K Kevinmon New member Joined Jul 19, 2007 Messages 4 Jul 19, 2007 #1 Find the difference quotient and simplify your answer. g(x)=1/(x^2) [g(x)-g(3)]/(x-3), x unequal to 3 ----- 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)
Find the difference quotient and simplify your answer. g(x)=1/(x^2) [g(x)-g(3)]/(x-3), x unequal to 3 ----- 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 Elite Member Joined Jan 29, 2005 Messages 11,978 Jul 19, 2007 #2 \(\displaystyle \L \begin{array}{rcl} g(x) & = & \frac{1}{{x^2 }} \\ \frac{{g(x) - g(3)}}{{x - 3}} & = & \frac{{\frac{1}{{x^2 }} - \frac{1}{9}}}{{x - 3}} \\ & = & \frac{{\frac{{9 - x^2 }}{{9x^2 }}}}{{x - 3}} \\ \end{array}\)
\(\displaystyle \L \begin{array}{rcl} g(x) & = & \frac{1}{{x^2 }} \\ \frac{{g(x) - g(3)}}{{x - 3}} & = & \frac{{\frac{1}{{x^2 }} - \frac{1}{9}}}{{x - 3}} \\ & = & \frac{{\frac{{9 - x^2 }}{{9x^2 }}}}{{x - 3}} \\ \end{array}\)
K Kevinmon New member Joined Jul 19, 2007 Messages 4 Jul 19, 2007 #3 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?
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?
D Deleted member 4993 Guest Jul 20, 2007 #4 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? Click to expand...
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? Click to expand...
