, 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
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
