Application of derivative

Additional Applications of the
Derivative
Mr. ABHISHEK SINGH

Increasing and Decreasing Function
Increasing and Decreasing Function Let f(x) be a function
defined on the interval a<x<b, and let x1 and x2 be two numbers
in the interval, Then
f(x) is increasing on the interval if f(x2)>f(x1) whenever x2>x1
f(x) is decreasing on the interval if f(x2)<f(x1) whenever x2 >x1
Monotonic
increasing
单调递增
Monotonic
decreasing
单调递减

Tangent line with negative slope f(x) will be decreasing
( ) 0f x′ >
Tangent line with positive slope f(x) will be increasing

( ) 0f x′ <


If for every x on some interval I , then f(x)
is increasing on the interval
is decreasing on the interval
is constant on the interval
( ) 0f x′ >
( ) 0f x′ =
0)( <′ xf

Increasing and Decreasing
Function
Procedure for using the derivative to determine
intervals of increase and decrease for a function of f.
Step 2. Choose a test number c from each interval a<x<b
determined in the step 1 and evaluate . Then
If the function f(x) is increasing on a<x<b.
If the function f(x) is decreasing on a<x<b( ) 0f c′ <
( ) 0f c′ >
)(cf ′
Step 1. Find all values of x for which or is
not continuous, and mark these numbers on a number line.
This divides the line into a number of open intervals.
)(xf ′( ) 0f x′ =

Example. Find the intervals of increase and decrease for the
function
Solution:
The number -2 and 1 divide x axis into three open intervals.
x<-2, -2<x<1 and x>1
(0) 0f ′ <
(2) 0f ′ > Risingf is increasing2x>1
Fallingf is deceasing0-2<x<1
Risingf is increasing-3x<-2
Direction
of graph
ConclusionTest
number
Interval
)(cf ′
3 2
( ) 2 3 12 7f x x x x= + − −
2
( ) 6 6 12 6( 2)( 1)f x x x x x′ = + − = + −
Which is continuous everywhere, with where x=1 and x=-2( ) 0f x′ =
0)3( >−′f

Absolute(Global) Maximum
Absolute(Global) Minimum
Let f(x) be a function with domain D. Then f(x) has an
absolute maximum value on D at a point c
if f(x) ≤ f(c) for all x in D
and
absolute minimum value on D at a point c
if f(x) ≥ f(c) for all x in D.

Relative Extrema
Relative (Local) Extrema :
A function f(x) has a relative maximum value at an interior
point c of its domain if f(x) ≤ f(c) for all x in some open interval
containing c.
A function f(x) has a relative minimum value at an interior
point c of its domain if f(x) ≥ f(c) for all x in some open interval
containing c.

Critical Points :
An interior point c in the domain of f(x) is called a critical
point if either or undefined. The corresponding point
(c,f(c)) on the graph of f(x) is called a critical point for f(x).
)(cf ′( ) 0f c′ =
The First Derivative Theorem
If f(x) has a local maximum or minimum value at an
interior point c of its domain and if is defined at c ,
then
)(xf ′
( ) 0f c′ =

Critical Points
Not all critical points correspond to relative extrema!
Figure. Three critical points where f’(x) = 0:
(a) relative maximum, (b) relative minimum
(c) not a relative extremum.

Critical Points
Not all critical points correspond to relative extrema!
Figure Three critical points where f’(x) is undefined:
(a) relative maximum, (b) relative minimum
(c) not a relative extremum.

Example
Solution
Find all critical numbers of the function
and classify each critical point as a relative maximum, a relative
minimum, or neither
4 2
( ) 2 4 3f x x x= − +
3
( ) 8 8 8 ( 1)( 1)f x x x x x x′ = − = − +
The derivative exists for all x, the only critical numbers are
Where that is, x=0,x=-1,x=1. These numbers
divide that x axis into four intervals, x<-1, -1<x<0, 0<x<1, x>1
( ) 0f x′ =
1 1 15
( 5) 960 0 ( ) 3 0 ( ) 0 (2) 48 0
2 4 8
f f f f′ ′ ′ ′− = − < − = > = − < = >
Choose a test number in each of these intervals
-1 min
-------- ++++++ -------- +++++
+0 max 1 min
Thus the graph of f falls for x<-1 and for
0<x<1, and rises for -1<x<0 and for x>1
x=0 relative maximum
x=1 and x=-1 relative minimum

