2/23 Families of curves (4.2) Review inflection points, First and Second Derivative Tests. Recall that a function f has an inflection point at x = a , if !!f a( ) = 0 and !!f x( ) changes sign at x = a .

E.g., f x( ) = x3 ! 6x +1 has an inflection point at x = 0 (i.e., the point 0,1( ) ), since !!f x( ) = 6x has a zero at x = 0 and !!f x( ) changes sign at x = 0 . SHOW.

Recall that the First and Second Derivative Tests can be used to determine if critical points (points p where

�

! f (p) = 0 or !f p( ) is undefined) are local maxima or local minima. The FDT states "If

�

! f changes from negative to positive at p, then f has a local minimum at p." (Or if

�

! f changes from positive to negative at p, then f has a local maximum at p.) The SDT states "If

�

! f (p) = 0 and

�

! ! f (p) > 0 , then f has a local minimum at p." (Or if

�

! f (p) = 0 and !!f (p) < 0 , then f has a local maximum at p.) Families of curves. We have studied the graphs of certain families of functions

already, e.g., the family of linear functions,

�

y = mx + b. Another family of functions is the quadratic family,

�

f (x) = ax2

+ bx + c . And another family is the family of sine functions,

�

g(x) = Asin[B(x !C)]+ D. The various constants m and b, a, b, and c, and A, B, C, and D are called parameters of their functions. Different values of these parameters give us different members of their function families. Calculus allows us to study families of functions in greater detail than previously possible. Parabolas. The graphs of quadratic functions,

�

f (x) = ax2

+ bx + c , are parabolas. The graph of every parabola has a vertex that is either a local maximum or minimum. DRAW both examples. We can find the vertex of a parabola algebraically by completing

the square, e.g.,

�

f (x) = a(x2

+b

ax) + c = a(x

2+b

ax +

b2

4a2) + c !

ab2

4a2

=

�

a(x !b

2a)2

+4ac ! b

2

4a and its vertex is

�

(b

2a,4ac ! b

2

4a). Since the vertex is either a local

maximum or minimum, we can also use the First Derivative Test or the Second Derivative Test to find it. We first find

�

! f (x) = 2ax + b and any critical points of f, i.e.,

we solve

�

! f (x) = 2ax + b = 0 . Therefore,

�

x = !b

2a is a critical point of f. SHOW on

graphs. To apply the First Derivative Test, we look at the sign of

�

! f on either side of

�

x = !b

2a. If

�

2a > 0, then

�

! f is increasing and changes from negative values to positive

values at

�

x = !b

2a; so f has a local minimum at

�

x = !b

2a. SHOW on graph. On the other

hand if

�

2a < 0, then

�

! f is decreasing and changes from positive values to negative values

at

�

x = !b

2a; so f has a local maximum there. SHOW on graph. Before we apply the

Second Derivative Test, we first find

�

! ! f (x) = 2a . If

�

2a > 0, then the graph of f is concave up and f has a local minimum there. SHOW on graph. If

�

2a < 0, then the graph of f is concave down and f has a local maximum there. SHOW on graph.

Graphs of functions

�

y = e!(x!a)

2b . This family of curves is related to the normal

density function (AKA the bell-shaped curve) used in probability. We'll assume

�

b > 0 .

Consider the case where

�

a = 0 and

�

b =1, i.e.,

�

y = e!x2

. DRAW graph. We have

studied this function before. The role of the parameter a is to shift the graph of

�

y = e!x2

horizontally: to the right if

�

a > 0, to the left if

�

a < 0. DRAW figure 4.24. Notice that the

graph of

�

y = e!(x!a)

2

is always positive (exponential graphs are always positive), has a

horizontal asymptote of

�

y = 0 ( limx!±"

e# x#a( )2

= 0 ), and a local maximum at

�

x = a (using

the FDT.) Since !y = "2 x " a( )e" x"a( )2 (Chain Rule), !y x( ) = 0 when

�

x = a. Since !y changes from positive to negative at

�

x = a, y has a local maximum at

�

x = a. Let's study the role of the parameter b by studying the family with

�

a = 0, i.e.,

�

y = e!x2b. DRAW graphs of

�

b = 2 and

�

b = 8 . Both graphs appear to have a local maximum at

�

(0,1), but the graph with

�

b = 8 appears to be dilated horizontally compared to the graph with

�

b = 2 . It appears that the parameter b affects the location of the inflection points. Let's see if we can show this using calculus.

First, we calculate

�

! y = "2x

be"x2

b. Critical points of the graph occur when

�

! y = "2x

be"x2

b= 0 , i.e., when

�

x = 0 (Why?) or at the point

�

(0,1) (Why?). Using the

First Derivative Test, we can show that

�

! y changes from positive to negative at

�

x = 0 , so y has a local maximum there. SHOW.

We calculate

�

! ! y = "2

be"x2

b+ "

2x

b

#

$ %

&

' ( "

2x

be"x2

b#

$ %

&

' ( = "

2

be"x2

b+4x2

b2

e"x2

b=

�

2

be!x2 b"

# $

%

& ' 2x2

b!1

"

#

$ $

%

&

' ' . Using the Second Derivative Test, we find

�

! ! y (0) = "2

b. Since

�

b > 0 ,

y is concave down and the critical point

�

(0,1) is a local maximum.

To find the points of inflection we need to solve

�

! ! y =2

be"x2

b#

$ %

&

' ( 2x2

b"1

#

$

% %

&

'

( ( = 0.

But this is equivalent to solving

�

2x2

b!1= 0 . Solving we get

�

x2

=b

2 and

�

x = ±b

2.

Therefore, the larger the parameter b is the further apart are the points of inflection. For

�

b = 2 the point of inflection occurs when

�

x = ±2

2= ±1, and for

�

b = 8 the point of

inflection occurs when

�

x = ±8

2= ±2 . SHOW on graph.

This is summarized well in the text at the bottom of p. 177.