Precalculus Algebra: Difference Quotient Examples for Functions √x and 1 - Prof. Robert Wa, Study notes of Pre-Calculus

Examples of computing the difference quotient for two functions: f(x) = √x and f(x) = 1. The difference quotient is a measure of the rate of change of a function at a given point. The examples demonstrate the process of computing the difference quotient using the formula f(x+h) - f(x) / h, where h is a small value. The document also includes the steps to simplify the expressions and obtain the final results.

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Pre 2010

Uploaded on 03/10/2009

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MA107 Precalculus Algebra
17 January 2008
Difference Quotient Examples
1. Compute f(x+h)−f(x)
hwhere f(x) = x
f(x+h)−f(x)
h=x+hx
h
=x+hx
h
x+h+x
hMultiply by conjugate
=x+hx
h(x+h+x)
=h
h(x+h+x)
=1
(x+h+x)
Done, because I could let h=0and not divide by 0.
2. Compute f(x+h)−f(x)
hwhere f(x) = 1
x
f(x+h)−f(x)
h=
1
x+h1
x
hEvaluated f(x+h)and f(x)
= ( 1
h)( 1
x+h1
x)Factored out 1
h
= ( 1
h)( 1
x+h(x
x) 1
x(x+h
x+h)) Multiply by 1 to force denominators to match
= ( 1
h)( x
x(x+h)x+h
x(x+h))
= ( 1
h)x−(x+h)
x(x+h)Subtracted the fractions
= ( 1
h)xxh
x(x+h)Distributed the negative
= ( 1
h)h
x(x+h)
=h
hx(x+h)Multiplied the 1
h
=1
x(x+h)Cancelled out the h
1

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MA 107 Precalculus Algebra

17 January 2008

Difference Quotient Examples

1. Compute

f(x+h)−f(x)

h where^ f(x) =^

x

f(x+h)−f(x) h

x+h−

x h

x+h−

x h

x+h+

x

h Multiply by conjugate

x+h−x h(

x+h+

x)

h h(

x+h+

x)

x+h+

x)

Done, because I could let h = 0 and not divide by 0.

2. Compute

f(x+h)−f(x)

h where^ f(x) =^

x

f(x+h)−f(x) h

1 x+h −^

1 x h

Evaluated f(x + h) and f(x)

h )(^

x+h −^

x )^ Factored out^

h

h

x+h

( x

x

x

( x+h

x+h

)) Multiply by “ 1 ” to force denominators to match

h

x x(x+h)

x+h x(x+h)

h

x−(x+h) x(x+h)

Subtracted the fractions

h

x−x−h x(x+h)

Distributed the negative

h

−h x(x+h)

−h

hx(x+h) Multiplied the^

h

x(x+h) Cancelled out the ‘h’