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 )
ba
Tangent parallel
to chord.
y
Slope of tangent:
f c
B
Slope of chord:
f b f a
ba
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
40
2c
20
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
3C
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
0C
2
s t 4.9t t
2
p