Transcript Ch.2 Limits and derivatives

What does
f 
say about f ?
Increasing/decreasing test

If f ( x )  0 on an interval I, then f is increasing on I.
If f ( x )  0 on an interval I, then f is decreasing on I.
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Proof. Use Lagrange’s mean value theorem.

f ( x 2 )  f ( x1 )  f ( c )( x 2  x1 )
Example
2
Ex. Find where the function f ( x )  x 3 ( x  5) is increasing
and where it is decreasing.
1
2
1

2 3
5
 Sol. Since when x0, f  ( x ) 
x ( x  5)  x 3  x 3 ( x  2),
3
3
and f is not differentiable at 0, we know
f is increasing on (-1,0), (2,+1); decreasing on (0,2).
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Example


10
Ex. Find the intervals on which f ( x )  3
is
2
4x  9x  6x
increasing or decreasing.
1
Sol. increasing on ( ,1)
2
decreasing on (   , 0), (0, 1 ), (1,   )
2


p
Ex. Prove that sin x  tan x  2 x when 0  x  .
2
Sol. Let f(x)=sinx+tanx-2x, x2I =(0,p/2). Then
2
2
f ( x )  cos x  sec x  2  cos x 
1
2
 2  (cos x 
cos x
f increasing on I, and f(x)>f(0)=0 on I.
1
cos x
) 0
2
Example


x ln x
Ex. Show that f ( x ) 
1 x
Sol.
1  x  ln x
f ( x ) 
(1  x )
is decreasing on (0,1).
 0?  0?
2
g ( x )  1  x  ln x  g ( x ) 
1
 1  0  g ( x )  g (1)  0
x

Ex. Prove that sin x  x 
x
3
when x>0.
6

Sol. f ( x )  sin x  x 
x
3
6
 f ( x )  cos x  1 
x
2
2
, f ( x )  x  sin x  0
The first derivative test
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


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A critical number may not be a maximum/minimum point.
The first derivative test tells us whether a critical number
is a maximum/minimum point or not:
If f  changes from positive to negative at c, then maximum
If f  changes from negative to positive at c, then minimum
If f  does not change sign at c, then no maximum/minimum
This explains why f ( x )  x 3 has no maximum/minimum at 0.
Example


Ex. Find all the local maximum and minimum values of
the function f ( x )  3 ( x 2  1) 2 .
Sol.
4x
f ( x ) 
3 ( x  1)( x  1)
 0  x  0.
3
All critical numbers are: 0,  1, 1.
Using the first derivative test, we know:
0 is local maximum point,  1 are local minimum points
Convex and concave
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Definition If the graph of f lies above all of its tangents on
an interval I, then it is called convex (concave upward) on
I; if the graph of f lies below all of its tangents on I, it is
called concave (concave downward) on I. The property of
convex and concave is called convexity (concavity).
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Definition A point on the graph of f is called an inflection
point if f is continuous and changes its convexity.
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The convexity of a function depends on second derivative.
Convexity test
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If f ( x )  0 for all x in I, then f is convex on I.
If f ( x )  0 for all x in I, then f is concave on I.
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Ex. Find the intervals on which f ( x ) 


x 1
is convex or
x 1
concave and all inflection points.
2
2(
x

1)(
x
 4 x  1)
Sol. f ( x )  
 0  x  1,  2  3
2
3
2
( x  1)
By convexity test, f convex on (  2  3,  2  3 ) and (1,   )
concave on (  ,  2  3 ) and (  2  3,1). the inflection points
are 1,  2  3 ,  2  3.
The second derivative test
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

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The second derivative can help determine whether a
critical number is a local maximum or minimum point.
The second derivative test
If f ( c )  0, f ( c )  0, then f has a local minimum at c
If f ( c )  0, f ( c )  0, then f has a local maximum at c
Ex. Find the local maximum and minimum points of
f ( x )  cos x 
1
cos 2 x .
2
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Sol. Local maximum x  k p
2
4
local minimum x  2 k p  p , x  2 k p  p
3
3
Before sketching a graph
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Using derivative to find the global and local maximum and
minimum values, and locate critical numbers
Using derivative to find convexity and locate inflection
points
Using derivative to find intervals on which the function is
increasing or decreasing
Find domain, intercepts, symmetry, periodicity and
asymptotes
Asymptotes

Horizontal asymptotes: if lim f ( x )  L or lim f ( x )  L ,
x  
x  
then y=L is a horizontal asymptote of the curve y=f(x)
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Vertical asymptotes: if lim f ( x )   or lim f ( x )   ,
x a
x a
then x=a is a vertical asymptote of the curve y=f(x)

Slant asymptotes: if lim[ f ( x )  ( m x  b )]  0 or
x 
lim [ f ( x )  ( m x  b )]  0 then y=mx+b is a slant asymptote
x  
of the curve y=f(x)
Slant asymptote

Since lim [ f ( x )  ( m x  b )]  0  m  lim
x 
x 
f ( x)
x
,
to find slant
f ( x) / x,
asymptotes, we first investigate the limit m  lim
x 
if it exists, then b  lim ( f ( x )  m x ), and y=mx+b is a slant
x 
asymptote.
Homework 9
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Section 4.3: 14, 16, 17, 47, 49, 70, 74