§3.1 Sketch the graph
A Procedure for Sketching the Graph of a Continuous
Function f(x) Using the Derivative
Step 1. Determine the domain of f(x).
( )f x′Step 2. Find and each critical number, analyze the sign of
derivative to determine intervals of increase and decrease for f(x).
( ) 0f x′ > ( ) 0f x′ <
Step 3. Plot the critical point P(c,f(c)) on a coordinate plane,
with a “cap” at P if it is a relative maximum or a “cup”
if P is a relative minimum. Plot intercepts and other key points
that can be easily found.
( ) 0f x′ =
Step 4 Sketch the graph of f as a smooth curve joining the critical
points in such way that it rise where , falls where
and has a horizontal tangent where

Example
Solution
Sketch the graph of the function
4 3 2
( ) 8 18 8f x x x x= + + −
3 2 2
( ) 4 24 36 4 ( 3)f x x x x x x′ = + + = +
( 5) 80 0 ( 1) 16 0 (1) 64 0f f f′ ′ ′− = − < − = − < = >
The derivative exists for all x, the only critical numbers are
Where that is, x=0, x=-3. These numbers divide
that x axis into three intervals, x<-3, -3<x<0, x>0.
Choose test number in each interval (say, -5, -1 and 1 respectively)
( ) 0f x′ =
-3
neither
-------- ++++++--------
0
min
Thus the graph of f has a horizontal
tangents where x is -3 and 0, and it is
falling in the interval x<-3 and -3<x<0
and is rising for x>0

f(-3)=19 f(0)=-8
Plot a “cup” at the critical point (0,-8)
Plot a “twist” at (-3,19) to indicate a galling graph with
a horizontal tangent at this point .
Complete the sketch by passing a smooth curve through the
Critical point in the directions indicated by arrow

Example
Solution
The revenue derived from the sale of a new kind of
motorized skateboard t weeks after its introduction is given by
2
2
63
( )
63
t t
R t
t
−
=
+
million dollars. When does maximum revenue occur? What is
the maximum revenue
Critical number t=7 divides the domain
into two intervals x<=t<7
and 7<t<=63
0 63t≤ ≤
2
2
63(7) (7)
(7) 3.5
(7) 63
R
−
= =
+
7
Max
++++++ --------
t
0 63

Concavity
The graph of differentiable function y=f(x) is
(1) concave up on an open interval I if is increasing on I
(2) concave down on an open interval I if is decreasing on I.
)(xf ′
)(xf ′

Concavity
A graph is concave upward on the interval if it lies above all its
tangent lines on the interval and concave downward on an
Interval where it lies below all its tangent lines.

Note Don’t confuse the concavity of a graph with its “direction”
(rising or falling). A function may be increasing or decreasing on
an interval regardless of whether its graph is concave upward or
concave downward on the interval.

The Second Derivative Test for Concavity
Let y = f(x) be twice-differentiable on an interval I.
1. If on I , the graph of f(x) over I is concave up.
2. If on I , the graph of f(x) over I is concave down.
Second Derivative Procedure for Determining Intervals of
Concavity for a Function f.
Step 1. Find all values of x for which or is
not continuous, and mark these numbers on a number line.
This divides the line into a number of open intervals.
Step 2. Choose a test number c from each interval a<x<b
determined in the step 1 and evaluate . Then
If , the graph of f(x) is concave upward on a<x<b.
If the graph of f(x) is concave downward on a<x<b.
0)( >′′ xf
0)( <′′ xf
0)( =′′ xf )(xf ′′
)(cf ′′
0)( >′′ cf
,0)( <′′ cf

-1
--------++++++ -------- ++++++
0 1
Type of concavity
Sign of

-------- -------- ++++++
0
No inflection
1
inflection
Type of concavity
Sign of
to be continued

-------
-
++++++
0
inflection
Type of concavity
Sign of

Note: A function can have an inflection point only where it
is continuous.!!

§3.2 Behavior of Graph f(x) at an
inflection point P(c,f(c))

-1.5
min
-------- +++++
+ 1
Neither
++++++
to be continued

-------- ++++++
-2/3
inflection
1
inflection
Type of concavity
Sign of ++++++
to be continued

b. Find all critical numbers of the function
c. Classify each critical point as a relative maximum, a
relative minimum, or neither
e. Find all inflection points of function
a.
d.
Review

-------- +++++
+
--------+++++
+
to be continued

++++++-------- -------- ++++++
to be continued

§3.2 The Second Derivative Test

Application of derivative

More Related Content

What's hot (20)

Similar to Application of derivative (20)

Recently uploaded (20)

Application of derivative

Editor's Notes