12.1 First Derivative and Graph

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12.1 First Derivative and
Graph
Example 1: Estimate the derivative graphically
To find the
derivative
at a certain
point
graphically,
draw the
tangent line
at that
point, then
find the
slope of the
tangent line
At x = -3, f’(-3) = -3/-2 = 3/2 (down 3, left 2)- or use slope formula
At x = -2.5, f’ (-2.5) = 0 slope of a horizontal line
At x = 1, f’ (1) = -1 (approximate)
At x = 2.5, f’(2.5) = ½ (approximate)
Example 2: Find the derivative f(x) = x2 + 1
Since f’(x) = 2x therefore
f’(-2) = -4
f’(-1) = -2
f’(0) = 0
f’(1) = 2
f’(2) = 4
F(x)
What can you conclude about the
relationship between the graph of
f(x) and the sign of f’(x)?
F’(x)
• What can you conclude about the relationship
between the graph of f(x) and the sign of f’(x)
• Note:
- If f is increasing: f’(x) > 0
- If f is decreasing: f’(x) < 0
Example 3
Given the function f(x) = x2 - 6x + 10
A) Which values of x correspond to a horizontal tangent line (hint: let f’ = 0).
f’(x) = 2x – 6 = 0
x =3
B) For which values of x is f(x) increasing? Decreasing?
To do this, we have to locate all points where f’=0 or f’ is discontinuous.
* From part A, we found out that x = 3 for f’(x)=0.
* Next, test some numbers on the left of 3 and on the right of 3 (or look at
the graph of f(x) to find the interval where f is increasing or decreasing).
x
1
2
f’(x) -4 -2
3 4
5
0 2
4
Left of 3
Right of 3
3
Therefore, f is decreasing on (-∞, 3)
and increasing on (3, ∞)
-------0 +++++
C) Sketch a graph of f = x2 - 6x + 10
X
F(x) =x2 - 6x + 10
1
5
2
2
3
1 important point
4
2
5
5
Definition: Critical values
The values of x in the domain of f where f’(x)=0 or where f’(x)
does not exist are called the critical values of f. The critical
values of f are always in the domain of f.
Thus, to find the critical values, set f’(x) =0, then solve for x. Also, find the
points of the discontinuity. If these values are in the domain of f(x), then
they are critical points.
To find the increasing interval or decreasing interval, find the derivatives of
the points on the left and the right of these critical points. If f’(a) is negative,
then that interval is decreasing. If f’(a) is positive, then that interval is
increasing.
Partition Numbers and
Critical Values
A partition number for the sign chart is a place where the
derivative could change sign. Assuming that f ’ is continuous
wherever it is defined, this can only happen where f itself is not
defined, where f ’ is not defined, or where f ’ is zero.
Insight: All critical values are also partition numbers, but there
may be partition numbers that are not critical values (where f
itself is not defined).
If f is a polynomial, critical values and partition numbers are
both the same, namely the solutions of f ’(x) = 0.
Example 4
Find the critical values of f, the intervals on which f is
increasing, and those on which f is decreasing for f(x) = 1-x3
Domain of f(x) is (-∞,∞)
F’(x) = -3x2 = 0
x=0
Since 0 is in the domain of f, x =0 is the critical value for f
x
-2
f’(x)
-12 -3
_
-1
_
0 1
2
0 -3
-12
_
_
Therefore f is decreasing on (- ∞, 0) and (0, ∞). Since
f is continuous at 0, f(x) is decreasing for all x
Example 5
Find the critical values of f, the intervals on which f is increasing,
and those on which f is decreasing for f(x) = (1+x)1/3
Domain of f(x) is (-∞, ∞)
2

1
F’(x) =
(1  x) 3 
3
f ' ( x)  0
1
3(1  x)
2
3
1
0
2/3
3(1  x)
1 0
F’(x) is continuous for all x except x = -1. Since -1 is in the domain of
F(x) so -1 is the critical value of f(x)
Therefore f is increasing on
x
-3
-2
-1
0
1
f’(x) .21 .33 ND .33 .21
+
+
+
+
(- ∞, -1) and (-1, ∞). Since f is
continuous at -1, f(x) is
increasing for all x
Example 6
Find the critical values of f, the intervals on which f is increasing,
and those on which f is decreasing for f(x) = 1/x
Domain of f(x): (-∞, 0) U (0, ∞) because f(x) is undefined at 0
1
 x 1
x
1
F ' ( x)  1x  2  2
x
1
F ' ( x)  0
0
2
x
F ( x) 
1  0
F’(x) is continuous for all x except x = 0. Since 0 is not in the domain of
F(x) so 0 is not the critical value of f(x). There is no critical value for this function.
