Transcript First Derivative Test, Concavity, Points of Inflection

First Derivative Test,
Concavity,
Points of Inflection
Section 4.3a
Do Now
Writing: True or False – A critical point of
a function always signifies an extreme
value of the function. Explain.
FALSE!!! – Counterexample???
As we’ve seen, whether or not a critical point signifies
an extreme value depends on the sign of the derivative
in the immediate vicinity of the critical point:
Abs. Max.
Local Max.
f 0
f0
No Extreme
No Extreme
f 0
f0
f0
Local Min.
f 0
First Derivative Test for Local Extrema
The following test applies to a continuous function f
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.
Local Max.
f0
f 0
c
f c  0
Local Max.
f0
f c
f 0
c
undefined
x .
First Derivative Test for Local Extrema
The following test applies to a continuous function f
At a critical point c:
2. If f  changes sign from negative to positive at c
then f has a local minimum value at c.
Local
Min.
f0
f 0
c
f c  0
Local
Min.
f0
f 0
f c
c
undefined
x .
First Derivative Test for Local Extrema
The following test applies to a continuous function f
At a critical point c:
x .
3. If f  does not change sign at c, then f has no local
extreme values at c.
No Extreme
f0
f 0
No Extreme
f0
f 0
c
f c  0
c
f   c  undefined
First Derivative Test for Local Extrema
The following test applies to a continuous function f
At a left endpoint a:
x .
If f   0  f   0  for x > a, then f has a local maximum
(minimum) value at a.
Local Max.
f 0
a
Local Min.
a
f0
First Derivative Test for Local Extrema
The following test applies to a continuous function f
At a right endpoint b:
x .
If f   0  f   0  for x < b, then f has a local minimum
(maximum) value at b.
Local Max.
f 0
Local Min.
b
f0
b
To use the first derivative test:
• Find the first derivative and any critical
points
• Partition the x-axis into intervals using
the critical points
• Determine the sign of the derivative in
each interval, and then use the test to
determine the behavior of the function
Use the first derivative test on the given function
f  x  5x
3
f  x 
2
2
x
5
3
 2x  3
5
Critical point: x = 0
(derivative undefined)
Intervals
x<0
x>0
Sign of f 
+
+
Increasing
Increasing
Behavior of f
Increasing on
  ,  
No Extrema
Can we support
these answers
with a graph???
Use the first derivative test on the given function
3
f  x   x  12 x  5
2

f  x   3 x  12
Intervals
Sign of f 
Behavior of f
Critical points: x = 2, –2
(derivative zero)
x < –2
–2 < x < 2
x>2
+
–
+
Increasing
Decreasing
Increasing
Local Max of 11 at x = –2, Local Min of –21 at x = 2
   ,  2  and  2,  
Decreasing on   2, 2 
Increasing on
The cubing function is always increasing, and never
decreasing… But that doesn’t tell the entire story
about its graph…
y  increases
y  decreases
Where on the graph of
this function is the slope
increasing where is it
decreasing?
This leads to our definition of concavity…
Definition: Concavity
The graph of a differentiable function y  f  x  is
(a) Concave up on an open interval I if y  is increasing on I.
(b) Concave down on an open interval I if y  is decreasing
on I.
Concavity Test
The graph of a twice-differentiable function y  f
x
is
(a) Concave up on any interval where y   0 .
(b) Concave down on any interval where y   0 .
A point where the graph of a function has a tangent
line and where the concavity changes is a point of
inflection.
Let’s work through #8 on p.204
y  2 x  6 x  3
3
2
y    6 x  12 x   6 x  x  2  CP: x  0, 2
2
y    12 x  12   12  x  1 
Intervals
x<0
Sign of y 
Sign of y 
Behavior of y
–
+
IP: x  1
0<x<1
1<x<2
+
+
+
–
2<x
–
–
Dec
Inc
Inc
Dec
Conc up Conc up Conc down Conc down
Establish some graphical support!!!
Let’s work through #8 on p.204
y  2 x  6 x  3
3
2
(a) Increasing on  0, 2 
(b) Decreasing on    , 0  ,  2,  
(c) Concave up on    ,1 
(d) Concave up on  1,  
(e) Local maximum of 5 at x  2
Local minimum of –3 at x  0
(f) Inflection point  1,1 