This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Set 04 Solution 2.7 Derivative as a rate of change 2.8 Derivative as a function 5x 2
− 20 5 x 2 + 20 x − 3
f ( x ) − f (a ) x − 3
()
m = lim
=
=
x →a
x−a
x−2
( x − 2) ( x − 3) lim f ( x ) − f (a )
3x − 3
3x + 3
=
x
x →1
x−a
x −1
3x + 3
3x − 3
3( x − 1) =
=
( x − 1) 3x + 3 ( x − 1) 3x + 3 m= ( =
m= ) ( ) = 3 ( 3x + 3 lim
x →1 ) ( 3x + 3 x→2 ) = 3 We need to find the point (x, y ) in order to use the point slope form. x = 2 ⇒ f (2) = y = −20 , so we have (x, y ) = (2 − 20 ) 3 a. f ( x ) = x − 2 x at P = (1, −1) Solution: You can use the power rule to differentiate the derivative: slope = f ' ( x) = 3 x 2 − 2 x =1 = 1 . You € 3 f (a + h ) − f (a )
(1 + h) − 2 (1 + h) + 1
m = lim
= lim
h→0
h→0
h
h
2
(1 + 2 h + h ) (1 + h ) − 2 − 2 h + 1
=
h
h 3 + 3h 2 + h
= lim
= h 2 + 3h + 1
h→0
h
m =1 f (x) = 5x 2
x −3 m = −2 7. The marginal profits of a company (in $/piece) can be described by the function: 2 at y = −0.01(x − 50) + 20 , where x is the number of pieces, and y is the marginal profit. a. Find...
View Full
Document
 Winter '06
 McAdam
 Derivative, Differential Calculus

Click to edit the document details