x
-1
f’(x) -1
__
0
1
Therefore f is decreasing on
ND -1
__
(- ∞, 0) and (0, ∞)
Example 7
Find the critical values of f, the intervals on which f is increasing, and those
on which f is decreasing for f(x)= 5lnx - x
Domain: (0, ∞) or x >0
F’(x) =
5
5 x
1 
x
x
5 x
0
x
x5
5 is in the domain of f(x) so 5 is the critical value. Also, f’(x) is discontinuous at 0,
but 0 is not in the domain of f(x) so 0 is not the critical value.
x
4
5 6
f’(x) 1/4 0 -1/6
+++
___
Therefore f is increasing on
(0, 5) and decreasing on (5, ∞)
Local Extrema
When the graph of a continuous function changes from
rising to falling, a high point, or a local maximum, occurs,
and when the graph changes from falling to rising, a low
point, or a local minimum, occurs.
Local
maximum
Local
minimum
Theorem 2: Existence of Local Extrema:
If f is continuous on the interval (a,b) and f(x)
is a local extremum, then either f’(c)=0, or
f’(c) does not exist.
How to find local extrema:
- Find all critical values for f
- Test each critical value to see if it produces a
local max, a local min, or neither..
From your textbook
From your textbook
Example 8: Which graph is the
derivative?
increasing
decreasing
Answer: Graph B
decreasing
Above 0
Below 0
GRAPH A
GRAPH B
Below 0
Example 9
Given f(x) = x3 – 9x2 + 24x -10
A)
Find the critical values of f(x)
F’(x) = 3x2 – 18x + 24 = 0
3 (x2 – 6x + 8) = 0
3(x-2)(x-4)
=0
So x = 2 or 4, both are critical values because they are in the domain of f(x)
B) Find the local maxima and minima
We need to test on the left and on the right of 2 and 4
x
1
2
3
4
5
f’(x) +
0
--
0
+
increasing
decreasing
increasing
Therefore there is a local
maximum at 2 and a local
minimum at 4.
Note: use your G.C to check by
looking at the graph
F(x)
Sketch a possible graph of f’(x)
Answer:
start with the critical points (min,
max): -6,2,and 9, then see if f(x) is
increasing (f’x >0) or decreasing (f’x
<0)
F’(x)
F’(x)
Sketch a possible graph of y = f(x)
Answer:
start with the zeros of the
function: -7 and 1, then see if f’(x)
is above 0 (f is increasing) or f’x is
below 0 (f is decreasing).
F(x)
Polynomial Functions
Theorem 3. If
f (x) = an xn + an-1 xn-1 + … + a1 x + a0, an  0,
is an nth-degree polynomial, then f has at most n x-intercepts
and at most (n – 1) local extrema.
The derivative is also an important tool for hand-sketching
graphs and for analyzing graphs and discussing the interplay
between a function and its rate of change.
Example 10
The given graph approximates the rate
of change of the U.S. share of the total
world production of motor vehicles over
a 20- year period, where S(t) is the
percentage of share and t is the number
of years.
A) Write a verbal description of the graph
Note that this is the graph of S’(t), not
S(t). The U.S. share decreases for the
first 6 years, increases for the next 10
years, then decreases again for the last
4 years. (Local min at 6 and local max at
16)
B) Sketch a possible graph of y = S(t)
Example 11
The graph of total revenue R(x) (in
dollars) from the sale of x desks is
given.
A) Write a brief verbal description of
the graph of the marginal revenue
function R’
The graph R indicates that the revenue
is increasing on (0, 450) with the
maximum at 450 and decreasing on
(450, 1000). Thus, the marginal
revenue must be positive on (0, 450),
equal 0 at 450 and negative on
(450, 1000).
B) Sketch a possible graph of R’(x)
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