f(a)=f(b),
then f ‘(c)=0 for at least one number c
in (a,b).
Mean Value Theorem (MVT)
If y = f(x) is continuous on [a,b] and
differentiable on (a,b), then there is
at least one point c in (a,b) such that
f (b)  f (a)
f ' (c ) 
ba
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
b
x

Example Explore the Mean
Value Theorem
Show that the function f ( x)  x satisfies the hypothesis of the Mean Value
2
Theorem on the interval  0,2. Then find a solution c to the equation
f '(c) 
f (b) - f (a)
on this interval.
b-a
The function f ( x)  x is continuous on  0,2
2
and differentiable on  0,2  .
Since f (0)  0, f (2)  4, and f '( x)  2 x
f (b) - f (a )
b-a
40
2c 
20
2c  2
f '(c) 
c  1.
Inc / Dec Functions
Example Determining Where
Graphs Rise or Fall
Where is the function f ( x)  2 x  12 x increasing and where is it decreasing?
3
Example
First Derivative Test for Local Extrema
The following test applies to a continuous function f(x).
At a critical point c :
1. If f ' changes sign from positive to negative at c, then f
has a local maximum value at c.
2. If f ' changes sign from negative to positive at c, then f
has a local minimum value at c.
3. If f ' does not change sign at c, then f has no local
extreme value at c.
At a left endpoint a :
If f '  0 ( f '  0) for x  a, then f has a local maximum (minimum)
value at a.
At a right endpoint b :
If f '  0 ( f '  0) for x  b, then f has a local minimum (maximum)
value at b.
First Derivative Test for Local
Extrema
First Dx Test
Concavity
Concavity Test
Inflection Point
A point where the graph of a function has a
tangent line and where the concavity
changes is a point of inflection.
Concavity
2nd Dx Test
Example Using the Second
Derivative Test
Find the local extreme values of f ( x)  x  6 x  5.
3
First derivative:
y is positive
Curve is rising.
y is negative
Curve is falling.
y is zero
Possible local maximum or
minimum.
Second derivative:
y is positive
Curve is concave up.
y is negative
Curve is concave down.
y is zero
Possible inflection point
(where concavity changes).

Learning about Functions from
Derivatives
Example:
Graph
y  x  3x  4   x  1 x  2 
3
2
2
Make a summary table:
x
y
y
y
1
0
9
12
0
4
0
6
1
2
3
0
falling, inflection point
2
0
0
6
local min
3
4
9
12
rising, concave down
local max
rising, concave up
5
4
3
2
1
0
-2
-1
0
-1
1
2
3
4
p
Antiderivatives are never
unique.
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.

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

Example 6:
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 .
Example 6:
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

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
9.8 2
s t  
t t C
2
s t   4.9t  t  C
2
The initial position is zero at time zero.
1 C
0  4.9  0   0  C
v t   9.8t 1
0C
2
s t   4.9t  t
2
p