Class notes Math111 Thomas' Calculus + MyMathLab with Pearson EText Access Card

, Pearson


f(x) = 10x3 + 18


658


E
a) [2 , 4]
AVERAGE RATS Of Change
=900 280
:




+ (4) = 101435 + 10

10(64) + 10
Go
=




=
648 + 18 j




= 658



f(2) = 10(2)3 + 18



=
10(8) + 10




= 80 + 18




= 98




b) [ -
1
, 1]

10(1)3 + 10
=
f(1) =

m


=
10(1) + 10



=
18 + 10


=
28



f( 1)
-
= 10) -

13+ 10

=
10) 1) -

+ 10



=
=
10 + 10




= O

, Not
2)R
R10 :
[0 ,


R(2) =2) + 1


=

= 3




R(0) : NATO)+ 1

=
1




a) find the slope of the curve y
=
x2-2x-3


at the point P(2 -3) by finding ,
the
limiting value of the slope of the secant lines thru point P




y =
X22x -3


P(2 , -3)



m =
f(x2) -
f(X)
V2 -


XI




y( 3) ( 3)2 2(3) 3y(2) (2)2 2(2)



i
-
: - - =
-
-


3



=
9 -

6 3 -
= 4 -
4 -
3



=
9 -

9
=
-
3



=
O




Msecant
::

N -

2X -
3 = 0


(x 3)(X 1) -

+




X =
3X =
-
1




are x 3, -1
limiting
=
the values

, instantaneous Rate of change

m + h)
(x
=

f(x) -




h




f(x + h) =
(X + h) 2 -
2(x + h) -
3


2
:
x + xh + n2 -
2x
-

2n -
3




him = (x2 + xh + h2 -
2x
-


2n -


3) -


(x2 -
2x -
3)
n> 0
-




u




um xixht--2R-3-ps
=




N




↳means
= N + 0
-
2



= X -

2