Difference Quotient

Difference Quotient

Postby nycmath » Sun Aug 16, 2026 7:51 am

Precalculus
Michael Sullivan
Edition 10
Chapter 2, Section 2.1

Find the difference quotient of f; that is find
[f(x + h) - f(x)]/h, h cannot be 0, for each function. Be sure to simplify.

81. f(x) = x^2 - 4

82. f(x) = 3x^2 + 2
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Re: Difference Quotient

Postby Eigenvalue » Sun Aug 16, 2026 11:29 am

[tex]\frac{(x+h)²-4-(x²-4)}{h}[/tex]

=[tex]\frac{x²+2hx+h²-x²}{h}[/tex]

=[tex]\frac{2hx+h²}{h}[/tex]

=2x+h

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Re: Difference Quotient

Postby Eigenvalue » Sun Aug 16, 2026 11:32 am

[tex]\frac{3(x+h)²+2-(3x²+2)}{h}[/tex]

=[tex]\frac{3x²+6xh+3h²-3x²}{h}[/tex]

=[tex]\frac{6xh+3h²}{h}[/tex]

=6x+3h

Eigenvalue
 
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Re: Difference Quotient

Postby nycmath » Sun Aug 16, 2026 7:09 pm

Eigenvalue wrote:[tex]\frac{(x+h)²-4-(x²-4)}{h}[/tex]

=[tex]\frac{x²+2hx+h²-x²}{h}[/tex]

=[tex]\frac{2hx+h²}{h}[/tex]

=2x+h


Very good. By the way, there is one difference quotient of f post that remains unanswered. Would you like to try it out?

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Re: Difference Quotient

Postby nycmath » Sun Aug 16, 2026 7:11 pm

Eigenvalue wrote:[tex]\frac{3(x+h)²+2-(3x²+2)}{h}[/tex]

=[tex]\frac{3x²+6xh+3h²-3x²}{h}[/tex]

=[tex]\frac{6xh+3h²}{h}[/tex]

=6x+3h


I hope you like my selection of questions. Some will be posted from the David Cohen textbook in just a little while.

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