Difference Quotient

[Kevinmon]

Find the difference quotient and simplify your answer.

g(x)=1/(x^2)

[g(x)-g(3)]/(x-3), x unequal to 3

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My work so far:

1/(x^2-9)
x-3

1/(x+3)(x-3)
x-3

___1____ times 1/(x-3)
(x+3)(x-3)

 

[pka]

\(\displaystyle \L
\begin{array}{rcl}
g(x) & = & \frac{1}{{x^2 }} \\
\frac{{g(x) - g(3)}}{{x - 3}} & = & \frac{{\frac{1}{{x^2 }} - \frac{1}{9}}}{{x - 3}} \\
& = & \frac{{\frac{{9 - x^2 }}{{9x^2 }}}}{{x - 3}} \\
\end{array}\)

 

[Kevinmon]

Ok, and if you change it to

(9-x^2)/(9x^2) times 1/(x-3)

you get

-(x+3)(x-3)/(9x^2)(x-3)

and the (x-3)'s cancel out, leaving you with

-(x+3)/(9x^2)

Is that right?

 

[Deleted member 4993]

Ok, and if you change it to

(9-x^2)/(9x^2) times 1/(x-3)

you get

-(x+3)(x-3)/(9x^2)(x-3)

and the (x-3)'s cancel out, leaving you with

-(x+3)/(9x^2) .................Correct
Is that right?