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Understanding the Difference Quotient for Calculus, Assignments of Pre-Calculus

An introduction to the difference quotient, a fundamental concept in calculus. It explains the setup and simplification of difference quotients, using examples with various functions including polynomials, radical functions, and rational functions. Understanding difference quotients is essential for developing derivative formulas.

Typology: Assignments

Pre 2010

Uploaded on 08/16/2009

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THE DIFFERENCE QUOTIENT
I. The ability to set up and simplify difference quotients is essential for calculus students. It is
from the difference quotient that the elementary formulas for derivatives are developed.
II. Setting up a difference quotient for a given function requires an understanding of function
notation.
III. Given the function: f(x) = 2
3x4x5
−−
A. This notation is read “f of x equals . . .”.
B. The implication is that the value of the function (the y-value) depends upon the
replacement for “x”.
C. If a number is substituted for “x”, a numerical value for the function is found.
D. If a non-numerical quantity is substituted for “x”, an expression is found rather than
a numerical value.
E. Careful use of parentheses is essential!
IV. Examples using f(x) = 2
3x4x5
−−
A. 2
f(4)3(4)4(4)527
=−=
B. 2
f(3)3(3)4(3)534
−=−−−=
C. f(a) = 2
3a4a5
−−
D. 2
f(2a3)3(2a3)4(2a3)5
=−−
2
3(4a12a9)8a125
=+−+−
= 22
+−+=−+
E. =
2
f(xh)3(xh)4(xh)5
+=+−+−
22
3(x2xhh)4x4h5
++−−−
22
3x6xh3h4x4h5
=++−−−
F. =
2
f(5h)3(5h)4(5h)5
+=+−+−
2
3(2510hh)204h5
++−−−
2
7530h3h204h5
=++−−−
2
3h26h50
=++
V. The difference quotient is so named because the operations involved are subtraction and
division. Common forms of the difference quotient are:
A.
f(xh)f(x)
h
+−
B.
f(ah)f(a)
h
+−
C.
f(5h)f(5)
h
+−
D.
f(xx)f(x)
x
+∆−
The purpose for simplifying the difference quotient is to get the “h” or the in the
"x"
denominator to cancel out.
pf3
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THE DIFFERENCE QUOTIENT

I. The ability to set up and simplify difference quotients is essential for calculus students. It is

from the difference quotient that the elementary formulas for derivatives are developed.

II. Setting up a difference quotient for a given function requires an understanding of function

notation.

III. Given the function: f(x) =

2

3x − 4x − 5

A. This notation is read “f of x equals.. .”.

B. The implication is that the value of the function (the y-value) depends upon the

replacement for “x”.

C. If a number is substituted for “x”, a numerical value for the function is found.

D. If a non-numerical quantity is substituted for “x”, an expression is found rather than

a numerical value.

E. Careful use of parentheses is essential!

IV. Examples using f(x) =

2

3x − 4x − 5

A.

2

f(4) = 3(4) − 4(4) − 5 = 27

B.

2

f( − 3) = 3( − 3) − 4( 3) − − 5 = 34

C. f(a) =

2

3a − 4a − 5

D.

2

f(2a − 3) = 3(2a − 3) − 4(2a − 3) − 5

2

= 3(4a − 12a + 9) − 8a + 12 − 5

2 2

12a − 36a + 27 − 8a + 12 − 5 = 12a − 44a + 34

E. =

2

f(x + h) = 3(x + h) − 4(x + h) − 5

2 2

3(x + 2xh + h ) − 4x − 4h − 5

2 2

= 3x + 6xh + 3h − 4x − 4h − 5

F. =

2

f(5 + h) = 3(5 + h) − 4(5 + h) − 5

2 3(25 + 10h + h ) − 20 − 4h − 5

2

= 75 + 30h + 3h − 20 − 4h − 5

2

= 3h + 26h + 50

V. The difference quotient is so named because the operations involved are subtraction and

division. Common forms of the difference quotient are:

A.

f(x h) f(x)

h

B.

f(a h) f(a)

h

C.

f(5 h) f(5)

h

D.

f(x x) f(x)

x

The purpose for simplifying the difference quotient is to get the “h” or the " ∆x "in the

denominator to cancel out.

VI. Examples using f(x) = [see IV for the “f(x + h)” substitution]

2

3x − 4x − 5

A.

f(x h) f(x)

h

2 2 2

[3x 6xh 3h 4x 4h 5] (3x 4x 5)

h

2 2 2

3x 6xh 3h 4x 4h 5 3x 4x 5

h

2

6xh 3h 4h

h

h(6x 3h 4)

h

= (^) 6x + 3h − 4

B. = [steps are identical to A, but using “a” for”x”]

f(a h) f(a)

h

6a + 3h − 4

C. =

f(5 h) f(5)

h

2 2

[3(5 h) 4(5 h) 5] (3 5 4(5) 5)

h

+ − + − − g − −

2 [3(25 10h h ) 20 4h 5] (75 20 5)

h

2 [3h 26h 50] (50)

h

2 3h 26h h(3h 26)

3h 26

h h

VII. Examples using radical functions

A. Given f(x) = x

f(x h) f(x)

h

  • − x h x

h

2. Multiply by the conjugate of the numerator to rationalize the numerator.

x h x x h x

h x h x

 (^) + −   (^) + + 

   

   

   

(x h) x h 1

h( x h x ) h( x h x ) x h x

B. Given f(x) = x

f(3 h) f(3)

h

  • − 3 h 3

h

  • − 3 h 3 3 h 3

h (^3) h 3

   

  • − + +

   

   

   

(3 h) 3 h 1

h( 3 h 3) h( 3 h 3) 3 h 3

D. Given f(x) =

4x

x − 5

f(x h) f(x)

h

4(x h) 4x

x h 5 x 5

h

 + 

−  

  • − −  

   

 

(x h 5)(x 5)

(x h 5)(x 5)

 + − − 

 

  • − −  

4(x h)(x 5) 4x(x h 5)

h(x h 5)(x 5)

  • − − + −

  • − −

2 2 4x 20x 4xh 20x 4x 4xh 20x

h(x h 5)(x 5)

− + − − − −

  • − −

20h 20

h(x h 5)(x 5) (x h 5)(x 5)

− −

=

  • − − + − −

IX. Practice problems: find and simplify for each function.

f(x h) f(x)

h

A. f(x) =

2 − 5x + 3x − 7

B. f(x) =

3 4x + 6

C. f(x) = 7x − 8

D. f(x) = 9 −5x

E. f(x) =

3x

4 −2x

F. f(x) =

4 2x

3x 1

X. Answers to practice problems

A.

− 10x − 5h + 3

B.

2 2 12x + 12xh +4h

C.

7

7x + 7h − 8 + 7x − 8

D.

5

9 5x 5h 9 5x

− − + −

E.

3

(2 − x)(2 − x −h)

F.

14

(3x 1)(3x 3h 1)