Download 2.9 Notes Increasing, Decreasing and The First Derivative

2.9 Increasing, Decreasing, The First Derivative Test
Y=f(x)
Open Intervals x values
Increasing: (A,B) tangent line has a positive slope
Decreasing: (B,C) tangent line has a negative slope
Constant: (C,D) tangent line has a zero slope
Increasing: (D,E) tangent line has a positive slope
Derivative =
0
+
+
+
-
+
0
0
+
+
A
Example:
1. Find the intervals on which
B
C
D
E
f  x   x2  4 x  3 is increasing and decreasing.
Work:
First find the critical numbers
f '  x   2x  4
0  2x  4
x2
Second create a table of values based on the critical numbers you find from the first derivative.
 , 2
 2, 
Test #
1
3
Sign f’
--
+
---
+
+
+
+
-Conclusions f is decreasing
Regarding
f
f is increasing
Finally list any relative max or min for f(x):
f(x) has a relative min at (2,-1)*
---
* (2,f(2))
+
This is the graph of
y  f '  x  for some function y  f  x 
Where does f have a relative max?
Where does f have a relative min?
2
* Since f’ goes from positive (about the x-axis) to negative (below the
x-axis) at x=-3 then f has a relative max at -3.
* Since f’ goes from negative (below the x-axis) to positive (above the
x-axis) at x=1 then f has a relative min at -1.
1
-3
-2
-1
1
-2
*The First Derivative Test:
Let C be a critical number where f is continuous on some open interval containing c.
If f’(x) changes sign from + to – at c then f has a relative maximum at c.
If f’(x) changes sign from – to + at c then f has a relative minimum at c.
Example 1:
Find the intervals on which
f  x   x3 is increasing and decreasing;
Work:
First find the critical numbers
f '  x   3x 2
3x 2  0
x0
Second create a table of values;
 ,0
 0, 
Test #
-1
1
Sign f’
+
+
Conclusions
About f
increasing
f is increasing from
increasing
 ,  
Finally list any relative maximums or minimums for
f  x   x3  NONE
2
3
Example 2:
Find the intervals on which
f  x   3x4  4x3 12 x2  2 is increasing and decreasing.
Work:
First find critical numbers:
f '  x   12x3  12x2  24x
0  12 x3  12 x 2  24 x
0  12 x  x 2  x  2 
0  12 x  x  2 x 1
x  0,  2, 1
Second create table:
 , 2
 2,0
Test numbers
-3
-1
½
2
Sign of f’
--
+
--
+
Conclusions
Regarding f
decreasing
increasing
Relative maximums and minimums=
 0,1
1, 
decreasing increasing
Relative min at  2, 30
Relative max at  0, 2
Relative min at 1, 3
Motion Along A Line
Example:
1)
s  t   t 2  7t  10
 7
 0, 
 2
7
2
7 
 ,
2 
Test #
1
4
Sign V(t)
--
V  t   2t  7
Position s(t)
t
moving left
+
* Sign of first derivative = right (+), or left (--) movement.
moving right
* The second derivative tells you acceleration.
For this example the second derivative is
a t   2
So if we add acceleration to our table it will tell us when we are speeding up and when we are slowing down. You have to compare
BOTH the sign of velocity and acceleration to make a conclution.
 7
 0, 
 2
7 
 ,
2 
Test #
1
4
Sign V(t)
--
+
Position s(t)
Sign of A(t)
moving left
+
moving right
+
Conclusions
slowing
down
speeding
up
How far has it travels in 3.5 sec?
s  3.5   s  0  * Final position minus initial position
9
1
  m  10m 12 m  12.25m
4
4
∞
-9/4
2m
0
10m
2)
s  t   t 3  20t 2  128t  280
V  t   3t 2  40t  128
A  t   6t  40
t
16
sec  5.3, 8sec
3
 3t  16  t  8 
40 20

6
3
 6.6 (Possible Point of Inflection)
 0,5.3
5.3, 6.6
 6.6,8
8,
Test #
1
6
7
9
Sign V
+
--
--
+
Sign A
--
--
+
+
Conclusion moving right
Slowing
down
moving left
Speeding
up
moving left
Slowing down
moving right
Speeding up
 
s  6.6   19.26m
s 5.3  14.5m
s  8   24m
∞
v=128 m/sec
-280m
t=0
-24m 19.26m
-14.5m 0m